Let
F
q
{\mathbb {F}_q}
be the finite field in
q
=
p
f
q = {p^f}
elements,
F
_
(
x
_
)
\underline F (\underline x )
be a
k
k
-tuple of polynomials in
F
q
[
x
1
,
…
,
x
n
]
{\mathbb {F}_q}[{x_1}, \ldots ,{x_n}]
,
V
V
be the set of points in
F
q
n
\mathbb {F}_q^n
satisfying
F
_
(
x
_
)
=
0
_
\underline F (\underline x ) = \underline 0
and
S
S
,
T
T
be any subsets of
F
q
n
\mathbb {F}_q^n
. Set
ϕ
(
V
,
0
_
)
=
|
V
|
−
q
n
−
k
\phi (V,\underline 0 ) = |V| - {q^{n - k}}
,
\[
ϕ
(
V
,
y
_
)
=
∑
x
_
∈
V
e
(
2
π
i
p
Tr
(
x
_
⋅
y
_
)
)
for
y
_
≠
0
_
,
\phi (V,\underline y ) = \sum \limits _{\underline x \in V} {e\left ( {\frac {{2\pi i}} {p}\operatorname {Tr} (\underline x \cdot \underline y )} \right )\quad {\text {for}}\;\underline y \ne \underline 0 ,}
\]
and
Φ
(
V
)
=
max
y
_
|
ϕ
(
V
,
y
_
)
|
\Phi (V) = {\max _{\underline y }}|\phi (V,\underline y )|
. We use finite Fourier series to show that
(
S
+
T
)
∩
V
(S + T) \cap V
is nonempty if
|
S
|
|
T
|
>
Φ
2
(
V
)
q
2
k
|S||T| > {\Phi ^2}(V){q^{2k}}
. In case
q
=
p
q = p
we deduce from this, for example, that if
C
C
is a convex subset of
R
n
{\mathbb {R}^n}
symmetric about a point in
Z
n
{\mathbb {Z}^n}
, of diameter
>
2
p
> 2p
(with respect to the sup norm), and
Vol
(
C
)
>
2
2
n
Φ
(
V
)
p
k
\operatorname {Vol} (C) > {2^{2n}}\Phi (V){p^k}
, then
C
C
contains a solution of
F
_
(
x
_
)
≡
0
_
(
mod
p
)
\underline F (\underline x ) \equiv \underline 0 (\bmod p)
. We also show that if
B
B
is a box of points in
F
q
n
\mathbb {F}_q^n
not contained in any
(
n
−
1
)
(n - 1)
-dimensional subspace and
|
B
|
>
4
⋅
2
n
f
Φ
(
V
)
q
k
|B| > 4 \cdot {2^{nf}}\Phi (V){q^k}
, then
B
∩
V
B \cap V
contains
n
n
linearly independent points.