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Lemma 8.
|
R d, 1 |
=
|
R d,− 1 |
.
F p such that its inverse element satis-
Proof. Let ( γ,δ )
R d and let u
fies η ( u 1 )=
1. By (8) and the analysis in Case 2, one has S ( uγ, uδ, 0) =
ω γ,δ ( x ) = ρ∈ F p
ρ =0 η ( u 1 ρΔ d ) ω ρ =
p− 1
N γ,δ, 0 ( u 1 ρ ) w ρ = p n + d 1
S ( γ,δ, 0) .
2
x∈ F p n
The above equality shows that for j
∈{
1 ,
}
,if( γ,δ )
R d,j ,then( uγ, uδ )
1
R d,−j . Thus, one has
|
R d, 1 |
|
R d,− 1 |
=
.
Applying Proposition 5 and Lemma 8, we have
= p m−d ( p n
= p m−d ( p n
Proposition 9.
|
R d |
1) and
|
R d,± 1 |
1) / 2 .
Proof. By (7), it is sucient to determine the number of ( γ,δ )
F p m
× F p n
\
such that Equation (5) has exactly p d solutions in
{
(0 , 0)
}
F p n . By the proof of
Proposition 7, this case can occur only if γδ
=0.
x p d 1
F p n }
be the set of nonzero ( p d
Let W =
{
|
x
1)-th powers in
F p n .
By Proposition 5(1), g δ,γ ( y ) = 0 has either 0, 1, 2, or p d +1rootsin
F p n .
F p n , by Proposition 5(2)
and Remark 6, if one of the solutions belongs to W , then all these solutions are
also in W and Equation (5) has at least 2( p d
When g δ,γ ( y ) = 0 has at least two different roots in
1) + 1 = 2 p d
1 solutions. If
none of these solutions belong to W , then Equation (5) has only zero solution.
Thus, in this case the number of solutions to Equation (5) can never be p d .
If g δ,γ ( y ) = 0 has exactly one solution in
F p n , by Proposition 5(3) and Remark 6,
δ
γ x , g δ,γ ( y )=0has
( p d
1) + 1 = p d solutions. Since y =
Equation (5) has 1
×
F p n if and only if Equation (4) has exactly one solution in
F p n . Furthermore, in this case the unique solution to Equation (4) belongs to
exactly one solution in
F p m
by the analysis in the proof of Proposition 5(3). When Equation (4) has exactly one
solution in
F p m , it has exactly one solution in
F p n since this equation has either 0,
1, 2, or p d +1rootsin
F p n and its solutions in
F p n
\ F p m occur in pairs. Therefore,
F p n if and only if Equation (4) has exactly
g δ,γ ( y ) = 0 has exactly one solution in
F p m .
By Lemma 4, the number of c
one solution in
F p m such that Equation (4) has exactly one
γ p m k +1
δ p m k ( p m +1) runs through
F p m is p m−d . For any fixed γ
F p m , c =
solution in
F p m exactly ( p m +1) times when δ runs through
F p n . Therefore, when γ and δ run
F p m and
F p n , respectively, there are exactly ( p m
1) p m−d ( p m +1) =
throughout
p m−d ( p n
1) elements in R d . This together with Lemma 8 finish the proof.
Proposition 10 describes the sums of i -th powers of S ( γ,δ, 0) for 1
i
3.
Proposition 10. (i)
S ( γ,δ, 0) = p n + m .
γ∈ F p m
δ
F p n
S ( γ,δ, 0) 2 =
p n + m ,
p
3mod4 ,
(ii)
(2 p n
1) p n + m ,p
1mod4 .
γ∈ F p m
δ∈ F p n
(iii)
γ∈ F p m
S ( γ,δ, 0) 3 = p n + m ( p n + d + p n
p d ) .
δ∈ F p n
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