<s>[INST] Prove that if $x$ is divisible by $4$, then $x=a^2-b^2$ 
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.
 [/INST] As $a+b\pm(a-b)$ are even,
$a+b, a-b$ have the same parity
Case$\#1:$ If $a+b$ is even, so will be $a-b\implies(a+b)(a-b)$ must be divisible by $4$
Let $(a+b)(a-b)=4m\iff\dfrac{a+b}2\cdot\dfrac{a-b}2=m$
If $m=pq,$  $\dfrac{a+b}2=p,\dfrac{a-b}2=q$
$\implies a=p+q, b=p-q$
Trivially choose $p$ or $q=1$
Case$\#2:$ If $a+b$ is odd, so will be $a-b\implies(a+b)(a-b)$ must be odd.
Let for odd $x(=uv),(a+b)(a-b)=uv$ where $u,v$ are odd
If $a+b=u,a-b=v;$
$a=\dfrac{u+v}2$ which is an integer as $u+v$ is even
$b=?$
Trivially set $u$ or $v=1$
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