Subset of $[N]$ with given density $\alpha$ contains at least $\Theta(N^2)$ 3-term arithmetic progressions Let $\alpha > 0$. Prove that there is a constant $b>0$ such that any subset $A$ of $\{1,\ldots, N\}$ of size at least $\alpha N$ contains at least $bN^2$ three-term arithmetic progressions (each with non-zero common difference, of course). (Of course, we also implicitly assume that N is large enough in terms of $\alpha$.)
The only related result I know (and we can use it without proof) is Roth's theorem - that any set of size at least $\dfrac{CN}{\log\log N}$ for some constant $C>0$ contains a three-term arithmetic progression. 
Is there a way to perhaps apply this theorem repeatedly or do I have to think of something completely different?
Any help appreciated.
 A: This can be proven using just Roth's theorem as you suggest:
Let $A \subset \{1,\ldots, N\}$ with density $\alpha > 0$. We assume throughout that $N$ is sufficiently large. (Also, throughout, "3AP" means a non-trivial 3-term arithmetic progression.)
Let $M = M(\alpha)$ be large enough such that, for any $M' \geqslant M$, any subset of $\{1, \ldots, M'\}$ with density at least $\frac\alpha 2$ contains a 3AP (such an $M$ exists by Roth's theorem$^\dagger$).
Now, for each $d \leqslant \frac{N}{10M}$, partition $\{1,\ldots, N\}$ into progressions with common difference $d$ and of length between $M$ and $2M$ $^\ddagger$, and let $P(d)$ denote the number of these progressions in which $A$ has density $\geqslant \frac \alpha 2$.
Then
\begin{align}
|A| &= \alpha N \\
& \leqslant 2M P(d) + \frac{\alpha N}{2}
\end{align}
so that $P(d) \geqslant \frac{\alpha N}{4M}$.
By construction, the intersection of $A$ with each of these $P(d)$ progressions contains at least one 3AP. Summing over all $d \leqslant \frac{N}{10M}$, we get
$$
\sum_d P(d) \geqslant \frac{\alpha N^2}{40 M^2}
$$
3APs in $A$. However, in summing over all $d$, we have over-counted - a given 3AP could be contained in multiple progressions of length $\Theta(M)$ with distinct common differences, $d$. But each 3AP in $A$ is contained in fewer than $M^2$ progressions of length $M$. (So each distinct 3AP in $A$ has been counted no more than $(2M)^2$ times.)
Hence, the number of distinct 3APs contained in $A$ is $\geqslant \frac{\alpha N^2}{160 M^4}$. Thus, since $M$ only depended on $\alpha$, we have that
$$
\# \{ 3 \text{-APs in } A \} \, \gg_{\alpha} \, N^2
$$

$\dagger$: Specifically, we need $M$ s.t. $\frac \alpha 2 > \frac{C}{\log\log M}$
$\ddagger$: We first partition $\{1,\ldots, N\}$ into $\sim \frac N d \geqslant 10M$  progressions with common difference $d$. We then split each of these progressions into blocks of length $M$ - there may be a small block of length $< M$ left over, which we just merge with the previous block, giving a progression of length $< 2M$.
