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Prove that for all $a,b\in\mathbb Z$ of opposite parity there exists a number $c\in\mathbb Z$ such that $c+ab$, $c+a$ and $c+b$ are perfect squares.

So we could prove that $c+ab=k^2$, $c+a=l^2$ and $c+b=m^2$, where $k,l,m\in\mathbb Z$.

This seems like an interesting problem and I can't see how I could solve it. I can't think of an idea for the solution, so some help would be great. Thanks.

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    $\begingroup$ Calm down, drink some water - you are asking too many questions in one day. Struggling is part of the process, let some time act on the problems! I'm not saying you didn't show effort, what I'm saying is: time and struggling are necessary. $\endgroup$
    – Ian Mateus
    Commented Dec 28, 2013 at 1:53
  • $\begingroup$ +1 to @IanMateus as for the OP: these questions have absolutely no context, so it is not clear whether these are trivial or open. Not a good way to get people to think about them. $\endgroup$
    – Igor Rivin
    Commented Dec 28, 2013 at 1:57

1 Answer 1

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Hint: If $a,b$ have opposite parity, then $a-b=2k+1$ for some $k\in\mathbb Z$. Now choose $c=k^2-b$.

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  • $\begingroup$ Any hint on how you arrived at your answer. After much trouble I got the same answer but you seem to have done it really fast. I think your approach will be very valuable for other problems too. $\endgroup$
    – user44197
    Commented Dec 28, 2013 at 4:32
  • $\begingroup$ At first, I noticed that $b-a=(c+b)-(c+a)$ must be the difference of two squares. As $b-a$ is odd and any odd number can be uniquely written as difference between two consecutive squares, this was a straightforward first attempt which luckily turned out to suffice, that is, after plugging in some values, I noted that $c+ab=(k+b)^2$ which can be easily proven. $\endgroup$
    – Tomas
    Commented Dec 28, 2013 at 22:06

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