Cryptography Reference
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Using the fact that
b
k
+1
=
(
p
2
1)
/
2
k
+1
−
=
2
((
p
−
1)
/
2)!
(
k
+ 1)!(((
p
−
1)
/
2)
−
k
−
1)!
=
((
p
−
1)
/
2)
−
k
k
+1
2
b
k
,
we find that the coecient of
T
k
is
4(((
p −
1)
/
2)
− k
)
2
−
(2
k
+1)
2
b
k
=
p
(
p −
2
−
4
k
)
b
k
≡
0(mod
p
)
.
This proves the claim.
Suppose now that
H
p
(
λ
)=0with
λ
=0,
we have
λ
=0
,
1. Write
H
p
(
T
)=(
T−λ
)
r
G
(
T
) for some polynomial
G
(
T
)with
G
(
λ
)
= 0. Suppose
r ≥
2. In (4.5), we have (
T − λ
)
r−
1
dividing the last term
and the middle term, but only (
T − λ
)
r−
2
divides the term 4
T
(1
− T
)
H
p
(
T
).
Since the sum of the three terms is 0, this is impossible, so we must have
r
= 1. Therefore,
λ
is a simple root. (
Technical point:
Since the degree of
H
p
(
T
) is less than
p
,wehave
r<p
, so the first term of the derivative
∈
F
p
.Since
H
p
(0)
=0and
H
p
(1)
H
p
(
T
)=
r
(
r
λ
)
r−
2
G
(
T
)+2
r
(
T
λ
)
r−
1
G
(
T
)+(
T
λ
)
r
G
(
T
)
−
1)(
T
−
−
−
does not disappear in characteristic
p
. Hence (
T − λ
)
r−
1
does not divide the
first term of (4.5).)
REMARK 4.39
The differential equation 4.5 is called a
Picard-Fuchs
differential equation
. For a discussion of this equation in the study of
families of elliptic curves in characteristic 0, see [24]. Once we know that
H
p
(
T
) satisfies this differential equation, the simplicity of the roots follows
from a characteristic
p
version of the uniqueness theorem for second order
differential equations. If
λ
is a multiple root of
H
p
(
T
), then
H
p
(
λ
)=
H
p
(
λ
)=
0. Such a uniqueness theorem would say that
H
p
(
T
) must be identically 0,
which is a contradiction. Note that we must avoid
λ
=0
,
1 because of the
coe
cient
T
(1
− T
)for
H
p
(
T
).
COROLLARY 4.40
Let
p ≥
5
be prime.Thenumberof
j ∈
F
p
that occur as
j
-invariantsof
su persingular elliptic curves is
p
12
+
p
,
where
p
=0
,
1
,
1
,
2
if
p ≡
1
,
5
,
7
,
11 (mod 12)
, respectively.
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