LECTURES IN MATHEMATICS Department
of Mathematics
KYOTO UNIVERSITY
15
Lectures on Algebraic Solutions of Hypergeometric Differential
Equations BY
Michihiko
MATSUDA
Published by KINOKUNIYA CO., Ltd. Tokyo, Japan
LECTURES
IN
Department
MATHEMATICS
of
KYOTO
Mathematics
UNIVERSITY
15
Lectures
on
Algebraic
Solutions
of Hypergeometric
Differential
Equations
By
Michihiko
MATSUDA
Published KINOKUNIYA
by
CO
•
,
Ltd.
copyright
-c
RIGHT
RESERVED in
Japan
Ltd.
to
his
teacher
who
taught
with
a
when
he
Hisaaki
the
author
perspicacious was in
a
Yoshizawa
kindly mind
student
Kyoto
University
Preface
proofs be
In
the
first
of
the
celebrated
devoted
Landau
and
chapter
four
to
Schwarz'
proof,
A.
Errera's
and
1873
Schwarz
solutions
Steiner's
classification An
which
in
transformations proof book
[9
the
reduced
succeeded
all
The the
will
by
the
[25],
the
be
treated.
last
of
all
equations of
the
Schwarzian
who
by
rotation derivative
considered
differential derivative
to
determination
subgroups
played
will
In
differential finite
first
fourth
Klein's.
in
equation
Schwarzian
the to
proof
those
Brioschi
[3],
II,
in
§10]
0.
rational
equations.
can
by
algebraic
functions,
Dirichlet's
theorem
function
fields
genus
of
algebraic
Fischer
[26],
metically
this
of
zero
and
Gesammelte
Landau
of
algebraic
description
it
and
to
hypergeometric
algebraic
solutions
and
second
third
several
Schwarz:
solutions
was
Kummer's
describe
be
found
A in
Goursat's
, Chap.VI]. An
plicit
to
hypergeometric
role
of
without
the
of
important
appears
the
[35] of
shall
due
liouvillian
algebraic
group.
we
theorem
transcendental
In
who
chapters
attempted
Eisenstein's
theorem where
theorem
on
was
accomplished
The
first
two
we
prime by
chapters
can
himself
to on see
—v—
power
algebraic
made
by
Ex-
Schwarz,
footnote
of
Schwarz' series
table expression application
arithmetical
on
[16,
arith-
progressions,
[4]. based
[16],
1922).
a beautiful in
Errera are
gain
Klein
equations.
(cf.
Springer,
numbers A.
was
by by
Kummer's
solutions Klein
given
defined
through
Abhandlungen, [27]
was
the
lectures
delivered
of of
by
the
author
Landau's due
at
second to
him. kindly
lectures
will
In
there
no
the
author
completed
S.
Zariski its
chapter
other
than
proof
is
of
by
of
the
is
lemma
no
theorem. Note Honda's [10,
P.
the
theorem
p.448]
comments
second
on
sufficiency
of
Gauss
a
this
equation
enabled
us
be
the
in of
improve to
Eisenstein's
[36],
theorem
use
to
offered
is
Setoyanagi
of
making
Klein
equation
M.
proof
(cf. that
Fuchsian
without
will
that differ-
theorem
Riccati's
have
chapter
the
of
algebraic
singularity,
treated:
hypergeometric
Recently
on
be
prove
case Kuga's §13. T.
criterion
§1]on
Liouville's
the
fifth
theorem
Bessel's
chapter
[29],
[30],
will [31]
be on
offered
to
elementary
prove solutions
in
equation. Goursat's
works
[7],
[8]
on
Kummer's
equation
will
not
be
here. For
problem
[30,
an
Uni-
chapters.
proved
by
is
Those
will
[15]
M. Kuga's
group
it Kyoto
fourth
considered
p.173].
gave
Rehm's
on
Note
treated
[24,
logarithmic
H.
in
that
work.
works
solutions
on
There
visiting
and
Kimura
monodromy
author,
third
those
1981.
Landau's
T.
based
group
Liouville's
there
the
of
be
following and
liouvillian
Picard-Vessiot student
[11]
of knowing
to
the
the
Ohasi
The
closure
by
Autumn without
informed
new
§61]).
stated happened
equations
[18,
was who
last -
are
ential
a
the
in
Okamoto,
be
Hukuhara
University
theorem K.
versity,
M.
Kyoto
an
confer
application with
F.
of
N.
Baldassarri
Katz'
general -
B.
Dwork
theory [1].
[14]
to
our
The K.
author
Nishioka November
and
would M.
Setoyanagi
like
to
express for
fruitful
his
sincere
gratitude
discussions
with
1984 Michihiko
—vii—
Matsuda
to them.
Preliminaries
The lowing a
analytic
existence linear
polynomials
the
note
is
complex
based
plane
equation
let
whose
variable
on
the
fol-
us
consider
coefficients
x:
n-1
xn+
P1(x)dn-1y++ dx
absolute
positive
the
independent
dyn 0(x)d
the
In
this
differential
of
P
If
in
theorem.
homogeneous
are
treatment
value
of
r,
then
constant
Pn(x)y
every a
root
of
formal
=
P0(x)
0. is
greater
than
solution
co a _ -
y(x)
which
is
=
E.0-kXk, k=0
determined dkv
by
(0)
=
for
lx1 < r.
ak'
the
0
initial
< k
values
at
x
=
0:
< n,
dxk
converges
dky
zk
we
have
- k dx
a
-
system
It
will
ak,0
of
be proved <
linear
as follows.
Setting
n,
differential
equations:
dz dxk-1
dzn -1n-1 dx
=
with
the
the
right
1
Pn-k(x) k0P0(x)
initial
zk(0)
For
-zk+ak'
n,
(zk
condition:
=
0,
0
hand
sides
< k
< n.
of
our
-IX-
=
+ ak)
equation
we
can
take
a
majorant
a
power
series: M
with
n-1
=
X L (Zk r k=0
positive
+ A) 1
constants
dZ
nM(Z
dX1
+ -
M
A)
E r-hXh(A
k=0
and
A
X
n-1 E
M.
Z(0)
+
Zk)
h=0
=
A
formal
solution
Z
of
0
'
r
is
given
by Z = A[(1
A
whose
radius
y(x)
converges
E k=1
of
- X)-nMr c(c+1)....
-
1]
(c+k -1)
r-kXk,
k!
convergence
for
Ix'
is
< r.
r.
Hence,
c
our
=
nMr,
formal
solution
Table
Chapter
I.
Schwarz'
§1.
Kummer's
§2.
Reducibility
§3.
Schwarz'
Chapter
Eisenstein's
§5.
Landau's
§6.
Rough
Note.
Reduction
§8.
Explicit
logarithmic
1
singularity
7 13
Landau's
criterion. theorem
first
20
and
second
theorems......
.....
.....
theorem
42 settling.
through
Kummer's
description
of
Landau-Errera's
§9.
Errera's
§10.
Two
§11.
Attainment
lemmata
Picard-Vessiot's
§13.
Liouville's
§14.
Kuga's
algebraic
......
48
solutions
57
..... ............
Schwarz'
Transcendental
§12.
Bessel's
...... of
equation..
theorem.
lemma............
V.
25 33
Klein's
IV.
Note.
transformations
estimation
§7.
Chapter
Gauss'
table
III.
Chapter
and and
Honda's
Chapter
contents
theorem.
table
II.
§4.
of
66 ........
..........
table................... liouvillian
theory
.
74
.....
78
solutions. 92
lemma
96
theorem
101
equation
106
Bibliography
110
—xi—
Chapter
§1.
Kummer's
Consider
x(1
(E)
a
-
a,
S and
S,
It
(1)
F(a,
unless
y
verges
in
has
the
is
If then
is
xl
< 1.
=
a
the
then
we
(E)
is
a
tion
of
Here
(E),
obtain
a
(3)
(1
solution
-
in
a
will and
be S.
indicated
by
A solution
is
x
not
greater
than
variable
E(a+1-y,
y
S+1-1,
2-y,
0,
into
which
z
2-y)
for
con-
by
y
=
x1-yz
z,
whence
x)
(E). independent
the
to
F(a+1-y, of
which
= 0
1 .2....n.y(y+1)....(y+n-1)
S+1-y,
change
solution
aSy
a(a+1)....(a+n-1)S(S+1)....(S+n-1)
to
transformed
z
',
equation
S)x]-
dependent
transformed
of
+ a+
integer
change
solution
(1
=40
x1yF(a+1-y,
If
is
rational
is
(2)
x)
a
we
(E)
y,
transformations.
series 00
S,
Gauss'
symmetry
hypergeometric
theorem.
differential
-
E(a,
a
and
[y
numbers
by
table
d2ydy x)---+ dx
complex y).
Schwarz'
hypergeometric
with
given
I.
E(a,
S+1-y,
of
x)Y-a-F(y-a,
x
into
1+a+S-y)
for
t.
a+13+1-y,
E(a+1-y, we
S,
variable
S+1-y, )lace
re
by
Y,
-1-
and
1+a+S-y
(E): E): -a ,
by
x
=
Hence,
1-x) 2-y),
y
t
x).
y and
=
xl-iz x
ty
is 1
a -
x ,
soluand
1-t
and
In
(3)
we
a'
= a
+ 1 -
multiply
it
by
x(1 1-y
is
a
and
-
solution
a,
IS and
y,
3'
=
xl-i
then
x) (4)13F(1-a y-a-
of
(1)
y'
= 1 + a
-
(4)
we
F(a,
13,
(1
-
x1-yF(l+a-y,
(8)
xl-Y(1
If (E)
is
1+a+(3
-y,
1
= 1 -
-
= 2 -
y
x)
x by x
x),
y-13,
1+3-y,
1+ a-(-Y,
1+a+3-y,
x)Y-a-F(1-a,
change
-
y'
(E):
the
1 -
1-3,
independent
transformed
t2(1
y,
2-y,
y and
x'
of
-
we
1-3,
replace
x) (6)F(y-a,
(7)
Q + 1 -
,
+ 3-y,
solutions
(5)
y by
(E).
In
obtain
then
replace
1 -
x),
x),
1+a-3-Y,
1 -
variable
x
x).
into
t
by
t
x
to
t)d
-
t[ot
-
1
-
(Y2)tldtdy
al3Y
=
0 .
dt
Here, Then and
let this
change
equation
the is
dependent
variable
transformed
to
E(a,
y
into
a-y+l,
3' multiply
=
(1) a
-
-
(8)
y
+
them
we 1, by
replace y'
x-a
=
3, a
-S+
then
-2-
y and 1,
x by x'
z a-3+1)
z. In
and
us
=
x-1
by
Y for
=
cc t —z . t
(9)
x-aF(a,
a-y+1,
a-6+1,
x-1),
(10)
x
6-y+1,
6-a+1,
x-1),
(11)
xa-Y(1
-
x)Y-a-6F(1-a,
y-a,
6-a+1,
x-1),
(12)
x6-Y(1
-
x)Y-a-13F(1-6,
Y-6,
a-6+1,
x-1)
(13)
x-aF(a,
(14)
x6Y(1
(15)
°B-1 xF,13+1-y,
(16)
xa1(1
are
and
13E(3,
a+1-y,
-
x)Y-a-F(y-6,
y!
=
obtain
1-6,
a+6+1-y,
-
1
y+1
-u-13,
1-x-1),
1-x),
1-a,
of (9)
1-x-1),
-
solutions In
a+6+1-y,
y+l-a-6,
1-x-1)
(E). (16)
4. a
solutions
we
-I- 6of
replace
y,
"°
y
and
=
1
x
-
by
x
(E):
(17)
(1
-
x)-aF(a,
I-6,
a-6+1,
(1-x)-1),
(18)
(1
-
x) -6F(6,
Y-a,
6-a+1,
(1-x)-1),
(19)
xl-Y(1
-
x)Y-13-1F(6+1-y,
1-a,
6-a+1,
(1-x)-1),
(20)
x1
-
x)Y-a-1F(a+1-y,
1-6,
a-6+1,
(1-x)-1),
(21)
(1
(22)
xl-Y(1
y(1
-
x)-aF(a,
-
y-6,
a-6+1,
x(x-1)-1),
x)Y-a-1F(a+1-Y,
1-6,
-3-
2-y,
x(x-1)-1),
(23)
(1
(24)
-
x)
x1-y(1
Thus of
we (E)
Then
(3F((3,
-
have
Y-a,
x)
13-a+1,
x(x-1)-1),
1F((3+1-y,
1-a,
2-y,
Kummer's
table
of
completed
x(x-1)-1). twenty
four
solutions
[25]. Let
us
X =
1
in
the
set -
y,
p
=
equation
a'
=
a,
xi
=
1
y
-
a
E(a', V
=
-
3,
v
(31,
y')
for
S,
=
a
+
-
a'
=
a
-
(3.
t
=
1
-
x
(3 +
1
-
1-1
=
X,
we
have
v'
=
and
In
the
_
yi
equation a"
=
=
p,
E(a", a,
p!
(3",
3"
=
a
=
y")
y‘
for
-
y
+
pn
=
y"
t
-
=
1,
x-1
y"
and
=
a
-
=
v,
z
=
S +
xay
a;
=
we
have
1
and Au
In
=
the
1
-
y"
=
equation a*
V,
E(a*,
=
a
+
-
y*
1
-
(3*,
-f311
y*)
for
=
+
y,
z 1
= xy-ly -
y,
_
(3*
=
we y*
=
a"
-
(3,u =
have 2
-
y
and A* Thus,
=
1
+X, Let
+I-1, us
differential z = ay
= +v
-X, are
p*
=
permuted
consider
Gauss'
equations
(cf.
y*
-
a*
on
Kummer's
=
+ xy'
-4-
9,
=
a*
-
of
hypergeometric
table.
transformations
Goursat[
-9*
§34]).
If
we
set
v.
A.
v.
for of
a non-trivial E(a
+ 1,
IS, y),
a(a For,
we
solution
+
1
have
y
which
-
y)
the
=
can
of
E(a,
5,
y),
then
z
be
zero
if
and
only
if
is
a
solution
+ a
-
1)x1Y,
0.
identity -
where
E =
(1
is
the
E
tion
y
of
z1
and
is
-
1)y
=
(y a
side
of
i)x-1lz
(E).
+ a(1
For
a
non-trivial
solu-
set a
z3
+ a -
hand
-
(y
only
-(1
(y
=
a
-(x)y
+
x(1
-
x)y',
+ xyl,
-
13)y
solution
+
of
(1
E(a
-
x)y'. 1,
13,
y),
which
can
be
zero
if
if
(1 z2
we
z2
is
x)z° left
(E)
z1=
Then
-
a
m a)
a)
solution
of
=
0;
E(a,
y
-
1),
=
0;
+
I),
which
can
be
zero
if
and
only
which
can
be
zero
if
and
only
if
(a z3
is
a
+
1
-
solution
y)
(3
of
+
1
-
E(a,
y) y
if
(y For,
we
-
have
a) (I the
E = z'1-
-
= 0.
identities: (1
-
= (1 - x)z2+
a)x-1z1
+
(1 -
(y - a -Q-
-5-
a) (y
1)z
-
a)x-1y
2 - (a+1-y)($1-1-y)y
= xz3+ yz If
we
-
3
a)
-
-
set
-a
y
=
x
Y2
=
xa-y(1
n,
Y3 =
x1-yn2,
-
(1
-
111'
x)Y-a-T-1
3
and
z
=
z2
then
we
x-c,
=
z
2-y_
x
=
1
2'
-
z3
=
(1
-
=
-
dt
''1'
x)Y-u-i3+1-
x)1+Y
3,
have
=
-
-1
--fl d t'
1
and
=
2
dx
2'
c
3
=
d dx
•3-
-6-
=
x
§2.
Reducibility
The tion
equation
y whose
of
x.
The
3,
y -
a,
that
a
is
solution
an
a, (4)
in
of
3,
has
our
that
of
x-1
If
0<
in
case
property. is
F(a+l-y, >
case
3,
case
case
a
given
y is
the
case y
<
a a
<
F(3,
ya,
<
0.
>
1
where a
not
an 3
(1)
in
a-3+1, and
integer. 3
<
polynomial
has
our
(23)
has
our
F(1-a,
)is
>
0
solution
2-y,
-7-
x)
has is
given
our
polynomial has
our so-
(1-x)-1in and
in
of
Secondly
(4)
solution
a
prop-
then
a polynomial
1-3,
our
polynomial
of
property.
by
polynomial
(9)
a
property
(1-x)-1))s has
a
First
given
solution
3
solu-
integer.
)is
a polynomial
(20)
the
If
the
is
Then
0.
case
where
x-1i
the
assume
other
an
x(x-1)-1i
I]).
a polynomial
the x,
a,
of
us
solution
a-y+1,
(1-x)-1))s
Then
in
of
solution
then
solu-
one
Art.
y is
solution
a
function
that
a
of
the
case
has
Let
< 0 and
y,
an
is
integer
that
integer is
it
rational
follows.
polynomial
F(a,
a
finding
an a
> y,where
(19)
y
0
a
not
is
a polynomial
Then
3-a+1,
solution
as
Suppose
solution
1-a,
by
case
is
then
a-3+1, the
that
in
is
if
(Schwarz[35,
proved
y
that by
F(3+1-y,
assume in
and
3
where
where
x
0,
a<
1-3,
integer
table
x)
> 0.
Suppose
lution
a
<
x(x-1)®1in
condition
property. y
a
y,
reducible
sufficient
be
(1)
be
171/y
If by
singularity.
derivative
Kummer's
2 -
to
rational can
given
case
erty.
a
integer.
1 -
assume
(1)
3, is
from
is
tion we
-
logarithmic
said
and
sufficiency
property
-
is
necessary
this
F(1
(E)
logarithmic
y
The
and
is
case the
(1
let a
-
x)-1,
us
polynomial
property by
other
(1).
of and
If
in a
<
y
then
F(a+1-y,
and if
the a
(3 is if
is and of
is
an
an
integer
is
in
the
if
that
one
a
-
1 -
case
y-a
order
to +
is
6 is
p
+
that
we may
assume
a
< 6, the
--+
is
-
non-trivial
A +-
that
it
odd
x)
solution
y of
(1+a+6)x x(1 - x)
z-
(E)
reducible
an
integer y-a-
has is
a
our
1-x) property,
polynomial
has
A -
to we
u -
1 2(1
-IS=— set
z
(E) been
sufficient since
show
have
v) ,
-++
v) .
= y'/y.
that
Then
2
+
y
a6 -
x(1
x)
-0,
is
z' If
z
the
function
+ z2 +
(1
+x
logarithmic of
x
- 1)a6(1-
derivative we
can
express
of z
our as
the
xx
) = O.
solution
y
sum
of
is
partial
a
rational fractions:
e..
z = P(x) + y y 13 i j (x - ci)j='
0 < i < n+1,
-8-
1 < j < r.,
is
proved.
satisfies z'+
1)-1
Then,
Therefore,
is
we
-
y-a-(3+1,
(6)
a
is
neither
y-13,
integer,
v) ,
x(x
is
sufficiency
necessity
of
integer. a
property.
v) ,=—2(1
a =—1(1 2
and
2-y,
The
an
an
solution
our
integer.
