If $\displaystyle n^2+1$ is prime, then $\displaystyle n^2+1$ can be expressed as $\displaystyle 4k+1$
I was thinking you probably take $\displaystyle n^2+1$ and maybe do $\displaystyle n^2+1=p$ and manipulate it from there?
If $\displaystyle n>0~\&~n^2+1$ is a prime number, then it is odd.
So $\displaystyle n^2$ is even. WHY?
That means $\displaystyle n$ is even, or $\displaystyle n=2k$ for some $\displaystyle k$
Therefore, $\displaystyle n^2+1=4j+1$ where $\displaystyle j=k^2.$