**Problem:** Let $p$ be an odd prime number and consider the $n$-dimensional vector space over the field with $p$ elements. I want to prove that the number of $3$-term arithmetic progressions in a subset
$A\subseteq\mathbb{F}\_{p}^{n}$ is $cp^{2n}$ for every $n\geq N\_{0}$, for sufficiently large $N\_{0}$ and a constant $c>0$.


**Fact 1:** I shall use Meshulam's theorem which states that if a subset $B\subseteq\mathbb{F}\_{p}^{n}$ has density bigger than $2/n$, i.e. $|B|> \frac{2p^{n}}{n}$, then $Β$ contains a $3$-term arithmetic progression. 


**Fact 2:** I also may use the following fact: First observe that if $U$ is a subspace of $\mathbb{F}\_{p}^{n}$, then there are $p^{n-k}$ distinct cosets of $U$.
Now, let $A\subseteq\mathbb{F}\_{p}^{n}$ be a subset of density $\epsilon>0$.
Then, there are at least $\frac{\epsilon}{2}p^{n-k}$ cosets $V$ of $U$ such that $|A\cap V|\geq \frac{\epsilon}{2}p^{k}$.


As an affine subspace of $\mathbb{F}\_{p}^{n}$ is just a coset of a subspace of $\mathbb{F}\_{p}^{n}$, we have that the above implies that there exist at least $\frac{\epsilon}{2}f(k,n)$ $k$-dimensinal affine subspaces $V$ of $\mathbb{F}\_{p}^{n}$ such that $|A\cap V|\geq \frac{\epsilon}{2}p^{k}$, where $f(k,n)$ is the total number of the $k$-dimensional affine subspaces of $\mathbb{F}\_{p}^{n}$. This number equals to
\begin{equation}
f(k,n)=p^{n-k}\binom{n}{k}\_{p}
\end{equation}
where $\binom{n}{k}\_{p}$ is the Gaussian coefficient and id defined as
\begin{equation}
\binom{n}{k}\_{p}=\frac{(p^{n}-1)(p^{n}-p)\cdots(p^{n}-p^{k-1})}{(p^{k}-1)(p^{k}-p)\cdots (p^{k}-p^{k-1})}.
\end{equation}




---


My attempt goes as follows: Let $n$ be large enough so that there exists $k<n$ such that $\frac{\epsilon}{2}>\frac{2}{k}$. Then, we consider $p^{n-k}$
$k$-dimensional affine subspaces $V$ of $\mathbb{F}\_{p}^{n}$ pairwise disjoint (we can take such by taking the $p^{n-k}$ of a $k$-dimensional subspace). We know that there $\frac{\epsilon}{2}p^{n-k}$ of them such that $|A\cap V|\geq \frac{\epsilon}{2}p^{k}>\frac{2p^{k}}{k}$. Then, considering each $A\cap V$ as a subspace of $V\cong\mathbb{F}\_{p}^{n}$, Meshulam's theorem implies that each one of them contains a $3$-AP and since they are pairwise disjoint there are $\frac{\epsilon}{2}p^{n-k}$ $3$-AP's in $A$. 


That is as far as I get. My idea is to use the more powerful statement of **Fact 2**, where by finding $cp^{2n}$ disjoint $k$-dimensional affine subspaces of $\mathbb{F}\_{p}^{n}$ it implies the desired. But I have no clue how to find them. Any ideas?



Asymptotic number of $3$-AP's in a set $A\subseteq\mathbb{F}_{p}^{n}$ of density $\epsilon$
I was going to comment with a link to where this Varnavides idea is written up, but to my surprise I couldn't find one simply done in the case of $\mathbb{F}\_p^n$, so I thought I'd sketch the idea here. (Of course none of this is original to me, it's one of those proofs that is well-known in the field, and is a routine generalisation of the Varnavides proof.)


Let $A\subset \mathbb{F}\_p^n$ be a set with density $\lvert A\rvert/p^n=\epsilon$. Let $k$ be some integer to be chosen later, and let $T$ be the number of three-term arithmetic progressions in $A$ (where here I'm only talking about genuine 3APs, i.e. $x,x+d,x+2d$ with $d\neq 0$).


Let $U$ be an affine subspace of $\mathbb{F}\_p^n$ of dimension $k<n$ chosen uniformly at random. We compute the expected number of 3APs in $A\cap U$ in two different ways.


Let $q$ be the probability that a fixed 3AP is in $U$ (this is clearly independent of which 3AP we're talking about). Then by linearity of expectation the expected number of 3APs in $U\cap A$ is just $qT$.


On the other hand, the expected density of $A\cap U$ in $U$ is $\epsilon$. We convert this into a lower bound for the expected number of 3APs as follows. Let $L$ be the total number of affine subspaces we're choosing from, and let $L'$ be the number of such subspaces $U$ where $\lvert A\cap U\rvert \geq \frac{2}{k}\lvert U\rvert$ (such subspaces are 'good'). In particular, by Meshulam's bound, any such $U$ has the property that $A\cap U$ contains at least one non-trivial three-term arithmetic progression


We know that $\sum\_{U} \lvert A\cap U\rvert \geq \epsilon p^k L$. The contribution from non-good $U$ is at most $\frac{2}{k}(L-L')p^k$. The contribution from good $U$ is trivially at most $L'p^k$. Therefore,
$$ L'+\frac{2}{k}(L-L')\geq \epsilon L,$$
and hence after rearranging, assuming $\epsilon\geq 4/k$ and $k\geq 4$, say, $L'\geq \frac{\epsilon}{4}L$, and hence the probability that $U$ is good is $\geq \epsilon/4$.


Since any good affine subspace contains at least one 3AP, the expected number of 3APs in $U\cap A$ is $\geq \frac{\epsilon}{4}$. Comparing this to the other calculation, we see that $T\geq \epsilon/4q$. We can calculate $q$ as follows.


We know that the total number of 3APs in $\mathbb{F}\_p^n$ is exactly $p^n(p^n-1)$. Similarly the number in any fixed affine subspace of dimension $k$ is exactly $p^{k}(p^{k}-1)$. Therefore


$$ q p^n(p^n-1)= p^{k}(p^{k}-1),$$


and so


$$ q = \frac{p^{k}-1}{p^{n-k}(p^n-1)}\ll p^{-2n+2k}.$$


Therefore $T \gg p^{2n-2k}$. Our requirement on $k$ was that $\epsilon \geq 4/k$, and so we can select some $k=O(\epsilon^{-1})$, and hence


$$ T \gg \epsilon p^{O(\epsilon^{-1})}p^{2n} $$


as required.


Notice that the type of dependence on $\epsilon$ was dependent on Meshulam's bound, but for a qualitative bound (so just $T\geq c(\epsilon)p^{2n}$ for some $c\_\epsilon$ depending only on $\epsilon$) any density result will do. Similarly, the more powerful result of Ellenberg-Gijswijt yield a correspondingly better dependence on $\epsilon$.