(E)
is
F(y-a,
6+1-y, has
v
y-a
integer
case
the
Thus
an
F(a+1-y,
an
polynomial
property.
x where
prove
A +
a
now
-
solution(2)
if
of
a
In
of
11 a =—2 (1 y
y -
integer.
the
our
Hence,
other
x where
In
has
integer.
an
x(x-1)-1))s
Suppose
polynomial
reducible
For
(22)
integer.
an
y a
2-Y,
solution
y is
nor
1-13,
it
where
P(x)
By the
is
a polynomial
x and
ci,
e1"are
complex
numbers
differentiation je..
-I1,1f--N z' =P'(x)
7 L 711 - X j (X
Comparing the order every
i,
pole
of
and
at
P(x)
that
C )1-1
of the pole =
0
point.
at
x
=
at x = ci we have ri = 1 for
co by
comparing
=
order
of
the
Hence,
e. z
the
e.
X
=
C
-
i
(x
-
2
c i)
Let of
us
set
c0=
0
(x-c.1)2we 2
of
every
x-2
i
and -
c1
=
1.
ei
=
1
Then
comparing
the
coefficient
have o,
1 for
and
different
(x
from
we
0
-
1)-2
e20+
(1
-
A)e0=
(1
-
u)e1
and
1.
Comparing
the
coefficient
have
e 0
+
l
+ el
0
and
Hence,
e
e0is
+ either
0 or
= 0. A and
e1
is
either
0 or
u.
Multiply--
2
ing
the
equation
x2z'
+
for
z
(xz)2+
by
X +
X2
+
we
(1--X+x
If we set X = X e.1then -
x-
(2
-
A -
have
x1-
for x =we 11)X
+
as
and
-9-
-
1—)xz
+
obtain =
0,
a(3 x
1
=
0.
x =2(X p - 1 ± v), which
takes
n,
one
of
n + A,
Hence,
one
Let
of
us a
the
following
n + u,
+X +p
is
not
at
x =0.
We assume
is
y = 1,
then
is
the
rational
4,(x) = F(a,
an
odd
integer
logarithmic
integer
is
A is
have
a
an
no
M. Kuaa[24,
of
of
logarithmic
integer.
solution
R, 1, x)log
(cf.
singularity
there
that
we
values:
n + X + p.
+v
consider
four
If
the
§19]).
(E).
If
singularity it
is
zero,
that
form
x + / A B xm, m=1
mm
where a(a+1)•
A= m
•(a+m-11iiO3+1)•
•((31-m-1)
-
_
(m:)2 which
is
m
the
coefficient
1+1+1 =--r---1-". a
of
xm in
F(a, 1 B 11
++,++
1, 1
x) •••
and 1
+
a+1a+m-134.1
(3+m-1
- 2(1 +2+ ••• +-)• 11,
The power series
AmBmxmconverges
sumed
that
a
a or
IS is B
where
neither an
m
=
N
is
(a
In this x = 0 is
integer
0,
m
the
+ m)(ii
caseYA a
nor
>
Q is
less
an
than
1
+ m)
of =
m such
we
less
than
1.
we
as-
If
either
set
that
0.
mBmxmis a polynomial singular
point
-10-
m
integer
Here,
N,
minimum
logarithmic
in Ix! < 1.
of x, and in each of
(E).
case
A
Suppose 1-y
that d1
x4)(x)
y
< 1.
Then
(E)
has
a
solution
of
the
form
y
dx1-y by
Gauss'
transformations,
and
we
have
d1-y
dxl-yF(a,0, if
and
1,
only
this
we
... 1 -.m.y.
forms
which
a
Suppose -
(1
have
a
system
that
y
> 1.
dy-1
(E)
0. of
the
0
solutions has
form
y,
with
a
solution
(2). of
the
form
f(1 - x)(14--10(x)} we
have
.„
y_lf(1
and
xm
of
Then
and
y-1
dx
solution
fundamental
transformations,
Q'
if
polynomial
.(13+m-1) .(y+m-1)
x)
Gauss' Gaus
=
.(a+m-1)13•
dx)-1 by
U
-(a-y)0(0+1)-
case
a-
=
if
a(a+1)In
x)
only
- x)u-l-f3-1F(a,
1, x)]. = 0
if
(1- a) (2-
a) •
• ••
• (y- a-1)
(1-13) (2-(3)
•
• ••
• (y-13-1)
= 0,
since
-
(1 by the
(3).
In
this
, case
R,
1,
we
have
x)
= a
F(1-a,
1-,
rational
1,
x)
function
solution
of
form for
X
with
x)a+0-1F(a
0
1-1 y (a+1-y). 1. <
m
< y -
••• ya+m-y)(13+17y): -m.(2-y).
2,
which
.(m+1-y)
forms
-11-
a
fundamental
-((3+m-y) xm system
of
solu-
tions
with
(1).
Therefore, is
no
logarithmic
are
such
(S,
13-1-l-y)
(cf.
singularity
that
consists
follows
the
condition
singularity
integers
Goursat[g It
mic
under
one
of
,
of
that
A is
at
x = 0 if
and
the
couples
(a,
numbers
h and
k
an
integer,
only
if
a+l-y)
satisfying
there a and and
h < 0 < k
§18]). that if
an one
irreducible of
A,
p and
-12-
equation v is
a
(E) rational
has
a
logarith-
integer.
13
§3.
We that It has
is,
X,
be
a
assume
neither
will
sume
shall
a,
proved
that v
13,
y
is
-
a
is
a
equation
nor
every
no
table.
our
y
-
solution
13 is
a
of
(E)
in
logarithmic
rational
(E)
solution
algebraic
there
1.1 nor
that
that
non-trivial
Schwarz'
is
irreducible,
rational is
§7.
algebraic
Hence,
singularity,
integer.
integer.
we
that
We write
(E)
if may
is,
as-
neither
as
(E) ddx22ydy + p(x)+
dx
q(x)y
=
0,
where
(x) Take we
a
_
y , 1 + a =+1-1-a+Yq(x) x
fundamental
denote
the
+
system
ratio
-
= aP1
1,'‘1x of
yl/y2
solutions
of
them
y1 by
z
- x11°1
and
then
y2 we
of
) (E).
If
have
z' = (Y1Y2Y1Y2)Y22' .
z"_Y1Y2Y1Y2Y2Y2 z' YiY2 yly2
2--
y2
d(z",171
=-
p(x)
= -
p'
Y21
.
_ + 2q
-
Y22Y2
= y1/
Y2
7p 11 22 ++
....
-PI
+ 2q+
Y2
n. 2
Y2z
2--,
y'2,2 J2l'Y2
dx'z''=-Iat-2-1----2-=
whence
-
y2
satisfies
dz"1z"212 (S) asi-(To -7(7T)p'
2q
-13-
-7/3
2P7-+
it
2-2-
7
_l1-X+1--p 2
The
left
hand
x2
(1-x)
side
is
the
2
2
2
X-
2
+
2
p x(1-x)
Schwarzian
v
-
derivative
of
The wronskian W = yiy2 - 17117satisfies
with
a
W =X-1
-
constant
C distinct
solution
Z of
(S)
1
z.
W' = - pW, and
1)p-1
from
0.
If
we
set
y = /W/Z'
then
yl =-12--(11471--111-)y = --1(p + ;1-)y, =
Zr'YI
c1(;)111
and
y"
= =
whence
1
=py'1-(-1)TT)
y
py' -
is
p177
a
Z"
(pTOY
Z"
1 d Z" qy74Ti(TT)
-
gy
1dZ" 4[p°a)7(i7)]y
-
1 Z" 7(fr)2
P'-2g27P21/1 1
,
solution
of
(E).
If
we
set
y3
= yZ
_1
yi = y'Z + yZ' = (y'y3 + W)y = (11%7 3 =
=
-
(py'
YiYi +
P(YIY 3
WI)Y-1
gy)y3y-1-
W)Y
Y317'17-1
pWy-1
- (1/73=
-14-
PY3
clY3'
then
for
a
whence the
y3
is
a
solution
of
Therefore,
Z
is
expressed
in
form
Z
with
C1y1 + Cy+ _112212
=
constants
At
C2y2 C' yC'z
C1z
every
of
theorem
We
shall
from
solutions
yl
and
W is
Schwarz
y2
our
z
point
0,
following
(x), coefficients.
such
we
can
that
take
a
y1(x0)
=
0
0 by the
have
(E)
are
has
only
rational
numbers.
algebraic.
For,
assumption.
algebraic
This
is
we
due
If
z
have
to
Heine
(cf.
1
and
E (0)
-
1)Up
(x-1)
x-vc(x-1),
n(x-1)
... we
can
take
yl
and
y2
such
that
form:
x-(x),
(x
=
co we
p.298]).
the
=
y
are
by
z1=
real
and
algebraic
the
and
y2
equation
13, and
= /W/z'
zo
where
u,
y2
At takes
yl
our
= zy2,
[35,
1
and
and
that is,
then
yl
0,
Hence, yl(x0)
solutions,
assume that
algebraic
C.
.d
singularities, is
C'
point x0.
of
dz (x)=0 x0'
+
different
and y2(x0) = 1 at this uniqueness
C2
12 and
point
system
+
-
C11C2'C'
fundamental
z
(E).
0,
,n(0)
0,
c(0)
and
0,
C(x-1)
Hence,
are the
line
-15-
convergent segments
power (-a.,
series 0),
with (0,
1)
and
(1,
+0.)
of
passing
real
through If
m and
n.
The
upper
rior
of
by
z.
<
a
are
0
(L)
<
f(x)
dicates
the
property
crossing
domain the
analytic
(0,
1)
the
circles
plex
and
numbers
is
> 1,
f(x)
is
<
Gauss'
trans-
all
integers
for
1.
transformed
may
the
to
angles
occurs
in if
the
Then the
only
origin of
if
V7
two
the
exterior
and
inte-
117 and
that
origin. or
the
X7,
suppose
holomorphic
at
x0
The = z(x)
of
real
corresponding is
by
the
the fol-
X + p > 1 + v.
symmetric
+c)
respectively.
p + v > 1 + X,
continuation (1,
lines
satisfied:
conjugate.
which
at
case
holomorphic
z(x)
z we
interior
former is
that
+v
vertical
function
is
<
with
v + X > 1 + p,
tion
0
of
of
that
triangle
the
z
property
is
in
The
a
1,
choice
X + p + v
If
<
segments
then
this
assume
u
and
numbers
lines
condition
has
may
z1
complex
either
circle.
lowing
1,
of
z0,
to
solution
13+m, y+n)
<
two
by
algebraic
suitable
contained
third
an
we
X
transformed
origin
circular
By
are
Hence,
plane a
circles is
has
E(a+k,
0
-
the
(E)
formations 9,
numbers
transformed
by
at
definition,
functions
z0,
z1
by
a proper
choice
in
the
of
of
z across
line
numbers to
them. to
-16-
the
x0
the
interior
the
func-
where
the
bar
and of
real
have
its
in
of
an
plane adjacent
a
Hence, (-00,
reflections lower
in-
the
branch
segments
the the
z
numbers.
line
induces Thus
then
0), in
of
comcir-
cular
triangle.
A many-valued singularities valued
is (cf.
is
[9
first
,
a
?r,
sition
of
the to
common the
tions
only
hedral
(cf.
tical
{1 A solution (1
-
u, -
X,
be
an
finitely
proved
many-
that
algebraic
the
con-
solution
proved
with
groups
yl,
in
here
by
(cf.
Landau's
x)1+a-1-(3-yF(a,Q+
-0,
y,
sphere
with
is
the
negative
vertical The
rotation
angle
have a
compo-
around
rotation we
we
equal
three
finite
rotagroup
solution. the
dihedral
group
and
third [40,
the
11, 1 -
{X, x).
-17-
the
tetra-
icosahedral the
group:
fourth
as
a
sub-
Chap.VIII]). four
values
-
group,
and
form
{1 to
Then
projection.
generate
the
1 -
1,
unit
they
following
corresponding
the
Thus
circles
1 -
satisfied.
angle.
Weber
{A,
is
the
are
octahedral
the
(L)
sides
algebraic
great
p,
two
and
instance
have
{X,
an
on
vertical
has
three
angles
it
algebraic
p.230]).
stereographic
them
contained
for
The
a
vertices,
the is
has
will
in
this
finite
group,
[2,
condition
V7 by
of
(E)
second
group
and
the
if
is
triangle
of
Such
The
the
vertex
around
and
It
reflections
twice
if
only
only
§5.
that
the
and
(E)
VI]).
in
iir
has
Schwarz
if
great-circular
angles
which
consideration
Chap.
Suppose
function
Bieberbach
satisfied
theorem
have
if
instance
geometric
(L)
Goursat
algebraic
for
By dition
analytic
A, 1 -
triangles
whose
multiplied
by
ver-
7:
v}, 1 p,
p, 1 -
v}. 0
is
given
by
if
We assume has
the
least
algebraic table
that
area
among
solution due
to
1
(I)
if
Schwarz
Dihedral
t2,
111
2 5'
(III)
triangle them
and
corresponding
and
only
if
1
2,
yl,
v is
u,
vl
is
Area/7
in
7'71'
Area/7=1=
A,
1 7'
1 =} 71'
the
= v.
1 (III) 7= 2A.
group:
=I2 = B,
111
{.5'71-'74-}'= Icosahedral
2B.6
group:
1 2'
1 3'
-2 5'
1 1 —3' —51'=
1 (VII) 1 15 2C,
1 -7'
1 1 51. 7}'
1 (VIII) { — 152C,=
(IX)
12 -S11 {.f''71'
(X)
{ -S.' 3 1 5'
11 (VI) 71'
Area/if
==
c,
--0= 1 1 }'
215 --
3C,
= 4C, {
-1' 5—1'55—1'-5}=
6C,
-18-
{A (E)
group:
111111
6
{A,
Then
arbitrary,
(IV)1--r7r741-J,Area/7 1
A > p > v.
group:
Octahedral
(V)
to
[351:
Tetrahedral
(II)f'
the
1 (XI)
P. has following
vl an
5
(XII)
{— 2 1 3 'T"5'
(XIII)
{S,51'-1-1-5-= 12
(XV)
and
60,
braic
z the
tion
n
to
z
in
these
case
are
denominator
z)
and
groups
=
0
of
has
its
3
70=7C, 1
of
In (I)
with
-19-
respectively v.
degree
N respectively,
triangle. the
6C,
10C.
the
F(x,
corresponding of
is
M-1N/2
{7''-3}
1
3'
N of
equation
equal
5'
order
where
1}Area/7 == 6C,
51 (XIV)
{11= 5' The
1
The
irreducible
with
respect
where particular v = 1/n,
2n,
M7 is x
is
(II),
the a
12 ,
alge-
to
area
rational (IV)
24
and
x
and of func-
(VI)
.
Chapter
II.
§4.
Landau's
criterion.
Eisenstein's
If a.convergent
theorem.
power series
y = X cxn
whose coefficients
n=0n
are
rational
the
field
numbers C of
distinct
which
field
as
see
n
y)
y
=
is
an
exists
integer
of an
for
Eisenstein
0,
algebraic Let
which
F(X, Y) = X Ak(X)Yk, k>0 We may assume that
there an
of
function
x
over
integer
every
(cf.
A
n.
Landau[26]),
follows.
numbers.
F(x,
is
theorem
as
algebraic
then
Anc
a
that
rational
that
that
proved
shall
an
numbers
such
known
be
Q of
such
0
is
will We
T
complex
from
This
expresses
we
function
F(X,
Y)
write
Ak(X) =
a
x
over
polynomial
the over
as
ak.X,
akj c C.
j>0
co = 0, and under this
yk = xkXcnkxn'
be
of
cnk E
assumption
k > 1.
n1 For
every
m
we
have
y a.cnk= We can that
take they
a are
0+ system
all
k and
of
linearly
a=Xb.,0 kjk for
k + n = m). finite
complex
independent
over
Q
jhehbkjhE j.
Then
y ehkjhcnk=
0
for
every
m we have
(j + k + n = m)
-20-
—
numbers and
eh
(h
> 1)
such
and
b.khcnk=0 for
every
(j
h.
Hence,
y
(Yb j khx)yk= for
all
h,
and
+ k + n = m) satisfies
0 k jj
one
of
them
gives
us
Thuswemayassumethatall.
a
non-trivial
ak ]are
that of
they their
are
integers
multiplying
denominators.
a00=
a02
0,
+
They a01
a10c2
relation.
+
+
=
+
Y)
by
numbers
a
common
and
multiple
satisfy
a10c1
a11c1
rational
F(X,
0,
a201c-=
2
0
and a.c.+ j10j11j-11,j-11
ac.++
a.c O
j -1 h) i>2
for
j
> 2. First
a10
h=0
we
assume
integers.
a
are
a10
A3r=
-
does
not
vanish.
If
we
set
a
then
01'--220 are
that
2k-3
integers,
If
ck -1°
we
2
assume
<
k
<
a2(a+ 02 c"-1- a11
1c+
a
c2) 1
that
J
then J'-J-'-
a 2j-1 cj = - a2j-2(a+Yac+x 0j h
=1
X y ckck...ck),k1 1
2
j-hi>2
ya.
h=0
ill
X
1+ -0-+ k.=j - h
1
-21-
.
ill
=
is
an
integer,
a
because
2-i-2 -
ck
ck
1
.
2j-2-2k1-•••-2ki+i =
2k1-1
a
2k--1 ck
...a
ck
1
.
and
2j
- 2 - 2k1 =
for
i
>
2(h
2.
b
n=0
that
b
whose
we
nXn
n
-
- 2ki
1)
+
i
Therefore,
Secondly B =
-
we
does
assume
A
vanishes. 0(B)
for
power
0(H)astheminimumof0(B vanish
as
a10
a10
and
are
take
order
vanish,
coefficients
can
that its
- 2 -
2(j
- h)
+ i
0
we
define not
>
+ i = 2j
a
as
of
our
For the
polynomial
series
in
X we
theorem.
a
power
minimum
of
H =
series such
BnZn
define
its
11).[RewriteyandA.which
n of
Z
order does
not
as
y = cxY + cXYc(i)+0,
0 < y < y'
<
a!
A.(X) and
= a.X
define
z
1 + a!X
we
replace
in
the
form
F(X,'
0,0
'=
Y in
Y) = F(X,
z = c'XYy+
we
F(X,
Y)
by
XY(C + Z))
Y = XY(C
= F(X,
set
-22-
2
< a!
<
cux''y + Z)
then
it
CXY) + FY (X,
+ 1-!FYY'(X If
< a.
by
y = cxY + xlz, If
1 +j)
is
expressed
CXY).XYZ
CXY) .x2yz2
E(X, then
we
C)
=
F(X,
CX')
= XYF
(X,
have
Ec(X,
C)
CXY)'ECC(X,
=
C)
X2yF
CXI)
(X,
YY
,
and
F(X, It
Y)
=
where
we
y)
we
write
we
(X,
c)
< 0(Er(X,
E(X, C)
C)Z
+1—TE 2 . CC(X'
C) Z
2
+ EC (x,
c)z
+
1
2!
