Prove that the product of some numbers between perfect squares is $2k^2$ [duplicate]

Here's a question I've recently come up with:

Prove that for every natural $x$, we can find arbitrary number of integers in the interval $[x^2,(x+1)^2]$ so that their product is in the form of $2k^2$.

I've tried several methods on proving this, but non of them actually worked. I know, for example, that the prime numbers shouldn't be in the product.
I was also looking for numbers $x$ so that between $x^2$ and $(x+1)^2$ there is actually the number $2k^2$ for some natural $k$. If we find all of these numbers, then we should prove the case only for the numbers which are not in this form.
These $x$s have this property: $x^2<2k^2<(x+1)^2$ leading to $x<k\sqrt 2<x+1$ and $x\frac{\sqrt 2}{2}<k<(x+1)\frac{\sqrt 2}{2}$. This means there should be a natural number between $x\frac{\sqrt 2}{2}$ and $(x+1)\frac{\sqrt 2}{2}$.
I've checked some of the numbers that aren't like that with computer, and they were: $3,6,10,13,17,...$. the thing i noticed was that the difference between the two consecutive numbers of that form, is either $3$ or $4$. I think this has something to do with the binary representation of $\frac{\sqrt 2}{2}$ but I don't know how to connect it with that. I would appreciate any help :)

Solution 1:

This is a community wiki answer to point out that the question was answered in comments by benh: This question is a duplicate of this one; the latter question was answered by Gerry Meyerson who found a proof in this paper of Granville and Selfridge.