For a pair of numbers whose difference and sum are square numbers, why is the lower number always even?

Question

I was looking for pairs of integers which satisfy $x>y>0$ and both $x+y$ and $x-y$ are square numbers.

I found that $y$ is always even. Why is this? Is this some special property of square numbers?

Answer 1

The two squares differ by $2y$, so must both be odd or both even. Even squares are always divisible by $4$, so their difference will be divisible by $4$, and odd squares are of the form $8r+1$, so always differ by a multiple of $8$.

Answer 2

You have $$2y=(x+y)-(x-y)=m^2-n^2.$$ Thinking about this modulo $4$, one sees that $m^2-n^2\equiv2\pmod4$ is impossible.