.,
CC
x
,
C
) z
2
=
C)
c))
< °(ECC(X°
c))
<
f3,1 E(X +
0 <
as
O(C)X
+
(1)1(C)X
13
have
1(c) cD(c) L
= E(x,
c))
,
and
+ E
have
0(E(X,
then
C)
satisfies
0 = F(x,
If
E(X,
=
= =
a,Ci 0.
Let
O(F(X,
= us
X(c
+
a.
+
i1)
set Z))
and G(X, We
Z)
=
X-LF(X,
XY(c
+
Z)
= X-L{E(X,
c)
+ E
Z)).
have
G(X,
(X,
c)Z
and
EC(X,
c)
= V(c)X13
+ 011(c)X+
ECC(X'c)=0"(c)X+cl,"1(c)X1
+
-23-
+
E(
X :CC' 2 c) Z 2 ±
•• •}
Suppose
that
c
is
have(s)(c)
c))
whence
L = a.
we
in
then an
0(C) index
> jy
c
the
first
can
not
j
is
0(C)
of
assume
multiplicity
that
order
M is
z.
of
If
less
0 and
> y.
+ Z))
root
for
from
s.
Then
we
of c
than
1 such
By the
0(C),
is
then
40(c)
a multiple three
that
terms a.
definition
0 and
root and
of there
+ jy
= g.
Q.
Then
define
zm
0(C), is
We have
J
= XL-YGZ'(XZ)
F(X,
FY'(XXcnXn) its
simple
consist
L =
X1 (c
a
case
distinct
FY'(X
and
of
= 0,
If
> y and
We may
root
0 and
0(=--E(X, acs
are
a
Y)
is
irreducible
over
we
have
0 finite.
Therefore,
if
we
by
(m-1) y = cx1 then the
there first
+ c+• is
case
an for
• •+ index z
m not
cxy(m-1)xy greater
. m
-24-
'
+ xymz than
M such
that
we
are
in
§5.
Landau's
We shall that
neither
pose
that
ery
solution
power has
assume A,
the
that
v,
a,
equation is
series its
p,
first
a,
3, (E)
Q,
coefficient
y,
f3, y
y -
x)
cnof
are
an
theorems.
rational y -
(E)
which
is
is
numbers
13 is
algebraic
since
the
second
a nor
has
algebraic,
F(a,
and
an
and
integer.
solution.
Then
irreducible
algebraic
Sup-
(§7).
by
our
evThe
assumption
form:
a(a+1)....(a+n-1)(30+1)....(13+n-1) n
-
We
a,
write
By
n
not
isfies
a
=
c,
m
are
the
v
+
<
natural
y
=
w
numbers
+
=,
and
u,
a,
v,
b,
w
are
integers.
form:
theorem Ancn
divide
M
c
-n m
l.2....n.(c+wm)....{c+(w+n-l)m}
Eisen Eisenstein's
does
as
a(a+mu)....{a+(u+n-l)m}(b+vm)....1b+(v+n-1)ml
=
tha that
ther
+
takes
such
the
b,
cn
c
u
y
m,
a,
Then
Q and
=
a
where
°
is
an
A.
there
is
integer.
an
integer
Suppose
Then
if
a
E 0
(mod
A distinct
that
a
non-negative
from
prime
integer
number
x
0 p
satisfies
cona congruence
e
is is
+
(w + x)m
a
non-negative
the fa
p),
integer
y not
greater
than
congruence +
(u
+ y)m}{b
+
(v
+ ym)}
-25-
E 0
(mod
p).
x which
sat-
We have
infinitely
form
p = 1 + km with
case
of
for
x = ck
-
prime
a positive
Dirichlet's
instance
many
theorem
Takagi
[37,
numbers
integer and
§10],
can
p which
k.
be
Landau
This
proved [28,
take
is
a
the
special
elementarily §108]).
(cf.
,If
we
set
w then
c + (w + x)m = c + ckm = cp = 0 (mod p). Here, IAL
Then
we may assume that
We
may
suppose
a
(u
we
+
than
lui + Iv! + iwl and
that y)m
=
0
(mod
p).
have
0 = a and
+
k is greater
+
u + y ak
(u+y)m ak
-
u
= a(p-km)
= ph >
-
with
an
ph,
whence h = 0 because
+
(u+y)m
integer
ck
+
u
E m(u+y-ak)
h. -
w
Since >
a
<
c
or
From
b
<
0 < y < x,
p) we
p > km and k > 1111+ lwl-
Therefore,
we
Thus we obtain
Then Here,
1-
Kummer's
YF(a
F(a+1-y, we 2
table
let
us
+ 1 (+1-y,
y,
8 + 1 -
2-y,
x)
is
have -
y
-m
ei-
c.
take
a
solution
of
(E)
of
of
x.
form: x
get
ph,
have y = ak + u and a < c by k > jul + M. ther
(mod
m
c+
1+
w
and
-26-
y,
2 -
y,
an
algebraic
x). function
the
a
in
case
+
a
1
>
a+
in
the
-
y
=
a
-
0
c
m
+
1
+
u
w,
u-
w,
1+
v-
w,
m+
v
<
y=
other
c+
a-m+
case.
m
0
case
b
1
>
+
in
the
are
-
y
-
-
y
-b
1
and
(a
+
c + M'
< 1)
Therefore,
mc+
-
a +
we
c m
+
Since
m
both
b
than
-1
b
case.
greater
have
(m -
-
c m
c)/m,
+
we
0 < A < 1,
<
-1
le
(3 +
-
c
+
m
c
<
1
c
+
m
<
1
<
1
>
c
m
<
b
m
<
1)
>
$
-
w,
0
b
<
v=
1+
v=
1
1-
<
1
y
+
1,
21
>
1
m
have
either
a
0 <
p
<
1
and
and
0
<
y
<
1.
a
>
+
.p.
obtain
we
and +
a
-
Q>
1
-
a
We have ab mm
(3 =—
-
-
m
get
p+
=_,
1
of
that
A+
a
<
c,
Suppose we
c
Similarly
-
other
a
a
<
0
in
m
c,
1-
13 +
a
1
and
-27-
-
IS3 =
A.
>
c
or
0
< v
b <
Since 0
>
Q by
1.
Then
a
+ a
$ -
< "Y 13 >
0.
-
1>
a-ab-+
Therefore,
a 1
+
1.
< b
and
A =
2-
p=
1
a y>
<
c
< b,
y
-
26
that =
p
is +
a
< y
<
1
+
Q.
Hence,
v
and 1+ Thus
the
condition
solution.
If theorem form
p =
then
we
is
e
can of
+
km
§198]).
our
discussions
are
infinitely
we
with
can
a
natural
prime
to
is
number
k
§1.
many number
has
an
algebraic
of
Dirichlet's
numbers
p of
(Landau a
> c
This
[28, or
remark
b
the
§108]),
> c
inde-
is
due
to
of
the
form
= a0e
number
a,
b and
+ hm,b
+ km such
Ihl + lil
c
k
a
integer
for
(cf.
k
+ Ijl
given
e
which
is
instance
is
+ im,c
relatively
= c0e than
there greater
+ lu
for
Weber
[40,
prime
to
m.
as
a0,b0,c0less theorem
that
p
e which
= b0e
numbers
By Dirichlet's
numbers
theorem
natural
write
prime
m.
Dirichlet's a
natural
p = e
prime
either
of
(E)
case
many
that
v.
[26].
special
natural
result
if
theorem
infinitely
the
Take
a
j.
satisfied
another
a
Q=
[27].
This
with
is
-y+a-
first
of
are
prove
relatively
Then
use
1
§3
Landau's
there
There =
in
1 + km with
pendently
p
is
we make that
Q>
(L)
This
Landau
+y-a-
is
a prime
than
Wand
+ Ivl + Iwl.
-28-
m and
+ jm integers
number
p of
h, the
form
If
we
set c
We
may
x = c0k
+
(x
+ w)m
suppose
that
a Then
-
we
+
(u
j
-
w,
= c0e
+ y)m
then
+
E 0
(j
we
have
+ x + w)m
(mod
= co (e
+ km)
= cop.
+
+
p).
have 0 ri a
+
(h
(u
+ y)m
+ u + y
= a0e
-
+
aok)m
(h
+ u + y)m
(mod
p)
and h
with
an
+
u
+
y
-
integer
a0k
H.
=
Hp
Since
a0k-h-u>
-
0
<
y
<
x,
we
get
h
+ 1j
Hp
and
c0k+h+u-j-w>Hp,
whence
H = 0 because
Therefore,
we
y
=
p > mk and k >
a0
<
a0k
-
c0or
If
we
h
-
b0
<
replace
u
+
+ lul
by
m -
(m -
e)
+ (a0+
(m -a0)
b =
(m -
b0)
c
(m -
c0)(m
+ iwl.
Thus,
e
then
co.
e
a =
=
wl.
have
and a0 < co by k > lh ther
u
(m -
e with e
-
< m,
m + h)m,
e)
+
(b0+
e
-
m + i)m,
e)
+ (c0+
e
-
m + j)m.
-29-
we obtain
ei-
Hence, is,
we
have
either
m -
a0
> c0or
b0>
either A
for
two
< B
(mod
a0<
m -
c0.
c0or
m
Following
-
b0<
m -
Landau
c0,that
we
write
of
m if
m)
integers
A and
B which
are
not
multiples
they
satisfy
AEA0'
BEBo
We have For have
(mod
obtained
every
the
integer
m) ,
0 < A0
second
p which
< B0
theorem is
of
relatively
< m. Landau
[27]:
prime
to
m we
either pa
< pc
< pb
(mod
m)
pb
< pc
< pa
(mod
m)
has
an
algebraic
or
if
(E)
Suppose
that
solution. A < C < B
A
(mod
C -A
(mod
(mod m),
m),
m).
Then
we have
A
B-C<
(mod
-C
m),
(mod
m)
and C-A
if
C -
a',
A and
i31 and
Landau's This
remark
(mod
C y'
B are
in
second is
not
Kummer's theorem
due
to
m)
multiples table
of
satisfies
independently Landau
m.
[27],
table:
-30-
the of
and
Hence,
we
the have
every
criterion
triple in
discussions the
following
in
§1.
1.
{X,
P,
v}:
{a, {c
2.
3.
fX, {A,
5.
{-A,
u,-0:
6.
{-X,
-u,
v,
-v
7.
9.
v}: ,
f-X, {u, X, y}:
-v, -A) :
c,
m + a -
{m -
b,
m -
{m -
b,
m + a -
00
c,
m -
{a,
-X,
17.
18.
b, -
{a,
{c
{- v,
u,-X}:
{-v,
19.
{-V,
20.
{-V,
c,
a,
a,
+ b -
-
c,
-
a,
m + c -
{m -
b,
m -
{c
a,
m + c -
b,
m + c -
{a,
{a, {m -
m + a b,
b,
a
c,
m -
m -
c},
c,
m -
m -
a,
-
c,
-
b,
m + a -
-31-
cl,
c},
cl,
c}, c, a
a
+ b
+ b
m + c -
m + c
c},
c},
m + a -
b,
-
b,
c},
c,
{m -
-A, )4):
a
m + a
{m
16.
b,
-
A,-yl:
{7u,,
-
00
13.
15.
m + c b,
{b
°:
m + c -
a,
f/14, V, A.):
{-P,
a,
-
12.
14.
c},
{c
{0
10.
11.
-
{a,
4.
8.
b,
b,
+ b -
-
c},
b
c},
c}, m -
a
-
a
-
b
+ c},
b,
m -
a
-
b
+ c},
b,
m -
a
-
b
+ c},
-
b},
-
b},
m -
m + a b,
b},
m + a
m + a c,
-
-
b},
m + a
+ c},
21.
{v,
22.
{v,
23.
{V, 4,
24.
{11,
-x}: p-}:
{b
-
c,
b,
{c
-
a,
m
{c
-
a,
b,
{b
-
c,
m
b -
-
al,
a , b b -
-
-32--
a,
a},
al, b -
a}.
§6.
For natural
a
given
numbers
satisfying prime
to
pa
(mod
m).
Then
a,
is
Suppose
co
an
p0
are
in
p0
(B)
There
is
such
{s,
2r -
{r
the
we
set
sixth
we
§5
we
s,
rl,
are
E prime
we
two
have
p0
co
X= p = 1/2
and
numbers have
with
the
{r
-
(mod
c,
m)
or
divisor to
of
p
m)
=
< pc
c
m such
1
relative-
pb
of
E b
to
co
and
and
< m.
that
pc
(mod
m).
satisfies
s
then
the
r'
=
s'
=
r
(B).
six
2s},
2s}.
case
r.
+ s,
-
first
<
r
2r
the
s
the
first,
the
m),
where
r
Then
from
triples
{s,
r
and the
+ s,
the
the
we
have
cases
In
the
fifth
case
co
table m =
2r:
fourth
and
third
are
r},
third
the
s
with
second,
and
-33-
(B):
m);
(mod
v = s/r,
{s,
s,
and
(mod
< a0p0
r},
-
from
c0
+ s,
2r
derived
<
(A)
r
s,
and
cases
s,
rl,
the
b,
integer
a,b0p
following
s,
in
a0p0
that
-
In
triple
that
2r
are
< pb
prime
s,
r
every
relatively
following
natural
respectively. and
the
that
-
(a,
a
n0),m=cn 0000°
every
of
for
common
relatively
For
end
m with
< pc
m,a0p
(A)
the
consider
greatest
assume
of
prime
than
pa
P
one
mutually
m we
that
the
b0<
(mod
Suppose
If
be
is
p0El
less
either
may
c0<
that
c
integer
We
< a0<
number
and
have
Let
m).
0
at
b
estimation.
criterion
m we
there
C0(mod
We
natural
Landau's
ly
Rough
= 2 and
and fifth
co
= r, we
are
in
the
case
We above
shall
stated
if which
Errera's
theorem.
m =
0,
(a,
if
r
prove
cussions,
Let
(A)
is
odd,
that
m is are
we
in
are
in
sufficiently
define
m) n1=
(b,
one
n1
in
order
and
n2
2.
The
(B)
of in
the the
to
if
of
is
three
even.
cases
following
dis-
establish
Landau-
by
of1nj number
r
m)n2.
13multiple 1,
case
great
unnecessary We
the
natural
numbers
with t
less
than
i,
j
=
m satisfy-
ing
t
E 1
(mod
is(m)/(n..13), natural
numbers
s
less
(mod
than
m),
(t,
m)
the
Euler
m
p0
= 1
such
E 1
function.
The
number
of
that
(mod
no),
(p0,
co)
= 1
flnol)/flno). Let
p'
13..),
wherecpis
E apo is
n
us
satisfy
the
suppose above
ap0 E ap6 where Hence,
e
is we
the
that
are
conditions
in then
the
case
we
(A).
If
p0
and
have
(mod e),
greatest
common
divisor
of
m and
(m,
a)n0.
Then
we
have
efflnol)/(1)(no) Suppose
we
that
-
1}
cp(nol)/cp(no)
21("n01)/"n0)
< co. is
- 1/
greater
C"n01)/"n0)
and
-34-
than
one.
have
efln01)/4)(no) Here,
if
we no
then
we
< 2co.
write
=
n0
and
n1as
df0,n1=
dfl,d
=
(n0,n1)
have
e =
(m,
(m,
a) n0)=
n01
= n1f0'm(m,
(m,
a) (n0,
nl)
=
(m,
a)d
=
a)
(m
n0
/f0
and
Hence,
we
a) nl(m,
a)nol/fo,
m
=
cdf00
'
obtain
el)(n01)
=
(m,
a) n0f014)(mf0-1(m,a))
> n0f-1
(mf
,
o)
and
4)(mf0) because
= f04)(m)
m is
divisible
Cb(m/no)
by
fo.
< 4)(m)/01-10)
Therefore, < en014)
we
have
(n01)"(nO)
< 2c0n-1= 2m/n0 Suppose ible Hence, ger p is
that by
"n01)"(n0)
(m, we
a),
obtain
p relatively odd
and
p /
pc0E
we nl
is
have
n0
= 2n0
prime
to
1,
3
2,
2c0(mod
equal < n1
and
and
a0
m),2c0=
/
a0,2a0,3a0(mod
m).
-35-
=
.
'
Since
= 2n0 a). p E 2
We have 4a0
and
pa0
(m,
that
n1).
one.
n01
=
m such (mod
to
< m,
with
There (mod
divis-
a0is
(no, is
n o) •
an
2) inteHere,
=
1.
Hence, of
pa0
natural
> pc0
(mod
numbers
p
p E 2 is
(mod
cp(m)/(1)(n0).
since c0/2.
-
p')
Therefore,
Let
e' us
is
the
than
these
p and
is
even
we
obtain -
and
11
that
< pc0(mod
m).
The
number
m satisfying
m)
greatest
suppose
pb0
n0')(p,
el{q)(m)/(1)(n0) where
and
less
For
n()1(p
m),
= 1 p'
b0
we
is
have
pb0
/
relatively
p'b0
(mod
prime
to
m),
a0
=
< 2c0, common
divisor
Om)/cp(no)
is
of
m and
greater
than
(m,
b)n0.
one.
Then
we
have „
,
(1)(m/n00) Thus,
we
have
< 4m/n-. either
Cb(nol)/4)(no)
= gi(m)/(1)(no)
2 or (1)(m/n0) < 4m/n;.
hold.
Then
and to
(m, 2,
b) then
we <
{2,
and
r
n0
pa0 Hence,
we
p2a0E
m = we
assume that
2n0=
n1and
obtain
(m,
the above equalities
n0is
b)
=
odd.Since
1,
2,
3.
If
c0= it
is
2
equal
have
2 + 2,
= n0.
= 1
We shall
have
2c0,
we
pa0E and
2
(mod 4}
is
Suppose E n0
+ 2,
n0))pb0E the that pco
2
fifth (m,
(mod
triple b)
E 4,
m).
given 2.
pb0
above
with
s
= 2
(mod
m).
Then, E 1,
3
(mod
m).
have
n0+
4,p2c0E
8,p2b0
-36-
E n0'+
2n0+
6
Therefore, Let ural
we
obtain
us
suppose
number
We have
p
the
number
than
that
of
yA. if
and
less
either
that
p
5
that than
and we
m < 10. are
in
m satisfying
<
c0or
of
p
for
for
a0p
we
we
case
p
1
have
(mod
the
the
n01).
efAcHnol)/0m)
-
11
< co
ef(nol)/qp(no)
-
2}
< 2c0.
We
latter.
Let
Hence,
a0p we
Take
n0),
former
We have
(mod
(B).
m).
have
1 itb q5(m)/q)(n0).
p E p°
the
> c0(mod
which
which
A>7
if
<
a0p
Then only
n0
a
(p,
may
nat-
m)
=
1
assume
is
not
us
indicate
E aop°
less
(mod
m)
obtain
and
Suppose
that
Onol)Pp(no)
is
3{C6(n01)/"n0)
-
greater
2}
than
2.
Then
we
have
(n01)/qb(n0)
and
Om/n002) By our such
assumption that
Hence,
there
a0p0>
T then
< 6m/n.
E
is
1
(mod
b0p0T
<
c0(mod
we
have
that
integer
c0p0Ec0(mod
m)
go(m)/q)(n01)
n01"m)/(Hn01)<
If
relatively
prime
to
T satisfies
1, and
-
p0
m).
n01))(T,=
n01{°m)/(19(n01) Suppose
an
they
11
< c 10.
is
greater
2c0.
