If $x=(n^2+(n+1)^2+1)/2$ and $a=x-1, b=x+1, c=n^2+1$ then $(a^2+b^2) = 2c(n^2+2n+2)$.

Take $c=n^2+1$, $a=n^2+n+1$ and $b=n^2+2$. Reasoning. Take $c=n^2+k$, $a=n^2+k_1$ and $b=n^2+k_2$. Using $n^2\equiv -k\pmod{c}$, we find $a^2+b^2\equiv (k_1-k)^2 + (k_2-k)^2 \pmod{c}$. The simplest choice is obviously taking $k=1$. With this, taking $k_1=n+1$ and $k_2=2$ works, completes a construction. Original Source. From St. Petersburg, 1998.