zakor wrote: Let $ a$ and $ b$ such that there exist $ c,d$ with $ a^{2}+b^{2}+c^{2}+1=d^{2}$. Assume that $ a$ and $ b$ have different parity. Then $ d^{2}-c^{2}$ is 2 mod 4, impossible. Now, for the converse, observe that $ a^{2}+b^{2}+1$ is either $ 4k+1$ or $ 4k+3$. Therefore we may take $ (c,d)=(2k, 2k+1)$ or $ (c,d)=(2k+1, 2k+2)$. Hope I'm right because this is my first post

Or just take $(a,b,c,d)=(m-n,m+n,m^2+n^2-1,m^2+n^2).$

first we prove the "if" part -- if possible , assume that $a,b$ are of different parity . then $a^2+b^2=1(mod4)$ so , $d^2-c^2=2(mod4)$ ,which is impossible.hence done then we prove the ""only if"" part -- a,b are of same parity .then take $d=\frac{a^2+b^2+2}{2}$ and $c=\frac{a^2+b^2}{2}$