-37-
are
distinct
than
from
1.
Then
each
we
have
other.
m
Since
("n01) we
= ("n0n01/n0)
(nol/n00)n0),
obtain
1
("m/n0)("m)/C"n0)n01n0("m)/"n01) Therefore,
either
(n01)/(1"n0) or
cb(m/no)
hold.
If
we
of
following
four
(ii)
m = n01
= 3n0,(n0,3)
(iii)
m = 2n01,
n01
(iv)
m = n01-=
2n0'n0E
in
is
the
the
case
fifth
we
have
either
Let
us
set
that
p = n0+ n0+
former
pa0E
cases
0 c0= at in
the
(n0,
6)
(mod
2).
are
in
case. the
beginning case
b0=
with
(i).
r
Then,
1 or
a0=
8
(mod
m)
2.Then,pc0E 2
(mod
m)
n0+
6
(mod
m)
n0+1. = n0
co
3,b0=
3n0+
6,pb0E Hence,
n0<
case
(ii).
Then,
5
= 4 and
2n0+
and
we
and
3.
have
and
m <
20.
Suppose
that
—
co
-38-
.
= 1;
1 and
—
we
2);
2,a0=
the
2n0+
2,pb0E
equalities
= 1;
then
1,b0=
above
possible:
(mod
= 3n0,
are
the
is
1
stated we
that
case,
3n0+
latter
(iv)
case
a0=
pa0E
the
the
= 1
assume
= 4n0,n0E
Suppose
in
We shall
m = n01
are
the
4(m)/(1)(n01)
(i)
s = 1.
in
= 2,
< 6m/n.
One
This
< 2m/no2
=
3
and
a0
=
1,
2.
We have
b0E in
case
case
(mod
1,
2n0
+
1
(mod
3) ,
and
n0E bo
in
n0+
E
n0
m).
2n0'+ E
1n0+
2
We
(mod
3).
2
(mod
m)
2
(mod
m)
Let
us
set
p
=
n0
+
Then
3.
9
pco
have
pa0E
n0+
3,
2n0
+
6
(mod
m),
6
(mod
m)
and b0E in
each
in
the
in
2n0+
case.
Hence,
case
(iii).
b0E
2n0+
case
n0E b0
in
case
n0
36
(mod
m). a0
E
2, (mod
4n0 E
n0
<
Then,
1
+
5
4n0
2,
5
and
co
6) ,
(mod
We E
3,n0+
=
6
m <
15.
and
a0
that
Suppose =
2,
4,
We
5.
+
4,
2n0
+
5
(mod
m)
+
4,
4n0
+
5
(mod
m)
set
p
=
no
+
30
we have
and
2no 6).
Let
us
+
Then
6.
c0
have
2n0+
12,4n0+
24,
5n0
4n0+
12,2n0
24,+24n0+
(mod
m)
and b0E in
each
case. From
c'
Let
n' 0
and
Hence, the
=
a
n" 0
table
+
be
b
-
the
c,
n0
<
at
the
c"
numbers
30
and end
=
b
defined
-39-
m < of
-
30
(mod
m)
174.
§5
a.
by
we
may
replace
c
by
are
m = (m, c')n6
= (m, c")n5,
and n be the least
commonmultiple
is
since
either
n
or
2n,
a =1(c By
the
+ c'
of n0, n6 and n5.
1
- c"),
b =7(c
+ c'
+ c")•
than
3,
Then,
inequality
3 > (1 + 1)N for
a
natural
number
N > For N
a into
N we
(N + 1)N/
given
(N+1)
natural
the
have
N > 2.
number
product
of
t
prime
greater numbers
as
let
us
decompose
follows:
h N
=
ilpHq-,
p
>
t
>
q.
Then,
(1)(N) = Hpa-1(p
- 1)Hqb(1
> Hpa-1+(t-1)/tHqbq
- 1) - 1
> Hpa(t-1)/tHqb(t-1)/t•q N(t-1)/t
If
we
't t
-- 2l't t
--
3
- 1 q 1 =-m(t-1)/t(t
have
Om/no)
< 6m/11'0",
then
(m/n
(t-1) o)<
/t2 6(t
-
-40-
2
1)m/no,
t
> 3
-
1)-1.
m
and
n0(t+1)/t
'
Let
us
and
n"0Then
< 6(t
suppose
that we
-
1)mt
this
> 3.
inequality
holds
for
each
of
n0,n 0'
0
obtain
(t+1)/t (nn'n")<
[6(t
0000 for
l/t
-
1)]3m3/t3(En<
[6(t
-
t > 3 with c = 1, 2, since m < En0n6n5. (Enn1/(t+1)3t/(t-2) 0'n0)<
[6(t
-
3/t
1)]n'n")
Therefore, (t+1)1/(t-2)
1)]
and
n<
< For
t
=
6(t
and in
n0< the
10
[6(t
-
t/(t+1)1/(t+1) 1)](Enn'n") 000
[6(t
-
1)] t/(t
we
-
2)21/(t
1) t/(t-2)=
Hence, (I)
of
545/4<
2).
if Schwarz'
(162)12,
m is
greater table.
-41-
"
-
have
167. case
-
21/(t-2)
than
2(167)-
= 21/8 3
then
<
we
9 8
are
Note.
Consider the
first
order
dx If
a
q(x)
homogeneous
over
= a(x)'
takes
theorem.
linear
differential
equation
of
C(x):
q
the
Honda's
E C(x).
form: e.
q(x)
= E x 1
then
it
has
an
=
11(x
-c
,c.e(rpe.e.CD
algebraic
solution e.
y
-
We shall ic
solution
an
irreducible
write
ci)
show
then
this
q(x)
converse
takes
polynomial
F(X,
Y)
then
the
=
differentiation
of
n
F(X,
Y)
(A' for
every
is
In
F(x,
above.
Let
C satisfying
F(x,
y)
An E C[X],
Am
0,
an
algebra-
F(X,
Y)
= O.
particular
= A'/A 0
y)
= 0 gives
as
- O.
n
we have 0 it
1 A' A' q =m_( A0 -_a), O Am
-42-
.
is
= E[A;1(x) + nAn(x)q(x)lyn
irreducible,
+ nAnq)/An n.
there
n
E An(X)Y-, n=0
E N I(x)yn + EnAn(x)yn-ly' Since
as
if
as
Y)
n
form
over
m
F(X,
the
that
holds
for
n
=
m,
and
we
obtain
be
We
which
takes
A
the
follows.
be
0
is
x
such
due
an
K
be
element
to
T. a
exception
algebraic
of
K.
of
Xcnx-
(0
the
Anc
is
is
an
[10,
a
Q(x)
rational
will
is
an
integer
for
and
whose homoge-
over
integer
§1]
< co)
linear
order
there
n
Honda
a
first
Q if
< n
every
be
n.
proved
as
degree
and
lemma:
an
number
If
ing the congruence a with
over
=
satisfying
of
that
He prepared Let
a
of
from
y
numbers
equation
function
This
series
rational
differential
distinct
above.
power
are
algebraic A
as
convergent
coefficients neous
form
there
field
is
a
of
rational
number
a (modP.%) for all
finite
number
of
finite
them,
a
satisfy-
prime ideals then
a
is
a
11 in K rational
number.
A proof
will
tional
integer
A distinct
we may
assume
and
n be
the
product
[k
that a W. of
be
a
given
as from
is
an
0 such
integer
A prime
prime
follows.
ideals
that
in
ideal
Since
K.
in
in
Aa
k
Let is
is
there
is
an
integer,
k denote
a
Q(a)
decomposed
into
K:
e2g = V11622T' g• Supposethateveryis is
not
an
exceptional
arationalintegera.satisfying
(mod
s.)
for
every
the i.
one. congruence
Then
there
ai
a
We have
e.
and
there
is
a .) 1 E 0
(mod p)
an
such
index
i
that
-43-
a
a.
(mod
) .
Suppose
ra-
that
r does not divide the discriminant
equation
defining
for
every
ing
the
prime of
integer
over
ideals
number
of
k
all
is
are
is
rational
with
the
numbers
p
integers.
a rational the
exception with
into
of
the
the
Then
integer
Hence,
decomposed
a
degree
of
finite
number
exception
product
satisfy-
of of
n
all
finite
prime
ideals
k:
P = M2.A
of
(mod
one
prime
them
ring
k there
w E a
in and
the
w in
congruence
them,
in
a
of an irreducible
fundamental
• 1n.
formula
in
lim s±1+
Ep-s/log
lim
E(Nr)-s/log
algebraic
s-1
number
theory
gives
us
= 1
and
-s1
= 1,
where p and T> run over all prime numbers and all of
degree
1 in
k
respectively
(cf.
[38, Chap.12 and p.255]). decomposition
of
p,
we
As -Artin's is bers
an
theorem immediate with
the
=
proof
there
lim nLp-s/log--of
by
(Takagi
our
s11
Honda,
of of
finite
-44-
=
n.
lemma.
if
we
[38 , Chap.16])
consequence exception
Takagi
= Nrn in the
s->-1+
the
stated
p.727],
obtain
s±1+
completes
[40,
Since p = Nli
1 = lim E(Iip')-s/log1= This
Weber
prime ideals
our
result number
make
use
then
the
that of
of
them
Tschebotareff
equality all
prime are
n = 1 num-
decomposed
into
the
product He
of
prepared
first
a
T
it
order
has
then
a
with
the
It
r
is
will
be
then
the
=
proved
if c
series
of
characteristic
solution
y
=
p:
Ecnxn
solution
follows.
of
If
we
over
write
r(x)
over F
.
as
y satisfy
b0(2c2)
0cN-1
0cn-1
+
-1
-1
+
b1c1
+
+ a1cN-2
b1(n-1)cn
we
of
polynomial
b(N-1)cN
b0(ncn)+ = a
F
equation
(x).
as
cn
+
= a
k.
E bnxn, n=0
a0c0,
b0(NcN)
in
N
E anxn/ n=0
=
1
differential
field
non-trivial
coefficients b0c1
Hence,
prime E r
N
r(x)
decree
linear
power a
of
lemma:
non-trivial
there
ideals
homogeneous
ddy = r(x)y, x If
prime
another
Consider
the
n
=
a0c1
+
a1c0,
+ bNcO
+ oa.
+ acN-10
++
bN(n-N)cn-N
+ acNn-N-1'
"'
•
such
that
ps
cn
O.
Then,
have
n-1===
cn-2cn-N-1
=
0
then
c0+ is
a
an
index
clx
solution. n
less
+
c2x
2
+
cn -N-2x
Take
an
integer
than
ps
for
s which
-45-
n -N -2
> N and if
we
there
is
write
Y
as
y
=
v0+
v1xPs+
v2x2ps
+
...
with
v0
-
c0-Fclx
+
ps
. . .+c
-1
ps-1x'
v1
then
=
there
is
vm
If
cps+
we
=
cps+1x
an
+
index
d0v0+
m
dlvl
...
+
such
+
...
-
dix
ps
c2ps
-1
-lxr-
-
that
+
dm -lvm-1'
diE(E'ID.
set
w = Y(1
-
doxmPs
(m-1) Ps
-...-
d
m-1
xPs),
then
w = v0+
... where
+
the
Hence,
x=
By
a
tional
function
q(x)
sum
of
have
splitting partial
of
[mps
x-i
dm-1v1-dm-2v0)x2Ps
...
< i
<
(m+l)ps]
+
,
vanishes.
polynomial
solution,
be-
0. lemmas
n
are x
we
rational
over
Q.
= P(x)/Q(x), field
shall
We write
P, K of
prove
our
numbers,
theorem
q(x) q(x)
= yt/y
as
fol-
is
a
ra-
as
Q E Cx].
Q(x)
fractions:
-46-
i
-
by
two
c
(v2
a non-trivial
x=p-1
all
+
-1vm-1-...-d0v0)xps+
solution
these
Since
the
dm -1v0)xPs
of
,, kxp)p
lows.
In
(vm-dm
w we
w is d d
-
coefficient
from
cause
(v1
let
us
decompose
q(x)
into
the
q(x)
=
R(x)
+e
x-c+
2 (x
where
R
prime
number
of
E K[x]
0(x)
and p which
and
one
congruence
y'
nomial
of E
by
the
of
powers
second
z(x)
Then
we
e',
does Ancn
over
lemma.
7
A,
from and
of
Let
us
'
of
the
0.
there
the
•
elements
divide
p)
+
c)
are
distinct
irreducible =
...
not
(mod
z(x)
of
e,
q(x)y
solution
p)
c,
-
leading y
a
non-trivial
congruence
divisors
over
satisfies
z' z(x)
Take
into
a
coefficient
Then is
decompose
K.
the poly-
E q(x)z
(mod
the
product
Y:
Rg(x)a.
have
z' (x) z (x)
Ea
g° (x) g (x)
'
Hence, for a prime ideal JA in K which divides p we have R(x) E 0,
e'
e" E
E 0 (modlt.)
and
e Ea Since
in
these
congruences
K with
is
e an
function
is
are
exception
of
=
e=e-_
R(x) and
(mod1)
a
0,_
rational
algebraic over
finite
(cf.
number, =
number function
0
satisfied
by of
x
§4).
-47-
-
for
all
we
obtain
prime
ideals
0,
the
first
over
C,
lemma. and
it
Therefore, is
an
algebraic
y
Chapter §7. We
neither of
Q,
Puiseux
for
Reduction
assume
a,
an
variant
and
y,
ay
a
over
(E)
is
irreducible,
a rational
C is
that
integer.
algebraically
p.64]).
Let
y in
C{{x}}.
C(x)
which
= yay[XL
solution
of
solution is
=
over
equation.
The closed
us
suppose
Then
there
does
not
field (cf.
that is
leave
is,
an
171/y
we isoin-
have
every
y
Kummer's
13 is
[12,
y)
- y(ay)'
There
y -
Iwasawa
C(x,
we
is
and
nor
settling.
equation
solution
of
y'ay
of
a
algebraic c
our
C{{x}}
K.
morphism
Hence,
y -
Klein's
through
that
series
instance
have
III.
a
(E)
of
natural
- a(17-1)]
z0+
which
is
is
algebraic.
(E)
number
1 n x-z1++
0.
n
such
linearly
that
independent
we
have
n-1 n x-
zn-1
with e.
00
zZ.=xEc..x3,
1]1]
e.e,
c..E
C.
j=0 Since
we
have
[xnz
each
yi
of
and
=nri-1 XL—XZ+
z.
is
y2
of
y.
= x
a
(E)
z']
solution
which
of
take
(E).
the
Hence,
there
are
two
form:
0.
with
p1p2.The
11Ec..xJ,p.EQ,cijEC, j=0 1J indices
c.00 p1and
-48-
p2
are
determined
by
solutions
p(p
-
Therefore, tinct
-
they from
Hence,
By
3
and
We
shall
of
p(p
-
and
y
are
1
A,
Kummer's
A is the
= 0
see
that y
yp
are
0.
a,
suppose tion
1)
+
and
that
=
0.
A is
table
u
rational
an
y)
a
and
rational v
are
number rational
disnumbers.
numbers.
A is
not
integer.
an
integer.
The
coefficients
+ n)
((
To
the
cn
contrary
of
a
solu-
form: cc
y
=
cnx
c0
n=0
satisfies (n and
we
an
integer.
which is
+
1)
have
(y
+
c -Y=
0
in
Therefore,
contradicts no
n)cn+,
our
solution
y
of
=
(a
case
y
we
get
< 0
the
because
c -y-1=
assumption
+n)cn,
n
neither
=
>
0,
a
c=-y-2
that
co
norQis c0
0.
Similarly
and
neither
=
0, there
form:
03
y
=
x-
cnx,
C00
n=0
in v
case is
y
an
>
1.
integer
The derivation
A is
isomorphism Take ratio
of
of
K over
a
fundamental = y1/y2
Weil is
uniquely
[41,
K of p.13]).
a differential
system satisfies
-49-
X
is
extension
A. C(x)
integer,
C(x)
algebraic
instance
z
an
u
nor
table.
d/dx
every
for
not
Kummer's
derivation
(cf.
the
by
of
degree
Then
Thus,
of
solutions
enlarged C(x)
of
Hence,
into
a
finite every
one. y1
and
y2
of
(E)
(S)d57(-f()=z"1z"2 11 -2A2+ 1 - u22 x2
where be
the
left
denoted
by
Morikawa
[33, The
every of
hand [z,
xl.
field
and Gz
which tion
it =
to
is
is
z)
confer
, 1)
of for
Let and
= zn an
Al(x)zn-1
z,
will
instance
H.
It
is
is
a
extension
of
the
an
element
T(x),
a
solution
since oz
Z)
= 0 be
an
be
of
form:
the
C(x)
+ A such
irreducible
n(X),
Ai
equa-
E C(X).
that
= T(E)
theorem.
For
gives
of
E t,
+
E of
Liiroth's
T(z).
C(x)
Ci
Z)
,(x)A n(x))
in
over
F(X,
F(X,
element
is
extension
+ C4) ,
by
Z)
-
form:
C(z).
C(A1'.
F(x,
x(x
derivative
calculus
a normal
T(x,
the
T(x)
Z)
normal
x)
Schwarzian
its
+ C2)/(C3z
z over
there
z)
o of
(C1z
F(X, Then
For
C(x,
takes
belongs for
the
-
Chap.3,§1].
isomorphism (S)
side,
(1
X2+p2-v2-1]
contained former,
We have
[T(z):
in
E(z)
since
and
every
T(E)]
= n
the root
and
latter of
T(z)(AC(x)
=
C(0.
vation
d/du
of
ddu-1 du = and
fw,
u}
The since
for
Tffx11
an
with
u'
0 we
define
a deri-
by
element
Schwarzian every
T(x)
d dx
(a7) for
u of
w of
E{{x}}
derivative
isomorphism
fz,
a of
-50-
T(z)
by
0
this
is over
an T()
derivation.
element we have
of
C(E),
C1z
c{z,
El
= {az,
El
-
+
El 3z
The
C2
{C
=
{z,
E}.
+ C4'
identity
fE, x} + {z, EA4)2 = {z, x} holds,
where
and
E is
{E,
x}
+
is
the
s(x)
equation er
in
a
{z,
[25]. this
solution
of
E1(al)2 right
The
the
differential
= s(x),
hand
side
coefficient
of
we
E
{z,
El
(z-0)eci)1(z)/(1)2
will
ramification
(E the
exponent
in
-
z0=
e,
function
0.
then
El
1
f(u)
Hence,
,
= iz
= is
we
-
a
E
=
co we
-
E00
convergent
power
series
of
u with
obtain
z0,
E -
E0=7(1 + co
For
C,
theorem
teflte),t
where
{z
lat-
E0)e = (z - z0) [yz)/(1)2(z)]e implicit
C{{t}},
i(0)
determined
1
1
z
be
Kummer's
C[z](z0)(1)2(z0)0 -1
1
by
is
,(z)eEg,E0'z0c (1). 1,
and
This
have
Eo =
the
(S).
section.
If
with
equation
have
-51-
1
+ clt
+
e-2)t-2
+ c -1t-1 ,
c.
E C.
