If $n$ is even, $a=\frac{n}{2},b=1,c=-\frac{n}{2}$ is a solution. If $n$ is odd, we take $a=\frac{n+1}{2},b=0,c=-\frac{n-1}{2}$. $n=a^2+b^2-(a+b-n)^2\Leftrightarrow \frac{n(n-1)}{2}=(n-a)(n-b)$ Clearly the number of solutions is finite, since $\frac{n(n-1)}{2}\geq 1\times|n-b|$ and each $b$ determines uniquely $a$ and $c$.

For $n=2k$ solution is $(k,1,1-k)$ For $n=2k+1$ solution is $(k+1,0,-k)$