Prove the following proposition: For any odd square number $x$, there is an even square number $y$, such that $x + y$ is a square number

Think of these as pythagorean triples. For every positive integer n:

$$(4n+1) + (2n)^2 = (2n+1)^2$$

So all we need to do is show all odd squares are of the form $4n+1$ for some n, and we are done. This is perhaps most easily done by showing that no odd number of the form $4n+3$ can be a square number (the contrapositive).

Hence, given an odd square x, one can write it in the form $4k+1$ for some k, and the even square you want, y, is simply $(2k)^2$.

Edit: Henry's answer was posted as I was writing this and I possibly got a bit carried away with detail as it was my first answer on the site, a hint might have been better.

Hint:

In case of $x=1^2$, the only solution is $1^2+0^2=1^2$, but if you are allowed $y=0^2$ in this specific case then you have the simple $(2n+1)^2+0^2=(2n+1)^2$ in general.

Otherwise, consider the Pythagorean triples:

3    4   5
5   12  13
7   24  25
9   40  41


Can you (a) spot the general expression for these and (b) turn that into an answer to your question?