{z, and
for
C4{z,
z
Hence,
=
E-1},
cc
{z,
El
= {z-1,
{z,
El
takes
E}. the
form:
a
(E where
c
runs If
over
Y
Y -
a E 0,
is
b,
c
E. IT,
Zip = 0,
- c) all
a
branching
rational
0(X) T(X)'
points
function
0,
of
of
E C[X],
E.
X of
(0,T)
the
form
= 1,
then
Here,
e
2N -
2 = E(e
x
the
is
= 0(x)/T(x). x by
Chap.V],
case
x = cc. is
a
it
= E0
and
E has
(cf. Eo
the
has
the
same
takes
the
exponent Y -
y by
This
is
special
elementarily If
N = max{deg
ramification
which
then
1),
We replace
X-1in
proved
x -
case Forsyth
of
for
at
X = x with
T(x)
= 0 and
formula
Riemann's
[5,
T}.
y
case
Klein's of
deg
Y -
Y-1in
[17, one
X -
Part
and
y
I,
can
be
§59]).
ramification
exponent
0,
exponent z = zi
e
which
at
z = z1,
satisfies
E(z.)
form: M
E -
Eo
=
II i=1
(z
-
zi)e/T(z),
zi
E CD,
T E C[z],
z.z.(ij),1-IT(z.) where branching
degT
= e(M points
+ 1) of
in
E are
case
degT
El,
> eM. , E
-52-
0,
r
and
Suppose that
that E -
Ei
the has
the
ramification vided
exponent
by
ei
and
n
ei. =
Then
eifi.
By
the
degree
Klein's
n
formula
of
E(z)
we
have
in
the
is
di-
r
2n
-
2
=
E f. (e. 1=1 1
-
1)
and
Here,
r
1
i=1
ei
we
have
^
>
2
= =r
-
2
r
0
if
=
=
>
+
and
only
if
n
=
1,
and
other
case
1.
n
Since ^1
E i=1i
'
we
-1=r, —E=
7
obtain ^
and
r
pose
<
-
2
3.
that
r
2
+
= n
<
We
shall
show
is
either
0
1
=_
z"
y
or
-
Y22[z° and
the
and
2,
e1
(1
We
0,
2.
To
the
contrary
sup-
have
+a+12,)x
-
1)
identity 7
d-z dE
dz dE
If r = 0 then d2z/dE2 =
r
2.
x(x
z"
r
that
we
n(1
+= 1
2
= n.
We may
= e2
E" E'
•
= 0 and 12/y2 is contained
then
ele2
'
in
C(x).
If
have
take
0 and
-53-
co as
E1
and
E2'
and
in
this
setting
E
= czn,
c
E C,
c
O.
Hence,
d-z
dz
dE2 and
17/y2
dicts
our
are
dE
is
in
n
C(x).
r ,1,12 + e2
in
one
+-
(E)
r
= 0.
1
e3
of
the
in
that
1 unless
1 e1
-1
Thus
assumption
3 because
we
/
1 -
=
each is
case
the
conclusion
irreducible.
Therefore,
four
cases
with
el
<
e1
= e2
= 2,
(ii)
e1
= 2,
e2
= e3
= 3,
(iii)
el
= 2,
e2
= 3,
e3
= 4,
n = 24;
(iv)
e1=
2,
e2
= 3,
e3
= 5,
n = 60.
1 and
{z,
El
-
-
(iv).
This
Conversely,
for
=
v,
following
0,
r
Since
(i)
We may take
contra-
oo as
e2
<
e3:
n = 2e• 3'
El,
n = 12;
E2 and
3.
Then
{z,
0
takes
the
form:
1-ei21 (K)
with the
(i)
2'
- e;2
+
-
1)
equation
eT2+e;2-e-32-1 2
CE -
will
be
solved
1) at
the
end
of
section.
formations all
-[
elements
of
z we of
C(z)
a
have each
given
finite
a subfield of
which
-54-
group C(E) is
of left
G of C(z) invariant
linear
trans-
consisting under
of the
1
action
of
G by
Liiroth's
theorem.
tained
in
CM
and
is
G is
the
Galois
determined clic
group
Therefore,
G is
is
that (z)
F(
we
-
case
case
the
case
11"(z)C ' F(z)
unless
four
groups
con-
Hence, CM G is stated
is a cyin
form:
(F,
F)
=
1
0
=
E C[zr,
F F
> =
F F
b.deg
b(deg b.deg a(deg >
= F
F
a.deg c(deg
>
+
1)
+
1)
F, 0 0 T
c.deg
1
and +
we
Klein's
1)
Y.
Let
2
us
that
assume
max{a.degF
n
,
C. deg
Y,
b.deg
F}.
have
(a - 1)R. + (b By
the
F E C[z]„
a++-1=n, Then
of
(C().
subfield
(i)-(iv)
a
=
b.deg b.deg
in
E takes
b"
z)
a.deg b.deg
in
(K) with
The
is
have
a.deg in
CM.
of
El
that
1
here
{z,
extension
over
one
Schwarzian
cyclic.
Suppose
such
1(z) of
it
_
of
a normal
by a solution
group.
§3 unless
C(z)
The
formula,
0,
F
and
+ (c
1)
=
F
not
possess
do
-55-
2n
-
2
multiple
roots,
and
there
are
Hence,
z
is
no a
branching
solution
Our
solution
z of
for
instance
Weber
points of
(K)
(K)
with
[40,
of
with
other
a
(i)-(iv)
= el, is
than
0,
b = e2
given
as
and
follows
1 and c
= e3. (cf.
Chap.9]):
2it (i) 2
z = e•, ( =,1-
E = sin
t,
i
= /1717,
22 1)(zr+
zr-
02
4zr
= z4
+ 2147.-5z2 + 1,
= z4
-
21/7-z2
121/-3f2 =1-3 (iii)
W = z8
W3 -
(iv)
f
The
Galois
the
octahedral
-
33z4
+ 1)
+ llz5
is group
the and
+ 1,
-
+ 228(z15
-
1),
f
= z(z4
-
1),
z5)
-
-
10005(z20
z5)
-
+ z10),
494z10,
E = 11 728T2f-5• dihedral the
group, icosahedral
-56-
'
-
1),
+ H3 = 1728f5,
group
= z(z4
=108'"v2,-4;
+ 1 + 522(z25
= z(z10
T2
33z°
(z20
f
+ 1,
K2 = 108f4'
T = z30
H = -
-
+ 1,
02' E = 12.1-3f2-13•
+ 14z4
K = z12
= 2r,
1)
4zr (ii)1
n
the group
tetrahedral respectively.
co.
group,
§8.
Explicit
We
description
shall
solve
of
Kummer's
algebraic
solutions.
equation
for
C:
{c, x} + {z, C1(4i)2= {z, x}. Here,
z is
the
solutions It
may
of
z
be 0,
A,
can
be
case
p
y2
1 and
v
(I) the
(E)
that
the By
have in
with
linearly
which
is
Gauss°
assumed
in
be of
C as
where
(IV), (XII)
has (VI).
and
a
assume and
been We
(XV)
function
may
table,
equation
(II),
irreducible.
we
Schwarz'
Our
and
cases,
to
points
values
g
algebraic
transformations
other.
v®1 E
independent
branching
their each
remained
of
of
and
permuted
of in
and
y1/y2
supposed
are
that
it
y1
ratio
they
solved
shall
in
solve
will
be
replaced
= T7
= 14C
by
(XII)° =- 45,u=_
A
7,
v = 7,
Area/7
and
(XV)'
By
A =
s(A,
1J,
v,
23'
s(x)j"[-L
x)
1-1 ==
we
we
for
a
of
C(x)
are
2
in
the
rational
+
the
tension
of
C(C).
is
invariant
left
case
(x of
function
by
2
(x)
the
we have
it
is
action an
-57-
„ -
222
+
x(x
of
theorem
Under
A
P2 - 1)
(i)-(iv)
translation
and
715=
140.
indicate
x22 If
2
the a
element
-
normal
the of
J.
1)
previous
because of
-
section, extension
C(z) Galois C(x).
is
then cI(x,
a normal
group
{z,
z) exx}
(I)
Suppose
. x = sin with we
a
that
we are
in
the
case
(i).
If
we
set
relatively
prime
to
r,
then
— x)K(x)2,
H,
K
H,
K E X[x]
2 t T
natural
number
s
which
is
have
= xH(x)2,
1 -
E = (1
E 7[x]
with
g in
case
s
H = is
1 de K =7(s
deg
odd,
E = x(1
-
1)
and
— x)H(x)2,
1 -
= K(x)2,
with
deg in
case
s
H = 7 is
deg Hence,
by
Klein's
there
are
The
ramification and
fore,
we
deg
In
=
K + s
no
(K,
Kx)
branching
of
E-1
-
case
1 = 2s H(x)
-
and
= 1,
of
at
2. K(x)
satisfy
H(0)H(1)K(0)K(1)
points
exponent
that
K = 7
each
formula
Hx)
and
s,
even.
H + deg
(H,
is
— 1,
E-1
z = 0 as
of
at
other
x a
= 00 as
function
0, than
0,
a
function
of
z is
1
and
of r.
have
11 7s For,
at
x
E-1
=
co we
T {z, x} = s(7,,, x).
get
= x—sci)(x-1)
= zrT(z),
-58-
0 E (r(x),
co.
T E T(Z)
x
There-
and 1
xr[0(x)]r By
the
1
-1
implicit
= z[T(z)]r.
function
theorem _
z
= x
r
j r,
E a.x j=0
a.
we
obtain
(E C,
ao
0
and
-11s-22 x1 =-2-41(T)lx+
{z, with
b.
E
b -1x
blx®1+
+b+blx 0
C.
J Suppose
that
E(x)
= xa(x
1 -
with
in
case
f.deg
f.deg
the
-
-
b,
F(0)F(1)
F case.
1 + b +
e,
e.deg
T > e.deg
other a
a,
T
c = max{e.deg in
form:
1)b(1)(x)d/F(x)e,
numbers
=
the
E = Y(x)f/F(x)e,
natural
c
takes
-
(0,
F)
=
(T,
0,
'Y,
F
€ C[x]
d
and
d.deg
We assume
that
(e
-
here
= 1,
we
set
b
and -
1 + c
f;
F)
F
F,
a
0,
-
1 + (d
1)deg
F +
0,
-
e.deg
F -
f.deg
T}
1)deg
(f
-
1)deg
T = 2n
-
2
with n = rilax{a Then,
by
Klein's
+ b formula
+ d.deg
0,
there
e.deg are
-59-
no
F,
f.deg
branching
T}. points
of
other
than
0,
1
(III), cases
and
co.
(V),
(ii),
(iii)
(VIII). and
(iv).
kz (2 - x) 2 4(1 - x)' Then
in
a
we write
function {z,
fz, then
we
z,
In
z(x),
are
(XI)
= s(1 21
pf
1
qf
x}
2 = s(=, p.
1 q
1 q.
following
not
E)
of
the
a branching
point.
If
(VII),
(IX),
1
and
(XIII)
x).
discussions we
are
in
for the
the
case
117285'H
,
El
5 -2
,
'117
11 = s(7,,,
1 71
cases
(iv),
{z,
where
E)
{z
,
7
E = -
H-T
-2
,
E)
(X) , and
E = 1728f5H-3,
(XI)
one
set
(VIII),
(IX)
in
xx2 - x)• -E
4(1
x = 2 is
we
as
the
{z,
in
E -
El
E = 1728f-T
in
us
1
17285' in
Let
that
have
{
(X),
El
Suppose
1 -
E = -
1728f5T-2,
fl1 S1T,7,7,,
r= (XIII).
(VII) E =
Let (2
-
us x) 2x-2,
set 1 -
-60-
E = -
4(1
-
x)x-2.
= s(7''1'
we
set
E) E-
Then
we
have
{z
,
x}
(IX)
= skT,7,7,
We
27x2(x Let
us
-
1
we
27x2(x
-
=
us
We
-
4)3=
(9x
-
8)2.
(3x
-
-
8)-2,
4)3(9x
-
8)-2.
have -
T1 the
1)
-
7
x).
identity:
(8x
-
9)3=
(8x22
-
36x
+
27).
set
=
1
we
64x3(x
-
------
(8x
1) (8x2
-
-
36x
9)3(8x2
-
+
27)-2,
36x
+
27)-2.
have
fz,
(2x3
31 = s(T,,,
x}
(XI)
us
(3x
1) (9x
21 = s(T,,,
64x3(x
Let
identity:
-
-
-
x}
(X)
Then
the
have
{z,
Let
1)
xl.
set
E =
Then
have
11,
We -
3x2
T1
have
the
-
+
3x
7
x). identity:
2)22+
27x2(x
set
27222 (x =71---x
-
1)(x
-
x + 1)-3,
-61-
-
I)=
4(x23-
x
+
1).
11 =—4(2x3
Then
we
-
3x22-
3x
+ 2)(x2-
x
+ 1) -3
have
22 T2 x} {z,= s(T,,, x)• (XIII)
(x3 Let
us
We have
+ 30x2
1 we
108x4(x
E =
+ 64)2=
(x2-
16x
+ 16)3
+ 108x4(x
-
1)
-
(x3
1) (x2
-
+ 30x2
-
1 T,
1 {z , x)•
16x
96x
+ 16)-3,
+ 64)2(x2
-
16x
+ 16)-3.
have
, x} Thus,
cases
96x
identity:
set
E = -
Then
-
the
of
4
= Kummer's
Schwarz'
equation
table.
has
These
been
are
solved
those
treated
in
the
twelve
by
Brioschi
[3]. The
icosahedral
CI
=
EC,
C'
=
(wc
group
E5
=
is
1,
C'
generated
=
-
by
C
and
if
we
consider
consists
+ 1)(C it
a
-
group
w)-1w2 of
= 1 -
linear
of
y
= Erc,
c,
=
(Erwc
c,
=
cr_g-1
cr-s)(c
E-sw)-1,
-62-
w,
transformations
of
C.
It
C' with
r
=(sr+sC
= 0, Let
z Then
we
1, us
2,
3,
-
Erw)(sswC
4
(cf.
+ 1)-1
for
instance
Weber
[40,
§74]).
set
= q(C)
= C2(C5
-
7)(7C5
+ 1)-1.
have
g(E0
= s2z,
g(-cl)
= -z-1,
and z
g(ug4-1) -
This of
is g(C)
wz
due
to
for
every
The
Gordan
[6].
Hence,
g(C')
transformation
roots
f(c)/(
14
of
is
the
a
linear
2,
3,
transform
group.
of
= c10
+ 11c5
-
1
are
r
w,
Erw°,
ww'
=
1,
r
=
0,
+
1)
1,
4,
where
W =
The
E
E4,
+
a multiple
z = g(0
E2
+
E3.
the
root
C2(1
+
cw.
Hence,
w)(7C5
the
branching
points
of
by
Klein's
are 0,co,
with
=
polynomial 2 (C5
has
w'
Era),
ramification
crw',
r
=
0,
exponents
1,
2
-63-
2,
at
3,
each
4
point
in
formula.
In and
the
(XV)'
following
we
discussions
suppose
in
,
(XII)'
1 -
,
If
we -
set
1) (2x
-
1
2
,
the
C(x),
since
group
is
=
1
translation
1 -
C(0 the
is
E =
(2x
-
1)-2,
icosahedral
the
group
{z,
{z, (XIV)
{z,
of
x}
x}
C(x).
74
Let
set
x} = {g(0,
z)
extension because
T(x)
and
= fg(c), Hence,
11 = s(7,, us
C(x,
group
between
element
(VII)
a normal
field
Galois
x)
theorem
termediate
an
1)-2,
have
{
is
E)
(XV)'.
E = 4x(x we
1 - C= H(C)3T(0,
11 = s(7,,, 71 7
El
(XII)'
by
(XIV)
and
(XIV)
and
(XII)1,
11 = s(7,1E)
EI
{,
then
cases
E = 1728f(C)5T(0,
= 1728f()5T(0,
in
the
that
C = H()3T(0, fc
for
is
a normal
of
C(E).
there
T(E). x}
is
left
we have
x/
x.
=11x).5'3'2'
-64-
Then
we
no the
invariant,
x). E =
The is
Under
extension
have
of
Galois
proper
in-
action
of and
it
(XV) '
Let
= 4x (x Then
we
us -
set
1) (2x
-
1
2
1)-2,
1
— E =
(2x
have {c,
x}
=
x) x)
(VIII)
and
z
,
=
{g(c)
,
= s4,
-65-
2
2
x)
,
.
-
1)-2
Chapter
IV.
Landau-Errera's
§9.
For an
E-number
to
c which -
1
vto
E-number k
If
-
1
to
If
p does k
is
equal
Hence, =
1
-
-
the is
v If
<
c = ap a,
divisor
r
(r,
a,
< r
then
< s
with
v of
relatively
This
(1)(c)
prime
notion
p prime
the c)
the
< — ap, For
k.
then
= cb(c)/(vp).
number
a
integers
integer
vp
0(a)/v.
equal
every
that
divide
and
number =
of
vp
is
number
of
r
a)
due
that r
to
a has
an
satisfying
1
Hence,
(r,
is
an
E-number
of
c.
satisfying
= 1
a multiple
r
= ps
of
p we
have
< —a . of
r
satisfying
our
inequality
with
(r,
ap)
to
p(P(a)/v Hence,
for v
r
1 a
of
call
inequality
p divides
1 ap
c we
— c
q(a)/v
to k
r<
c<
not
number
the <
lemma.
number
the
Suppose
vp
equal
if
(1)(c)/v
v.
is
c
c
[4].
Errera's
natural
satisfy
equal
Errera
given of
k
is
a
theorem.
is p
-
(a) /v
an
E-number
is
prime,
=
(p
of then
-
1)cp(a)/v
c.
This
for
every
= (P(c)/v.
is
due
to
integer
Errera k
the
[4]. inequality
k - 1 137—Tp
satisfied
only
by
r
=
k.
Hence,
-66-
p
-
1
is
an
E-number
of
p.
The
number
of
r
satisfying
k k-1 c2 < r Ec2 is
equal
to
Suppose
(1)(c)
that
c/v
= Pp
(r,
c)
c
and
her
=T
is
pe-1(p
an -
q
be
s
that
1,
the
< c.
number
the
following
with
r
satisfying
c.
For
an
E-number
prime
of
numbers
our
c2.
p.
inequality
Then,
with
= (P(c)/v. of
three
natural
< c,
(k,
number
of
natural
r
E k
(mod
number
of
In
following
r
the
= 2,
eleven c
numbers
r
r
(n,
our
c we
numbers
=
2,
c)
have
n,
another
E-num-
k
and
c
are
given
satisfying
r
(r,
c)
satisfy case
= 1,
(i)
we
(mod
our
= 1, condition
have
s/q
= 1:
k = 1; with
(ii)-(xii) 0
satisfying
additional
condition
cases E
1.
n) ,
our
n
=
numbers
which
2)
satisfying
we 2) ,
k
condition
=
r
< cn
have
s/q
is
1.
In
= 1/2:
1;
with
r
= 1,
2;
<
cn
are
1
and
1.
(iii) our
of
k
c
(ii)
+
(p)
be r
>
the
n
v = Hpe-1
is
to
that
(i)
the
c
1).
the
r and
= 1
that
n Let
and
E-number
Suppose such
c)
Hence,
number
equal
¢(11p) v
= Ppe
the
= 1 is
Hence,
= (c2)/c.
