I encountered the ring $\mathbb{Z}/(n^p+1)\mathbb{Z}$, where $n$ is some positive integer and $p\in\{2,3,5,...\}$ a prime and I am wondering whether there is a difference in the structure when $p=2$ and $p$ an odd prime?
Independently of $n$ and $p$, if $n^p+1$ is prime, then we obtain a finite field but I am interested in the difference when $p=2$ or $p\neq 2$. My hunch that there should be a difference comes solely from the fact that the polynomials $x^p+1$ exhibit a different symmetry structure in their zeros over $\mathbb{C}$ if $p=2$ and $p\neq 2$.
If $p=2$, then $n^p+1=n^2+1$ might be a prime number (take $n=2$) or not (take $n=3$). It is not known if there are infinitely many primes of the form $n^2+1$, so nothing much more can be said apart from: there are some values of $n$ for which $\mathbb{Z}/(n^2+1)\mathbb{Z}$ is a field, and some values for which it is not.
For $p$ odd, it is easier in some sense. Indeed, $n^p+1=(n+1)(n^{p-1}-n^{p-2}+\cdots +1)$, so your ring is NEVER a field since $n^p+1$ is always composite.