If $x\in Z$ and divisible by $4$, then there are existing $a,b \in Z$ , so: $x=a^2-b^2$
I had been trying to figure out which way this should be solved. First I found out that if $x=20$, then $a^2 = 6^2$ and $b^2 = 4^2$ (of course $20=36-16$ ).
now I'm not sure if i could place those examples as an answer, or I need to prove it with some "proving tricks". anyway, I thought about that $a^2 - b^2 = (a+b)(a-b)$ and $4\mid x$ is also $x=4k$ for some $k$ integer.