< (r,
r
are
c k
and
= 3, k + jn,
(n,
3) where
= 1, j
-67-
is
k
either
1 or
2 and
k + jn
/
0
(mod
3). (iv)
our
r
are
k
(v) our 0
r
= 4,
and
k
k and
is
tively c if
-
then
prime
to
case
is
r
r
even
are
are
r
are
1,
1,
1,
r
are
then
(xi) our
r
4 and
r
due
to
3;
is
5;
either
2 or
4 and
k + jn
/
are
are
7,
1,
5 and
11
and
Errera
[4]:
Either
s/q
and
(k,
for
a
2)
natural
is
less
than
c we
an
odd
number
c
k
c = 8,
= 1; number
have
+ p in
an
p rela-
even
case
number
p is
even;
E 1
(mod
3);
k E 1
(mod
3);
19. c
= 10,
13,
k
E 1
(mod
3);
19.
c = 20, 19
and
k 31,
c = 6, 13,
n = 5, 1,
2) ,
13.
13,
13,
(mod
= 2C(c).
n = 3,
7 and
= 1;
c = 5, 7,
7 and
2)
= 4(c)
= 3,
n = 4,
(xii) our
k = 1,
k = 1,
j
E 0
qb(2c)
n = 3, 1,
c
odd,
n = 3,
(x) our
(c,
c which
n
(ix) our
= 1,
where
C(2c)
p is
(viii) our
6)
n = 2,
(vii) our
(n,
= 2,
odd
p in c
2) ,
3).
(vi)2 c
(mod
+ 2n.
k + jn,
(vi)1n
if
n E 1
c = 6,
are
(mod
c
< 1/2
(mod
37,
43,
E 1
(mod
3); 49. 4);
17. c
31,
k
E 1
= 12, 41.
or
k
E 1
We shall
s/q
-68-
= 1.
(mod
5);
prove
The
the
latter
following
case
occurs
lemma
only if
if
we
we
are
are
in
in one
Before pose cy
(mod
k
n)
and
the
(i).
the
(c,
n)
= 1.
Then
(mod
n).
If
r
be
written
it
can
(r,
c)
= 1 if
only
if
The
above
begin
and
We have if
of
case
we
that E -
the
eleven
proof there
as only
= 1/2
note
the
following:
an
integer
r
E k
(mod
n),
-
with
= nx (x,
occurs
only
(ii)-(xii).
is
r
if
s/q
cases
we
satisfies
and
case
c)
Sup-
y satisfying
cy = 1,
then
r
an
and
0
E -
cy
integer < r
x.
< jc
17-c
for
n
every
integer First
and
n
is
this ber
E
1.
of
r
k
r
(mod
is
s
Suppose
< q/d
prime
equal
to
divisible A number
r
relatively
divisor
d of
c
c)
by
=
1,
r°
= r
condition and
c to
s/q
+ cn/d.
with < l/d
is
divisible
c,
and
r
the
< cn/d
is
If
s/q
< 1/2. by
4.
number
Hence,
Then, of
the equal
to
= 1/2, r"
our
num-
r
= r
then + cn/4
satisfying
4,
for
every
then
we
j. can
Therefore, write
c
n = 2. as
c
=
2c'
If
c
with
is
(c',
not n)
=
satisfying
^ E k is
common
(j + 1)4n
q/4
by
greatest
(r,
our
jin
,
satisfied
that
relatively
the
satisfies
n)
satisfying
We have
d = 2.
that
If
condition
q/d.
is
suppose
not ^
j.
(mod
n) ,
prime
to
(r, c,
c')
since
-69-
= 1 it
is
odd.
If
it
is
less
than
1.
c'n
then
c'.
it
Then
is we
(v
-
less
than
2c'.
Suppose
that
n/2
< 1 + 2/(v
v
is
an
E-number
of
have
2)/v
< 2/n,
and v If
v
c'
= 3,
to
3,
< 2n/(n
> 3,
-
then
n
then
two
since
(n,
v
of
c
(h
we
-
< 6.
If
of
k + n and
k,
c')
= 1.
let
us
Secondly ber
2) ,
have
1)/v
a
n = 4,
v
k + 2n
Hence, that
natural
2).
< 4 and are
(n,
number
h
c'
= 1,
relatively
n < 6 and
suppose
< 1/n
then
-
=
1.
If
prime
n = 2,
c)
3.
4. For
an
E-num-
= 2,
3,
4,
such
that
only
if
that
n > 6.
< h/v.
Hence, s/q We have
<
assume
v
> 6,
we
of
c
is
less
5 can
We have case greater
c
than
is
-
than 120 30
'
6.
60,
1 < c/n,
even,
we
40 14'
> 2.
c,
and
40,
30,
and have 30 6
2].
'
24 6 '
us
suppose number
24,
15,
20,
< 1/2 -
1)
if
Then,
every
10,
and
the
above
20 6 '
15 3 '
12 2
'
10 2
-
s/q c
these
'
8 2'
< 1/2 values 5 1'
2,
5.
n > c/[4)(c)/2
< c/n,
if
from
8,
6.
E-number
different
12,
For
-70-
c
that
a prime
s/q 2(s
and
Suppose
Let
Then,
c/Wc) 60 14'
< 2 if
< 1/2.
divide
= 120, s
cp(c)
c(c)
s/q
not
c
and
that
have
21 <7+F.
+ 1)/v
q = cp(c),
We may
and
(h
1]. if
In n are
is
3
They
are
prime
to
If
v
less
than
6.
6 except
Thus,
> 7 then
if
s/q
12/2,
but
n > 6 then
< 1/2.
12
s/q
is
not
< 1/2.
relatively
Suppose
that
n
= 5.
We have
1<1<22<3<3<4<4<5 56<56565
Hence,
s/q
number
of
2 and
3 can
We have less
c
< 2/6 c
E 2
less
which
5,
7,
As
(mod
5),
then
Suppose
former
that
c,
are
If
13,
above s/q
define
= 1/2.
In
occurs
= 8.
s/q
s/q
< 1/4
< 1/2.
If
Our
is
if
then
every
natural
to
24
E-
different prime
the
that
c
y by
cy
the
from to
5.
numbers
are
1,
k
(mod
case if
k
3,
5,
numbers
we E 1
n).
If
y
have
s/q
<
(mod
7.
are
5).
We have
7. that
by
Our
-
only
are
Suppose
= 12.
other
and
numbers
divisible
number
relatively
5 <<
< 1/2. c
is
that
23.
Suppose we
suppose
prime
prime
19,
1 < 8162432 T< 3 <7-<7-< Hence,
c
c = 24,
17,
us
a
relatively
case
c
Let
Then since
< 1/2.
11.
The
6.
8.
11,
< 2/8
v = 6.
than
12,
7,
1/2.
if
divide
24
s/q
5,
is
= 24,
1,
1,
< 1/2
not
than
Hence,
6'
n = 4.
7,
then
If
v
> 8,
6 is
an
E-number
by
5,
then
then of
and 1<1<22
_
3<4<3<5
46<4-66<
Hence,
s/q
< 2/6
E-number
of
c
and
E-number
of
c
is
Then
c = 9.
The
6'
< 1/2. s/q less natural
If = 1/4 than
c
is
divisible
< 1/2. 8 and
numbers
-71-
Let c
is
less
us
assume
relatively than
9 which
that prime are
4 is
an
every to
35.
relative-
c
ly
prime
to
Hence, s/q c
are
1,
2,
2 <<
91827< 4 <4<
s/q
< 2/6
< 1/2.
4,
c
is
7,
8,
7
5 < 4
< 1/2.
If
5,
Suppose
divisible
and
.
that by
n = 3.
11,
then
If
10
v
is
> 12,
an
then
E-number
of
and
Hence,
3< 10
1 , 3
s/q
< 4/10
E-number ber c
9
and
of is
c
c
=
= 2/6
less
Here,
80,
21
s/q
80,
is
<< 80160 3 is
< 1/2.
relatively 1,
20,
8
30
7,
16,
an
,
then
to
are
13,
divisible us
is
5,
assume
7,
that
then
6 is
every
prime
to
an
E-num77.
Then
and 8,
5.
of
c
and
60. to
the
17,
by
relatively
<-T<
= 40,
11,
is
E-number
If
9,
c
10,
50
40
7 17'
Let
2 and
prime
c
c
and
relatively
prime 3,
If
12 by
40,
then
2 , 3
< 1/2.
than only
=
20
6< 10 < 1/2.
s/q
divisible c
If
is
4 17'
19,
80
and
less
numbers
21,
than
less
23,
27,
80/3.
than
29,
40
31,
Hence, which
are
37,
39,
33,
and
13<4017,23
<82
33
Hence, bers
s/q are
< 6/16 1, <<
3, 2040 3
< 1/2.
7, 7,
9,
11, 13
< 27.
Suppose 13, <7-
17, < 17.
-72-
that 19,
c and
= 20.
Then
our
num-
If
y
The c
E 1
(mod
former
3),
occurs
= 16.
Then
if
our
5 <16< Hence, bers
s/q are
then
3,
y E 1
The
latter
c = 5.
(mod 3) occurs Then
our
k
E 1
numbers
are
1,
3,
5,
32 9 < 3<
9,
The
y
E 1 former
(mod
3)
occurs
s/q
if
and
numbers
that
only are
s/q and
In if 1,
k 2,
9,
11
,
=
Suppose
that
13,
and
15,
1/4.
c
= 10.
Then
our
num-
= 1/2. only
the
other
E 1 3,
(mOd 4,
case 3).
s/q
= 1/2,
Suppose
that
and
4.
if
-73-
33
7,
3).
s/q
7.
= 0.
3 < 12<
if
(mod
case
and
then
then
other
11.
Suppose
5 < 16<
1 < 5 7 < 2, If
the
if
1 < 13- < 3, If
In
only
< 1/2. 7,
= 1/2.
and
7,
< 3/8 1,
s/q
In k
the
E 1
other (mod
case 3).
s/q
= 1/4.
§10.
Given following
natural
Two
numbers
lemmata.
c,
k
and
n,
we
shall
prove
the
lemma: Suppose
that
for
(mod
n) ,
a
divisor
e
of
n
every
integer
r
sat-
isfying
r is
less
= n.
E k than
We have If
d
satisfying
is
ec. c =
d
= 3,
(c,
our
equal
Then
(n,
< r
n)
is
to
each
1 d.n
<
c)
= 1,
0 < r
= 1 and
if
e/n
greater
c = 1,
< cn 2,
3,
6 unless
e
= 1/2.
than
1
then
the
number
of
r
with
+1 other.
<
c)
6 only
condition
cn
(r,
dcn,
0 < j
< d
Hence,
e
and
e< Therefore, c
then
d4.
e we v
=
our
-
2
less
v
Suppose
d
=
1.
If
v
is
an
E-number
of
n
note
in
the
previous
v2
2'
> 4 then
n/e
than
Then,
-
that
<
-e- < 1 + If
n.
have
v
by
1
dl-i'dl=
4.
< 2.
section.
Suppose
c = 6,
4,
that 3,
<
(c)
-
Hence,
1
-74-
2.
every We have
E-number
of
c
is
and
n < 1 e 2 O(c) in
case
c
= 4 then
is
even.
n/e
< 2.
We
shall
< k
<
of
If +
k
is
us 0
=
is
an
assume
< r
1,
k
=
1;
(ii)
c
=
2,
k
=
1,
(iii)
n
=
3,
c
=
4,
(iv)
n
=
3,
c
=
10,
(v)
n
=
5,
c
=
6,
1
then
r
is
2
then
n.
In
case
even
k
then n
>
odd is
we 3.
2c.
=
case
they
are
less
than
2c.
In
they
are
less
than
2c.
Thus,
d
(d
=
-
(n,
1)cn/d
the
c)
case
is
k
= k
k
our
than
If
In
the
and
than
1
if
c
r
2c.
and
satisfying
Then
we
our
are
in
one
3;
r
the r
integer
than
with
and
have
In
r
odd
< 3,
(i)-(v):
less
cur.
less
cases
our
n/e
greater
every
is
=
=
6 then
integer
that
< cn
five
= 3,
odd
c
c
c
lemma:
(i)
If
because
if
another n
following
c
k
Let with
the
Hence,
that
2c.
condition
1
prove
Suppose
0
-
1, =
7; 1,
=
1,
0
<
+ n
r
with
(iii) our
the
case the
greater
than
< 2c
and
-75-
2c our
(iv)
<
on
0
<
is <
13,
19;
11.
latter k
7,
r (v) above 1
is r
<
even
k,
which
cn
is
and
r
is either
=
k
<
=
4.
It
is
impossible,
r
are
1,
7,
which
are
1, our
7, r
five then
13, are
cases we
have
1.
2c.
or If
are
19, 1,
k
11, actually
and and oc-
Hence d=
,
1 , a
1
d
2.
2 =
It
We may
1.
1
n
3.
impossible
,
since
that
>
1.
is
assume
C(c)
n If
is v
odd. is
Suppose
an
E-number
1]
in
that of
c
then v
-
2
_ V
2
<
,2n vs — n -
n
2
and
< 1 +
7vWe even.
Suppose
than
5,
and
vide
c.
A
tains
2'
n < 2c/[b(c)
have
is
v
that a
have
and
r
=
5
then
c
=
r in
case
= k
r in
case
k
Then
case v
<
1,
4, 1,
= 4
7,
2,
and
to
is
10.
and not
5
c greater
can
less
Hence,
is
not
di-
than
c/3
4
con-
< 4 and
c
=
4
then
k
=
3.
we
have
These
r
=
cases
5, are
11
in
case
k
=
impossible.
2
If
13
11,
14
and 6,
3.
length
2
of
c
4, 8,
3,
c
case have
=
k
in
2,
=
If
we
=
=
r in
9
E-number
case
8 3 < 3,
1
8. 3, 3.
no from
whose
prime
-
Since
8
we
is
different
relatively
-
n < c/[b(c)
There
interval
12.
g5(8)
and
number
closed
than
less
1],
n = 3.
prime
number
a
-
9,
12
Hence, there
it is
is no
impossible.
E-number
-76-
Suppose of
c
greater
that than
n 3.
>
5. Hence,
c = 6, then
4,
We have
n = 5, r
These
cases
odd.
We r
It
3.
is
and
=
we
7,
17
3,
-
c/Wc)
ll
= C,
7,
13
3,
8,
9,
19
4,
9,
5,
25
5,
10.
in
case
k
=
imnossible.
If
c
have =
c
4.
If
c
=
6
have 2,
are
and
1,
11
in
case
k
=
1,
2,
7
2,
8,
13
3,
4,
14
4,
5,
10
5.
impossible.
-77-
=
3
then
n
=
5,
since
n
is
§11.
In a a,
=
From
end
of
values
If
b,
all
and
are
of
a
them
of
triple
common
tion m
=
we 2r
are
=
2r
(2) m
=
=
(4)
satisfying
triples
by from
(a, the
each
table other.
b,
c,
at
the
For
m)
the
elements:
m -
3,
a m/2,
2},
following
b + c,
m + a
-
b,
b
-
a.
then
criterion,
v = 1/2. c
-
m = 4.
Landau's
c0of
even
24
distinct
to
with
the
numbers
and table
and
We shall
it
assume
m is
not
m/2.
of
{a,
b,
belongs that
Under c}
with
to the
this m
(I)
greatassump-
0
(mod
r:
+ 1,
2},
(r
2)
(I)
with
odd
r:
{ 1,
r
{ 1,
3,
2},
{1,
5,
2},
(III)
{ 1,
3,
2},
{l,
9,
2},
(XIII)
+ 1,
2},
(I)
6:
(3) m
c,
= {1,
divisor
r
be
equal
table
with
(1) m
+ b -
c}
have
set
have
six
satisfies
Schwarz'
est
not
table.
y = c/m,
we
there
Schwarz'
natural
triple may
m -c,
we
m are
which
b, This
table
each
c
of
= b/m,
c
§5, of
c,
Schwarz' = a/m,
where 1.
Attainment
10:
-78-
c):
m =
12:
{1
5,
3
1,
1
10,
3},
{2
5,
3
},
2
10,
3},
(V)
(6)
{2
5,
4},
2
11,
4},
(v)
(7)
{1
7,
4},
1
9,
4},
{3
7,
3
9,
4},
{,
7,
1
11,
3
},
{2,
7,
2
11,
3
},
{1
13,
5),
4
13,
(5)
m
=
{1,
(9)
=
n n Z. V
(10)
=
m
(12)
=
(13)
7,
3}, 3},
5},
{4,
7,
5},
51,
{1,
11,
4},
{1,
13,
4},
{3,
11,
4},
{3,
13,
4},
{1,
13,
6}
{1,
17,
61,
{5,
13,
6}
{5,
17,
6},
{1,
13,
8}
{1,
19,
8},
{7,
13,
8}
{7,
19,
8},
{1,
11,
6},
{1,
(X)
(X)
_
(IX)
24:
(11)
m
(II)
15:
(8)
m
4},
e
30:
25,
6},
-79-
(IV)
(IV)
{5,
11,
6},
{5,
25,
6},
(VIII)
(14)
{3,
13,
6},
{3,
23,
6},
(XII)
(15)
{1,
19,
10},
{1,
21,
10},
(VII)
(16)
{3,
13,
10
{3,
27,
10},
(XII)
(17)
{5,
11,
10},
{5,
29,
10},
(18)
{5,
17,
10},
{5,
23,
10},
(XV)
(19)
{7,
13,
10},
{7,
27,
10},
(XII)
(20)
{9,
19,
10},
{9,
21,
10},
(VII)
{1,
31,
12},
{l,
41,
12},
,
( VIII)
m = 60: (21)
{11,
31,
12},
{11,
41,
12},
(VI)
(22)
{1,
31,
20},
{1,
49,
20},
(VI)
(23)
{7,
37,
20},
{7,
43,
20},
(XIV)
(24)
{7,
37,
20},
{7,
43,
20},
(XIV)
(25)
{13,
37,
Given a< we
20},
natural
{19,
49,
numbers
a,
c<
b<
m,
that
for
every
ap
< cp
< by
(mod
m)
by
< cp
< ap
(mod
m).
assume
m=
20},
0
b,
(mod
integer
or
-80-
(VI) c c),
and
m satisfying (a,
p relatively
b,
c) prime
= 1, to
m
either
If
m/c
is
ble.
not
equal
We shall
Let
us
set
a If
p'
=
the
(A)
an
E
in
m'
that
p
§6.
one
of
Errera
m =
then E
If
p0
the [4]
p'
(mod
p'
m'd
there
above
as
ta-
follows.
,
is
m'
an
m').
=
c'n.
