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Proof. By Theorems 1 and 2 we have that p k
( a ) if and only if p k
|K
|
#
E p ( a ). Recall
F p m the Abelian group structure of
that for any a
E p ( a )is
E p ( a )
Z n 1 × Z n 2 ,
p m
E p ( a )= n 1 n 2 . Suppose that p k
with n 1 |
n 2 and n 1 |
1, and we have #
|
#
E p ( a ).
n 1 , it follows that p k
Since p
|
n 2 and
E p ( a ) contains a subgroup G isomorphic
Z p k .Ageneratorof G is a point of order p k
to
in
E p ( a ). Conversely, if
E p ( a )
contains a point of order p k ,then p k
|
#
E p ( a ) by the Lagrange Theorem.
2
Divisibility of Kloosterman Sums
We say that a
F p m is a zero of the Kloosterman sum, or a Kloosterman zero
for short, if
(0) = 0, and we will often implicitly
exclude the case a =0anddealonlywith a
K
( a ) = 0. It is easy to see that
K
F p m . There is a lot of interest
in finding zeros of Kloosterman sums, see for example [3] for an application to
construction of bent functions.
If we can find an integer s such that s
=0.Thisgives
one of motivations for studying divisibility properties of Kloosterman sums. The
following two results are well known and easy to prove.
K ( a ), then
K
( a )
Lemma 2. For each a
F 2 m ,
K
( a ) is divisible by 4.
Lemma 3. For each a
F 3 m ,
K
( a ) is divisible by 3.
The following theorem was first proved in [7]. There are two proofs given in
[7], and one proof is given in [4]. We give a new proof as an illustration of the
methods applied in this article.
Theorem 4. [7,4] Let m
3 . For any a
F 2 m ,
K
( a ) is divisible by 8 if and
only if Tr( a )=0 .
F 2 m .Since m
Proof. The result holds for a = 0, so let us suppose that a
3,
by Theorem 3 we have that 8
|K
( a ) if and only if
E 2 ( a ) contains a point of order
E 2 ( a 2 l ) for any positive integer l ;let
8. By Lemma 1 we can replace
E 2 ( a )with
E 2 ( a 8 ): y 2 + xy = x 3 + a 8 . By Lemma 7.4 in
[12], an x 0 F 2 m is the x -coordinate of a point of order 8 on
us for simplicity take the curve
E 2 ( a 8 )ifandonly
if X = x 0 is a root of the polynomial ( X + a 2 ) 2 + aX . This happens exactly if
Tr(1
a 4 /a 2 )=Tr( a )=0.
·
Lemma 7.4 of [12] used in the previous proof belongs to the theory of division
polynomials for elliptic curves. A detailed treatment for characteristic 2 can be
found in Chapter 7 of [12]. The following theorem is new.
Theorem 5. Let m
( a ) is divisible by 16 if and only if
Tr( a )=0 and Tr( y )=0 where y 2 + ay + a 3 =0 .
4 . For any a
F 2 m ,
K
F 2 m .Since m
Proof. The result holds for a = 0, so let us suppose that a
4,
by Theorem 3 we have that 16
E 2 ( a ) contains a point of
order 16. As in the previous proof, by Lemma 1 we can replace
|K
( a ) if and only if
E 2 ( a )with
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