Finding solutions with distinct variables to systems of linear equations over $$\mathbb {F}_p$$ F p

Abstract

Abstract Let us fix a prime p and a homogeneous system of m linear equations

$$a_{j,1}x_1+\dots +a_{j,k}x_k=0$$

a

j , 1

x 1

+ ⋯ +

a

j , k

x k

= 0

for

$$j=1,\dots ,m$$

j = 1 , ⋯ , m

with coefficients

$$a_{j,i}\in \mathbb {F}_p$$

a

j , i

∈

F p

. Suppose that

$$k\ge 3m$$

k ≥ 3 m

, that

$$a_{j,1}+\dots +a_{j,k}=0$$

a

j , 1

+ ⋯ +

a

j , k

= 0

for

$$j=1,\dots ,m$$

j = 1 , ⋯ , m

and that every

$$m\times m$$

m × m

minor of the

$$m\times k$$

m × k

matrix

$$(a_{j,i})_{j,i}$$

(

a

j , i

)

j , i

is non-singular. Then we prove that for any (large) n, any subset

$$A\subseteq \mathbb {F}_p^n$$

A ⊆

F p n

of size

$$|A|> C\cdot \Gamma ^n$$

| A | > C

·

Γ n

contains a solution

$$(x_1,\dots ,x_k)\in A^k$$

(

x 1

, ⋯ ,

x k

)

∈

A k

to the given system of equations such that the vectors

$$x_1,\dots ,x_k\in A$$

x 1

, ⋯ ,

x k

∈ A

are all distinct. Here, C and

$$\Gamma $$

Γ

are constants only depending on p, m and k such that

$$\Gamma <p$$

Γ < p

. The crucial point here is the condition for the vectors

$$x_1,\dots ,x_k$$

x 1

, ⋯ ,

x k

in the solution

$$(x_1,\dots ,x_k)\in A^k$$

(

x 1

, ⋯ ,

x k

)

∈

A k

to be distinct. If we relax this condition and only demand that

$$x_1,\dots ,x_k$$

x 1

, ⋯ ,

x k

are not all equal, then the statement would follow easily from Tao’s slice rank polynomial method. However, handling the distinctness condition is much harder, and requires a new approach. While all previous combinatorial applications of the slice rank polynomial method have relied on the slice rank of diagonal tensors, we use a slice rank argument for a non-diagonal tensor in combination with combinatorial and probabilistic arguments.