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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 :)

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Good work. But I'm just wondering, have you invented this by yourself or have you found this problem somewhere? – barto Feb 9 '14 at 12:37
@CODE Many (most?) consider $0$ a natural number. Better to say "positive" to avoid possible misunderstandings. – Daniel Fischer Feb 9 '14 at 12:42
@CODE,"come up with" and "come across" a problem have different(entirely opposite) meanings.I tried using induction but haven't gotten anywhere.We may have to consider the prime factorization of k since there are never two nunbers k and 2k between two consecutive perfect squares,but I don't think that will get us anywhere. – rah4927 Feb 9 '14 at 12:43
Okay if there is no stipulation on $k$ and if it can be equal to $1$, then doesn't the problem reduce to finding an even number between $x^2$ and $(x +1)^2$?? Any even number, $y$ between $x^2$ and $(x +1)^2$ is representable in the form $2m * 1^2$ innit.. – Ishfaaq Feb 9 '14 at 13:27
This should answer your first question:… – benh Feb 15 '14 at 23:54

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.

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