$ (x^2-y^2)=m$ $ (x-y)(x+y)=m$ Because $ x-y\equiv x+y(\mod 2)$ so $ m$ is not in the form $ 4k+2$ Now we prove that if m is not in the form the equation has solution. Case 1 $ m\equiv 1(\mod 2)$ Suppose $ m=pq$ then $ x=\frac{p+q}{2},y=\frac{p-q}{2}$ is solution. Case 2$ m\equiv 0(\mod 4)$ $ m=4pq$ then $ x={p+q},y=p-q$ is solution. Other problem: Find the number of pair $ (x,y)\in N$ $ x^2-y^2=m$

If $a^2-b^2=m$, then $(a+b)(a-b)=m$, where the two factors have the same parity, hence either $m$ is odd or $4|m$. And note that $(n+1)^2-n^2=2n+1$ gives all the odd numbers and $(n+1)^2-(n-1)^2=4n$ generates all integers divisible by $4$.