0
integer
Suppose
p rel that
we
satisfies
such
=
1
=
1
that
m'),(P0,m)
c
(mod
m),
r
(mod
E a'
m°).
set
k
the c'
=
For,
+ n0x
(mod =
p6 satisfying there
is
an
a',
(i),
2,
a'
(q),
relatively condition
integer in
c')
prime
y
such
satisfying
then
= 1.
have
m'). c
case
(mod m'),
is our
and if we set p6 = 1 + yno no
we
< c'
If r = a' + xn
is an integer
assumption
r
are
is
to
c),
n0'0)(p''c')
(mod
a'p6
we
due
(a,
to
integer
a'y E x (mod c'),
If
=
p6a° < c' (mod m°).
p'a'
our
d
of
(mod
to c' then there
By
c}
have
is,
that
dc',
m such
=p'
p0a
that
b,
theorem
prime
1 is
P0 we
=
to
E
there
and
c
case
p' then
fa,
m = cn0and
prime in
2 then
this
relatively
atively are
prove
da',
is
to
=
c'
since =
1,
and
n
s/q (no,
and
-81-
=
n0
in
= 1. 2)
Errera's
Therefore, =
1,
lemma, we
have
then
we
-
c
= 2a,
Here,n0>
m =2an0,(a,
1 because b = d'b', m"
=
Here,
d'
=
There
is
an
and
p1
c
< m.
=
2,
and
p1
(b,
such
2
n0),)(p1,
with
us
latter
integer
= 2 + xn0
=
1. set
c)
= d',
m"d'.
the
(mod
Let
= d'c",
c"nm 0'
1,
p1E
c
b)
an
happens
only
if
c"
is
is
even
2c"
.
an
integer
odd.
that
odd
c)
= 1,
integer
x,
since
c
.
We
have
p1a
=c+xfc+f
plc
E 2c
m
(mod
=_m
m),
and
because
n0> plb
that
define
Then, +
2c
< c
+
< m,
(mod
m) ,
is,
k
< k
k=
k
m) ,
Hence,
< 2c
p1b' We
2.
(mod
is
xn0is
relatively
2c"
m").
by
p1b'
(mod
m"),
0<
k<
relatively
prime
to
c"
and
relatively
prime
to
c"
then
prime
p"1b'
(mod
E
to
k
+ xn0
c"
such
that
(mod
m").
-82-
m". less there
than is
If
r P 1
Hence,
we
r Let
us
have
<
set
c
n
we
are
in
(i)
c"
= 1,
k = 1;
(ii)
c"
= 2,
k = 1,
(iii)
n0
= 3,
c"
= 4,
(iv)
n0
= 3,
C"
= 10,
(v)
n0
= 5,
c"
= 6,
last
four case
(i) c us
=
have
2,
Let
We
= us
n0in
one
d'
c =
2 =
the
second
the
following
of
lemma
of
the
five
previous
cases:
3; k = 1,
7;
k = 1, k = 1,
= 1,
> 1,
1,
b
E
we
since
0
have
7,
13,
19;
11.
c"
1
+ +
=
it
n0'cp1E
is
= 2,
2) ,
m =
is
relatively
4,bp1E
n0.This
We
d'
(mod
p111=2+-0°Then
(ii) c
=
even.
We shall
dis-
separately.
a
set
b
cases
Since
ap1E We
m").
and
each
Let
=
Then
the
cuss
(mod c"
section.
In
2c"
is
and 2n0,n0E prime
2
the
case
1
(mod
(2)
of
(mod
2).
to
m and
(mod
2).
to
m and
m). our
table.
have
2,
a
1,
set
pl
api
E 2 + n0,cp1E
=
b 2
E
1
+ n0.
(mod
2),
Then
m =
it
4,
is
2n0,
relatively
bp1
E 1,
n0
E
1
prime
3
(mod
m).
have
ap2 111E 4 + n0,cp2 Therefore,
if
n0
> 6 then
E 8,bp2 each
-83-
E 2 + n0,6 of
4 + n0,2
+ n0(mod + n0and
m). 6 + n0is
greater tion.
than
8 and
Hence,
n0
2_ 5= 1 case
(mod
6).
of
our
(3)
7.3
E 1
(4)
of
(mod our
and
10).
and
b = 11,
our
table. (v)
n0
= 3.
either
Suppose
that
b is
either
Hence,
11.
m = 12,
This
is
a
= 5,
E 1,
7,
13,
17,
23,
We
have
6,
a
=
b = 13, Let
a
b is
a = 2,
= 10,
Then
3 or n0
5.
= 5.
3 or
the
pl
= 5,
case
(6)
m = 30, 19
29.
=
23. us
we
that
contradicts
our
assump-
pl
= 5 and
This
Then
9.
is
pi
This
the
= 7 and
is
the
case
of
bpi
E 1,
our
table.
7
(mod
12),
We have
bpi
above
Suppose
It
We have
(iv)
and
2n0.
table.
b = 5,
c
< 5.
table.
= 4,
c
than
Therefore,
(iii)
c
less
3,
assume
(mod These
m =
This
is
pi
that
30), are
30,
the
the
pi
case
we
= 11,
are
=
cases
7,
(14) in
bpi
of
the
(17)
our
case
E
and
1,
11
(18)
(mod
of
30),
table. (B)
of
d =
(a,
§6.
As
set
= da',
c
= dc',
m'
= c'n0,
m = mid,
c)
and
b = d'b', We
say
that
a
c natural
=
m" number
= c"n0, p
-84-
less
m = m"d', than
m belongs
d'
= (b, to
C if
c). it
is relatively
prime
of C we write
p E C1 in case ap < c (mod m) and p e C2 in case
by < c (mod m). absolute
value
natural
number
ly
prime
write C'I/ is
to
By our assumption indicates p'
m'
less
and
an
p°
number
tively
of
of
such
r
/ C" .
that <
an
if
it
< my,
than
b'p"
that
is
element
a
relativep'
of
C'
CI =
E r
(mod
and
s'
m'). be
that
m" belongs
to C" if For
< c" (mod m").
Let
q'
be
of
such
it
is
rela-
an element We have
p" of
Ic2Mci
lemma we have
positive
by
our
assumption.
If
an
integer
to
d',
and
satisfies
then
we
E
ap1 r
=
(mod
m"),
(p1,
d')
=
1
have
cpi and
1
E c,
< c a p1
+
bpi
(mod xm"
satisfying
is
E b
m).
(mod
Here,
a
relatively our
m) is
relatively
prime
to
condition
-85-
such
prime
d'
then that
there api
r
We say that
= C2I/ C" are
we
prime to c' then there
a°10' r
C'
We have IC1/
relatively
0
By Errera's
sides
pi
teger
with
p" e C° if
both
(mod m').
such
p" less
C°1/IC`
if
C'
For
to m" and p" E 1 (mod no).
C" we write
since
< c'
no).
We say
Then we have s°/q° = Ici / c' .
number prime
C
(mod
to
p
lc2 , where the
number.
belongs
r = a' + xn0is
element
a natural
p1
E 1
m'
For an element
1cl = Icil
cardinal
than
p'
with 0 < r < c°.
=
the
p' E CI if a'p' C°1-If
the
to m and p E 1 (mod no).
E r
is (mod
an
inm).
Hence,
we
m"
e
and
Since
have =
c"
our
our
Then
We
(mod
the
first
we
d')
=
d)
either (a, we
d'
that
n0
in
us of
Let one
1,
set
the
c
=
us of
c)
= 1.
set
the
we
= 1 we We may
k = a
and
following E 0
ten
= 3,
(n0,
(iii)
c = 4,
(iv)
c
(v)
n0=
3,
c = 5,
a
= 1,
4;
(vi)
n0=
3,
c
a
= 1,
7;
(vii)
n0
(viii)
= 6,
3)
= 1,
E 1
(n0,
2) ,
2),
= 1,
= 8,
in
a a
= 1,
a
= 1,
3,
7,
n0=
3,
c = 20,
a
= 1,
7,
13,
(ix)
n0=
4,
c = 6,
(x)
n0 them we
set
no
p
=
1
even. is
If
a
by
n0
2:
3;
9; 19;
5;
= 1,
11.
separately.
n0is
b = 1 + n0.This
c = 12,
lemma.
5;
c = 10,
= 5,
Errera's
that
2;
= 1,
a = 1,
d = 1,
= 1;
= 3,
(ii)
=
= 1.
(i)-(x)
a = 1,
(mod
6)
a
d')
that
cases
c
n0
(mod
(d,
n = n0
(ii)
m = 2n0,since
n
have
assume
n0
If
a,
section.
have
= 2,
(i)
=
2.
b,
d'
Similarly
c
tat
k
previous
(i)
shall
d',
2
2.
(a,
d = 1 or
are
=
d = 1,
that
= 1.
Let
lemma
1,
= 1,
c)
m).
have
assumption
Hence, is,
in
< c
assumption
(ml, By
c" m",
(m", by
r
the E 1
+
n0
then
We have case
(mod
(1) 3)
-86-
then
it
p2 of
is
E 1,
our we
relatively
cp
E 2
(mod
prime
to
m),
and
table. set
p = 1 + n0.
We have
p2 3
E +
1,
cp
no aT
we =
E
3
(mod
m).
Hence,
b
=
1
+ n0,2
+
2n0.
If
3
+
no
n0
(mod
E
n60'0'0'0
<
2
m)
+
2nCT
E
9,bT
6
and
n0=
4.
It
is
(mod
3)
then
we
set
a
and bT
E
Hence,n0<
b
=
1
3
+
2n0,6
6
and
n0
For
p
(iii) b
since
d'
=
1
+
=
1
by
+
E
2
+
no,6
bT
E
2
+
3n0,6
Hence,
n0
< 6.
is
case
(10)
(iv)
If
2
+
and
p2
E
3n0
n0
1
we
=
If
3
+
2n6
+
+
3
+
2n0,
d')
=
6n0)
2n0
we
have
(5)
2n0.
3no). of
We
T =
3
+
our
have
table. a2
n0
we
(8)
of
our
E
1
(mod
T
=
2
case p2
we E
n0(mod
E
1
get
set
+
table,
4n0)
no
and
then
8, 4n0).
is
the
case
(7),
and
if
n0
= 5 it
table.
(mod and
b
6) =
then 1
+
we 4n0,5
set
p = 1 + 4n0. +
2n0.For
T
We =
get
E
2
+
3n0,10
+
3n0,c
b
E
2
+
5n0'10
+
n0(mod
This
is
E
impossible.
12, 6n0),
If
-87-
'
the
If
a
< 4.
+
is
1.
= 3 it
E 1
1 For
3n0,cT +
case
n
3n0).
It
1
our
n0
(mod
5.
+
of
=
no(mod
=
n0
the
+ n0. +
(m",
E
have
2n0,2
2n0,
aT
the
T =
have
(mod Hence,
For
no
E 5
(mod
6)
then
we
set
P = 1 + 2n0. For
We have
T = 2 + 3n0we b
and'
4,
b = 7,
13.
(vi)
We
Hence
get
b
pc
7,
= 13,
19. We have
p E 1,
7,
the
(mod
13,
the
(mod
+ 4n0.
(13)
(„nlv
of
our
table
if
of
if
our
and
table.
only
if
(12)
of and
our
table.
only
if
30)
if
,
(19)
and
(20)
of
if
and
only
if
30).
table:
11
13
pa
1
7
11
13
3
21
3
9
7
19
17
1
9
3
9
27
pc
10
10
20
10
pb
13
1
23
19
19
13
29
7
21
27
21
3
27
9
27
21
the
cases
We have 7,
case
and
(9)
24)
(mod
7
p E 1,
if
case
1
(viii)
= 1 + 2n0,5
24).
E c
19
following
have
b
the
case
(mod
is
pc
is
15)
p
, we
and
15).
E c
This
(vii)
the
is
pc
19
(mod
(mod
This
13,
This
E c
13
We have
p E- 1,
6n0)
6n0),
5.
We have
p E 1,
(mod
+ 5n0(mod
10.Hence,n0=
(v)
Hence,
E 1
get
E 2 + n0'10
n0<
Hence,
p2
13,
(15) pc
31,
E c 37,
,
(16) (mod
49
-88-
60) (mod
60)
our
table.
and
the
get
Hence,
following
table:
p
1
7
11
13
23
29
pa
1
7
11
13
23
29
7
49
17
31
41
23
13
31
23
49
19
17
19
13
29
7
17
11
pc
20
20
40
20
40
40
pb
31
37
41
43
53
59
37
19
47
1
31
13
43
1
53
19
49
47
49
43
59
37
47
41
we
obtain
the
cases
(22),
(23),
(24)
and
(25)
only
if
of
our
ta-
ble. (ix)
Hence,
We have
p
=
1,
b
=
13,
(x)
5,
Hence,
We
p
E
1,
b
=
31,
pc
13,
17
17.
This
have
pc
11,
24)
if
and
24).
is
the
case
c
(mod
60)
(mod
and
(mod
(mod
E
31
41,
E c
(11) if
of
and
our
only
table. if
60).
we
get
the
case
(21),
filling
up
our
table.
Thus, note
that
we
if
m/c
d'
=
(b,
c)
is
d
nor
d'
is
1
vious
section
=
Let
3c"
2. and
b
have
proved
= n0is
equal
Landau-Errera's equal
to
1.
To
to
the
contrary.
we
have
either
us
suppose
is
odd.
that If
we
to
the set
2 then
prove
d
=
-89-
=
either
it
we
By
the
first
2
and
d'
former p
theorem.
c
1,
3
or Then
is
(a,
c)
assume
lemma =
it
d =
shall
occurs. +
We shall
odd
neither
of d
or
the
pre-
=
3
and
d°
c
=
2c'
=
and
relatively
prime
to
m =
op which
2c.
We
a,
Cp
contradicts
have
E C, our
Hence,
either
our
we
in
rived m =
m
=
from
m
=
=
{s,
r
{s,
m -
=
d one
=
of
1
c
(mod
hat
they
d'
or
the
=
2c),
satisfy If
1.
following
Landau's
m/c
=
which
cases,
2
then
are
table:
+ s, s,
r},
(s,
r)
= 1
(I)
r},
(s,
r)
= 1
(I)
12:
9,
6},
{3,
7,
6},
{3,
11 ,
{1,
13,
10},
{l,
17,
10},
{3,
11,
10},
{3,
19,
10},
{7,
11,
10},
{7,
19,
10},
{9,
13,
10},
{9,
17,
10},
{1,
17,
12},
{l,
19,
12},
{5,
11,
12},
{5,
23,
12},
{7,
11,
12},
{7,
23,
12},
{5,
6},
9,
6},
(II)
20:
(IX)
24:
{11, m
Schwarz'
-
2r:
{1, m
are
E b
assumption
criterion. theorem
by
17,
12},
{11,
19,
12},
(IV)
60:
{1, {11,
41, 59,
30}, 30},
{1, {19,
49, 31,
30}, 30},
-90-
{11,
31,
30},
{19,
59,
30},
by de-
{29, {7,
In the
this
41, 43,
30} , 30},
{7,
(13,
43,
30) ,
{23,
43,
30} ,
case
assumption
{29,
where that
47,
{17, {23,
m/c either
49 , 30}, 30},
47 ,
= 2 it d
{13,
30},
37,
301,
37,
{17,
53
,
30}, (XIV)
30}.
was =
(VI)
1
-91-
proved or
d'
by =
1 .
Landau
[27]
under
Chapter
V.
Transcendental
§12.
Let
k
Kolchin
be
[22]
which
has
a
Picard-Vessiot
differential
proved
the
liouvillian
theory.
field
the
of
existence
following
solutions.
of
property.
characteristic its
0.
universal
Suppose
that
K is
extension
of
k in
generated
differential
extension
of
K.
Then
L has
The
proof
is
based
Let
II be
following
image
theorem
differential and
ideal
K be
a
generated If
an
k
posed
u of
P is
transcendental
then
there
all
is
k which
vanishes
of
Then
is
it
not
contain
y -
generic
zero
then
it
gives
be
K{y}
k,
Let
in
a
II'
u is is
differential
not
that as
a
k is
in
R{y}. P
we
is
a
to
(cf.
algebraic with
au
com-
indeterminate closure
set
K = k ideal over
and over
a
We may
ideal
prime
u,
is u
k(u)
the
transcendental equal
au
algebraic
If
Kfyl
there
single
the
prime , Yn
u is
o of
a
the
follows.
if
isomorphism
-92-
on
closed.
such
II denote
by
differen-
ideal
k then
since
finitely
y1,
in
with
where
u because v of
k
0,
finitely
algebraically
proved
polynomials u,
a a
the
isomorphism
generated
does
au
will
in
ideal
a
over
ideal
the
is
k
one.
at
Then
contained
a prime
then
k
algebraic
differential
k.
over
differential
y over
Hence,
It
an
of
not
L is
indeterminates
if
o of
u is
a
p.51]:
the
ideal
that
is
with
a prime
[19]).
of
= v
k{y}
p.25],
it
k.
book[34,
isomorphism
u and
0.
extension
element
Kolchin[20,
over
in
II is
differential
K in
Ritt's
differential by
assume
in
over
P and
a
differential
isomorphic
R.
extension
generated
tial
E.
if R.
and k. we
set
Let ential
II be
prime
polynomial
zero
u of
(cf.
over
y (n)
k.
assume
closed. system
Then,
by
nant
field
the
y
n)
k{y}.
This
k
is
there of
due
to
differis
a
k
is
Kolchin[21]
is
not
equation
a.
E
k0of
theorem
nn>
in
k
Q such
that
k0,since
the in
over
k:
k.
there
contained
degree
the
n
is
algebraically
is
a
fundamental
the
field
of
wronskian
the
determi-
differential
ideal
Picard-Vessiot tells
group
of
K
G is
is
an
subfield
morphisms
in
a
G0
we
normal
of leave
is
an
that
every
have
-93-
equation.
properties.
of
P
algebraic which to
algebraic
such
our
automorphisms
is k.
of
left
The
consists
element
of
E
invariant
of
G0
G of
there of
E =
group
component
G then
K
of
matric
subgroup
subgroup E
for
fundamental
G belongs
algebraic E
G which
it
element in
its
differential
Then An
extension
us
all
k.
k0.
in
ferential
component
n
automorphism
identity If
a
over over
every
index.
<<
work[20]
,n>
under
the
0,
constants
nn
is
1
G denote
the
11 then
is
a
by
Kolchin's
of
=
existence
,
7a.vn-i) L 1,]
Let
If
differential
any
nl,
,
generated
of
k
W(yi,
k{y}.
constant
theorem
linear +
solutions of
in
every
existence
ay1+ (n-1) the
constants
in
contained 0 and
homogeneous
that
of
not
D(u)
This
ideal
pp.108-109]).
a +
is
that
M. Matsuda[32, Consider
in
differential
D(y)
H such
algebraic
We
a
all
invariant.
is
finite a
dif-
autoFor
of
[G : •
where
G0l
E0is
if
every
called
a
of
that
degree
Li
and
k then
is
we
,
n
G has
its
Extension
E
of
extension
L of
the
ficient
il)
under
is
G0.If
constant,
k
is
the
that
determinant
is
equal
to
if
and
algebraic
al
=
1, only
in
to a
a
liouvillian
is
reducible
tion
of
our
element
if
constants
there
is
the
a
fi-
form.
k
for of
our k. form.
such
E k, of
The
2 and
our
the
belong
to
k.
