There exists a pair of positive ints $(m,n)$ such that $(m,n$ and $m+n)$ are all perfect squares

Question

The only thing I figured to do was just to express these three guys..

$m =a^2$ ,$ n=b^2 $, and $m+n = a^2 + b^2$

Answer 1

There are an infinite number of those. Consider the recursive definition of $a_k = k^2 $: $$x_{k+1} =x_k + 2k + 1$$ or in other words: $$ (k+1)^2 = k^2 + (2k + 1)$$

You can prove that for every $n$ odd square there's at least one $m$ even square number such that $m+n$ is also a square:

$$n = 2\sqrt m + 1$$ $$m = (\frac{n-1}2)^2$$