If
there
is
a
of
k then
is
Go
is
a
G0is
normal
ir-
n'/n Then,
of G can
non-trivial
there
which
There
coef-
equation
derivative
component
of
form.
is
equation
is
a
Picard-
is
contained
of
the
in
identity
non-singular
solu-
that
0
G0.Since
-94-
i.
extension
triangular
logarithmic
finite each
liouvillian
that
not
of
EforL.
equation
extension
equation
a to
n does
triangular
T
in
our
L. 1-1(ui)
u'/ui i
We assume
triangular
c
Li -1or
of
is,
E of
to
k
of
k:
of
reducible
order
extension
TE = c, every
G0is
liouvillian
extension
of
contained
solution
reduced
Vessiot
is
that
non-trivial
solution
u!1E
vanishes.
k,
field
of
extension
either
the
y'
over
be
extension
algegraic
have
of
whose
n = L
an
that
k
extensions
component
a1
reducible
liouvillian
•••,fln>
Suppose
for
invariant
of
LlL
E = k
not
left
•••
differential
k = L0c
every
a
deta•W.
chain
If
W(111,
differential
k0is
such
E
finite.
A
nite
of
element
oW = G is
is
subfield
determinant
then
since
[E0'ki
the
wronskian 0,
=
in
G,
we
have
T(O for
)
every
= c'aE,
c'
element
cs of
over
k0,because
G0is
reducible
to
algebraic
solution.
is
no
consist
of
gebraic
G is
a
single
group, '
a,
it
e
k
G.
0
Here,
aE
not
reducible
diagonal
form. Then
element,
linearly
to
triangular
We shall G is
the
consists
is
not
independent form.
suppose finite
identity.
that
and
Since
of Hence, there
Go
does
G0 is
an
not al-
of
0 a a'
E
k0,a
G
of
0.
3, For
every
element
GT)1 = c1n2, Therefore,
[G
an2 a G0]
G
which
= c2nl
= 2.
This
is
Cl, is
-95-
not
c2 due
contained
in
G0we
have
e k0. to
I.
Kaplansky[13,
§19].
§13.
We
shall
suppose
homogeneous over
Liouville's
that
linear
k
lemma.
=
C(x)
differential
with
equation
x'
=
1
of
the
and
consider
second
a
order
k:
d-Y dx2xprx0117 'd If
we
set
Y = y/i171
Y" + by
W'
of
(s/2)Y
= -
solution
that
+ v2
Let
us
it
has
a
is
sion
if
Y is
of
algebraic
algebraic:
our
is
equation
shall
assume
tive equation
in
is
in a
the
G'
that
left over
p2/2
a
of
Y is
a
a
liouvillian
A and
not
invariant
extension
of
extension
in
is
liouvillian A of
k
k.
E'
k
for
The
of
k
for
and
y extenY.
automorphisms of A' over k and G6 be
identity
finite
over
extension
Picard-Vessiot
is
irreducible
extension
p
G'.
are if
an
1n
and
algebraic
T
k:
-96-
hvpergeometric
and
numbers,
only
if
G
case and
is
W is
finite.
extension
is a solution
under
our
rational
of
We k .
Then
Y such that TY/Y is a con-
for every automorphism T in N. v
satisfies
v = Y'/Y
a Picard-Vessiot
[G' : N] = 2 and there stant
-
is
solution
then
Hence,
+ 2q
it
equation:
that
contains
component
p'
W then
derivative
Let G' be the group of all the
q E12(X).
wronskian
s = -
non-trivial
there
P,
s/2.
contained
E(^R)
the
Riccati
= -
= 0,
logarithmic
assume
Then
which
The
the
v'
with
= 0,
pW. of
k.
q(x)y
Its it
logarithmic
satisfies
a
deriva-
quadratic
v2 The
+ av
+ b
= 0,
coefficients
a'
= a2
because
of
+ s
v'
a,
a
and
-
2b,
b
b E T(x). satisfy
b'
+ v2
= -
+
-
= a(b
s/2.
As
+ s/2) the
compatibility
condition
we
have a"
which
=
is
3aa'
due
to
treatment
confer
following
lemma:
The of
the
I.
D'
=
a D
a3,
§25].
For
Chap.IV].
is
the
of
Due
half
the
Liouville's
of
the
quadratic
to
him
original
we
obtain
logarithmic
equation
the
derivative for
v.
have
(a2
-
=2a (a2
4b)'
our to
be
assumed
are
equal
to
1/2.
2aa'
2b)
m 4b)
to
will
=
+s
= 2a(a4 Return
-
Watson[39,
discriminant we
2as
Kaplansky[13,
coefficient
For,
given
s'
-
4b'
4a (b
+ s/2)
= 2aD.
hypergeometric
differential
be
over
irreducible
Then
a
k.
fundamental
equation, Suppose
system
which
that
of
A and
solutions
p
is
by ivt-ivt2 Y1
which
=
e,
lie
y2
in
dx2 (IT)=
Here,
the
p
v
=
/-1,
extension
x
of
k
=
sint,
because
4x(1 - x). on
existence and
e,i
liouvillian
problem
Kolchin's A,
a
=
are
the theorem
half
integers
field
of in
the then
-97-
constants previous our
equation
can
be
solved
section. has
If a
by two
solution
of
in
a
liouvillian
is
algebraic
rational
extension if
and
number.
only We
where
our
equation
under
the
assumption
of if
by
the
shall
has
k
transformations.
remained
prove
a
Gauss'
one
that
this
transcendental
that
of is
A, the
It p
only
liouvillian
neither
A,
p
nor
v
and
v
is
case
solution is
a
rational
in-
teger. Let tial
us
express
the
coefficient
a
of
c1
=
1,
a
half
v
in
the
ci
e
sum
where
=
ei
n+1
e.
i =0ci x
1 -c0=
is
0,
either
an
integer
or
-e.
a'=1X
(
Comparing
the
condition
we
x
-
2
c.)
'
an
coefficients
(x
of
(x
C,
integer.
Then
-
-
1
c. )
ci)-3
3
•
in
the
compatibility
get
1.
Hence, e.
1=
Comparing
-1,
-2,
those
i
of
2e0
= -3e- 0'
2e1
= -
0,
x-3 2P
1.
and -
(x
2Pe00-
-
1)-3
we obtain 1
e3
P =-(12
_2
-
X2)
-
p2).
and
3e1
-
2Q -
20e1-
e11Q
= -(1
Hence, e0= us
-1,-1±A,e1= multiply
-1, each
term
in
-1±p. our
-9
we
2e,
- X
2e.1=3e.2ei,0,
1
par-
fractions:
a
Let
of
equation
8-
by
x3:
have
a
3
2e.xx3a"
=
(
x
-
3(xa)'x-c)
c.)
1 2
eix-e
xax2a° =)(I
e.x
31
1
lx
x-c
i
(
x
-
2),
c.) 1
=
-
2P
2Qx3
(x
xax2s
e.x 1 (7)[P x-
=
-
R(2x 1)3
(x Q
C. 1
-
+2
(x
-
1)x x3s' 1)2
x2 -
+
1)
1Rx
x]'
where
1R =-2-(1
For
x
=
+
00 we
2X
=
v2
-
X2
u2)
have
-3X2
-
2S
-
2SX
-
X3,
S =
P
+ Q
-
R,
where n+1 X
=
ei. i=0
Hence, X
=
-1,
-1±v,
1 because S -= v2). -2-(1 then
X is
and or
v is a half
pose
an
negative
integer
an
integer.
If
integer, X is and
integer
each
a
that
integer
If
v or
a is a
half a half
less
e0
since
2e0
is
an
-
,
3
e1is
-1.
If As
the
if el
X is by el
to
-1,
X = -1
either our
an
lemma.
± v integer Sup-
= -1,
X is
a
then
v is
either
us
assume
X -1 last
equal
Hence,
integer
Then,
-99-
1L
and
± X then
integer.
integer.
en than
= -1
integer.
half
of
case
let
half
that
e0=
half
integer. This
-1
and
e1
If proof
p is
=
is due
-1
u.
a
half
to ± M.
Then
1J is
integer, Setoyanagi[36].
-100-
either
then
an
v is
integer
a
half
or integer.
a
§14.
Take not
Consider
a
a
connected
simply
contain
and It
over
has
an
matrix
p
of
equation
ement
f(x)
of
continuation x.
the
around Suppose
form
f
m
m'
h=1
j=1--I
13.
is
of
E over
origin
that
= u/v
and
a
the bh
= f
a1
We may
=
we •
assume
form. can
..•
•
that
is
a
j(x)
we
is
singular
is
the
ecruation. a
regular
group We
is
to
show
by
every
a
H of
shall
prove
equal
to
H if
that
an
el-
analytic
rational
function
point.
Then
expressed
in
holomorphic
<
are
If
c
analytic
for If
ak
itself
v are
the
group
points
u and
every ah
suppose =
to
is
p-1Gp.
sufficient
integer.
anu/anv
to
our
f(x) the
takes form:
x)O,ahEC '
non-negative
from
that
dif-
solutions
for
there
k. does
the
of
k
Q and
monodromy
returned
that
h
in
which
Then
of
such
singular
k induced
then
k
over
plane
system
equal
the
x = 0 is such
above
fundamental
is
It
a,13A (11_,.4(x)x''(log
then
takes
k
of
the
attheorigin,wherea. and
over
is
complex equation.
constants
Fuchsian. E which
equation
extension
of E
the
our
E over
closure
is
a
Picard-Vessiot
field
Zariski
of
by
image
the
differential U in
point
isomorphic over
the
our
of
a
automorphisms
that
domain
generated
U is
theorem.
linear
singular
field
x
all
homogeneous
any
ferential
Kuga's
0 and the
continuation n.
= ah
at denotes
Let
+ ibn
us with
a hj .(0)
around suppose real
0
automorphism
that numbers
the v ah
that ah'
in
h
one
-101-
>
of
Z.
the
following
three
cases:
'•
(i)
h)0,< 1bhl=
(ii)
b1
(iii) We may
< 0,
b1=
suppose
v(x)
bh
that
g1(x)
g2(x)
is
is the
argument
x)
x
[qi(x)
sum of
is
2 < h < t:
1. (31 > bj,
the sum
2 < h<
> bl'
0,Q=
= x11(log
where
the
b'
of
the
not
the
other greater
2 < j
< n.
Let
us
set
+ (42(x)], terms
with
terms. than
1
If 7
then
< h < Z,
the
j
absolute
we
= 1 and value
of
have
.
lim x+0 for
c
every
g2(x)
n.
=
Let
0
us
set ibh
q_ (x)=XcPhl(0)x+
g3(x).
h=1
Then
under
the
above
limng3(x) x+.0 for
every
n.
=
If
condition
we
get
0
we
set
P
ib-
L(x) = Ig'111(°)xh1,
i
h=1 then
we
obtain
(b
L(x)[111(0)1
- Xkbh1(C1)1 e1
-b
)arg
h]
x
-b
argx
x + -...:
In
e1
h=2
In
the
second
first case
case (ii),
(i) L(x)
we
have
+ 0. if
L(x) arg
-102-
a
-- co if x + +0:
arg In
the
third
the
case
(iii), may
L(x) = iq1(0)1 be
negative
lim
such
> 0.
Hence, there
is an integer
n which
that
xNanu(x)/anv(x)
= 0
x->0
for
a
sufficiently
absolute
value
origin is
a
is
tion
that
a
+
if
under
the logarithmic
is
a
of
OR= the
has
the
of
is
that
be
the
Thus,
due
to
given.
not
the
Therefore, f(x).
is
proof can
=
0,
X E
f(x)
M. Kuga The
condi-
removed.
For
C
that
which and for
it
is
of
X,
point
irreducible
if
of
it and
is
analytic
here
H is
if
-103-
half
in-
solution k
and
there
one
there of the
X,
p
and
automor-
Picard-Vessiot
ex-
and
e
is
continuation a
domain
1)(1.1-1)/2,c( identity
are
By
irreducible,
only
from
complex
the
v
aco denote
induced
a
and
over
Let
Y over
u
liouvillian
Since
respectively;
H0
two
equation.
transcendental
point.
is
cx(X-1)/2(x component
differential that
a
integer.
k
7.
theorem
Fuchsian
A2)y
singular
1
This of
prove
equation
E 0,
than point
hypergeometric
singular
of
tension
-
shall
rational
around
(x2
our
logarithmic a
x.
sketch be
assumption
no
Then
+
we
our
greater
condition
example.
to
theorem
not
the
equation
xy'
Return
of a
N under
singular
equation
counter
teger
x is
essential
Bessel's
gives
number
function
our
x2y"
phism
arg
where
instance,
is
an
rational p.173],
v
of
not
[24,
this
great
0)
E C.
in
H can
not
contain
all
of
by
van
and
a0'a1and
Kampen's
M.
one
is
and
(cf.
only in
of
they
for
p.293]).
contained
solutions
since
theorem
Sugawara[23,
contains a1
a.,
Y"
-eXTri
our
instance
A.
They
one
of
(s/2)Y
satisfy
them.
H0.Then
+
generate
is
= 0 such
that ,
A
G0=
a
-ep7i
group
Komatsu,
M. Nakaoka
aao=e,and 01cc'H0
Suppose
there
monodromy
that
H neither
fundamental
ao
system
nor of
0
u1= 0
here AB
-e-X7iB
0 because
-e-wri
Each of 002 and Gel
of the irreducibility.
-1
(=
a.-)is
containedin
H0which
is
reducible
to
diagonal
form;
here
-A(eX7i
+
e-Affi) e2X7i
2 CT_ = U
0
e-2X7i
e (A+p)Tri
+ AB
-Ae-p7i
a001=
-(A+p)Tri
-BeATri If
the
then
eige eigenvalues we
Hence, p
is
T.
have hay
A is
of
either
Kimurarl Kimura[15]
if
assumption
ion
tion.
SUDIO Suppose
=
an integer
distincttinct,
are
0
,
or a
a
half
half
e
our
2Xri
e-2XTri
assumption.
integer,
iinteger.
note our that hat
that
there
equation it that
has
is
is s a
and
This
le.
irreducible. x=
0
is
a
no
similarly
proof
transcendental
logarithmic loc There
logarithmic
-104-
'
is
contradicts
which
integer or
that
is
due
to
•
We We shall sha solution
a-2
-Be-ATri -Be-A
an
either
e-(A+P)
liouvillian
singularity is
no singular
under
algebraic point.
the soluThen
there
is
o2Y./Y. 0 1
1
tion holds
that for
a
fundamental
is
a
x an
system
constant
=
0
is
equation
for
a
of
each
i.
logarithmic which
is
not
-105-
solutions
Y1,
This
contradicts
singular
point.
Fuchsian.
Y2
such
that our
Our
assump-
statement
Note.
Consider
Bessel's
Bessel's
equation.
equation:
x2 dx2d2yydd(x2 A
solution
is
Jv(x)=X unless
v
mental
system
=
0
we
is
given
by
negative of
a
1
solutions
If
V
is
a
is
by
not J
v(x)
an
integer
and
a
J_v(x).
fundaFor
form:
x + y (-1)k-1
positive
c.
n+2k
v
given
the
1
k=1 If
(y) x
is of
v
function:
integer.
solution
Jo(x)•log
= 0,
Bessel's
k=F(k+1)F(n+k+1) 0(-1)k a
have
_ v2)y
22k2(1 +1•
• +1) (=i)2k.
(lc!)
integer
n
then
by
the
recurrence
formula:
dy
yn+1(x) we
have
a
xyn(x)-dTc solution
Jn(x)•log where
01) n(x)
of
the
form:
x +n(x),
is
defined
inductively
by d(ID
n+1 (x) Thus
•
a
integer.
= 1Jr-1,1)n(x)
logarithmic Let
- Jn(x)]
singularity us
set
- din appears
Y =y.
Then
if we
have
d2Y dx-
The
-
2 + [1 - 12(v2 x
logarithmic dV di
2121 + v+_2(v
derivative
x
- 1)]Y
= 0.
V = Y'/Y
_T)
= 0.
-106-
x
4
satisfies
and
only
if
v
is
an
v
If
we
set
Y = then
we
P(t)eix,
t
=
x-1,
i
=
get 2
t2 dt2+ The
- 2i)dpdt
(2t
coefficient
ck
of
4-v.2)P
P(t)
=
= O.
cktkis
determined
by
k=0 2ik
Hence,
ck
nal
(k
P(t)
is
In
this
teger. tion
=
is
11 7+
a
case
we
x.
Then
us
set
x
e co+ j
t
if
-1. and
a rational
it
if
function
let takes
only
us the
v is of
suppose
a
x and
that
half
in-
our
equa-
V is
a ratio-
form:
e.
=0
x]
c
,0=0,c.
j
have
e2 Let
of
V is
n
and
v)ck
Conversely
of
V(x)=
-
polynomial
reducible.
function
v)(k
= -1,
ej
multiply =
a>.
= 1,
each Then
j
term
we
in
0,
e02=±
Riccati's
v. equation
for
V by
x2
and
obtain
n ec°L
e. =0
and
=
0
3
hence
1n+ 7 ± v = 0. Therefore, We pose
shall that
v
is
show v
a
half
that is
a
there half
integer is integer.
if no
our
equation
algebraic Then
-107-
is
solution. every
solution
reducible. First is
supexpressed
in
the
form
Y(x) If
Y(x)
= f(x)eix would
sible,
that have
braic
function
integer
is
rational
our
=
+
integer.
since
our
a
which
-
2as
is
this
if
is
x.
our
would
impos-
Secondly equation be
an
irreducible is
prove
that
liouvillian
solution.
satisfies
-
It
function
We shall
has
so. of
_v(x)
equation
polynomial.
s'
be
Then
J .9(x)J
because
a(x)
3aa'
would function
solution
equation
function a"
a half
impossible a
eix
g E C(X).
transcendental
x,
not
f, then
algebraic
is
if
a
not
of
it
which
is
v is an
However,
algebraic
eix
would
one
be
since
suppose
+ g(x)e-ix,
the
differential
alge-
over an
cC(x).
integral
v is
a half
There
is
a
equation:
a3,
where
s
= 2 -
=
1 2(2v-
-7).
x
It
takes
the
form
n
a =
Here,
we
ej Let x
j =0
us and
e.
x
-C
.
ic(21=0,C.€C
7
have
=
-2,
multiply set
x
j each
=
co.
0, term
Then
e0= in
we
our
obtain
n
-4
and
e . = 0 j=0 3
hence
-108-
-1,
-1±2v.
differential
equation
for
a
by
-
1 ± 2v
+
e . = 0, j=1
because
e0can
integer
or
logarithmic theorem
not
a
half
be
equal
integer.
singularity is
due
to
to
-1
The
Therefore,
former
appears Liouville[29]
.
in
,
Michihiko Matsuda Department of Mathematics Kyoto Sangyo University Kamigamo, Kyoto 603, Japan
-109-
is
this
[30]
v impossible
case.
,
[31]
either ,
This
.
is
since
epoch-making
an a
Bibliography. (The
numbers
in
brackets it
1 .
F.
Baldassarri
ential
and
equations
101(1979), 2
L.
3
F.
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6
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66,
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Dwork,
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42-76.
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11.
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II,
lineare
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Ann.
11(1877),
iiber
das
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iiber
die
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Leipzig,
[52].
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18.
Springer, R.
Math., 20.
[iii].
97].
Hodge
entialgleichungen,
E.
Japanese),
104].
17.
19.
to
[95,
the
which
[48].
solutions
and
quadratures,
16.
1957.
Japanese).
(in
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"c
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