TITLE: Prove that for all $n\ge2\in\mathbb{Z}$ and $p$ is prime then $n^{p^p}+p^p$ is composite.
QUESTION [5 upvotes]: Prove that for all $n\ge2\in\mathbb{Z}$ and $p$ is prime then $n^{p^p}+p^p$ is composite.

This question has been on my mind for some time now, and I needed some help.
What I have attempted so far:
If $p=2$ then
$$n^4+4=(n^2+2n+2)(n^2-2n+2).$$
Since $n\ge 2$, both parentheses are larger than $1$ and therefore this number is composite.
If $p\gt2$,
$$n^{p^p}=(n^{p^{p-1}})^p.$$
Let $p^{p-1}=x \therefore n^{p^{p-1}}=n^x.$
Now we get that
$$n^{p^p}+p^p=(n^x)^p+p^p.$$
Since $p$ is odd, then we can say that that equals
$$(n^x+p)((n^x)^{p-1}-(n^x)^{p-2}p+\dots+p^{p-1}).$$
Now $n^x+p\gt1$, but I got stuck with trying to prove that the right paranthesis is greater than $1$. Any help would be appreciated.

REPLY [6 votes]: For $p=2$ we obtain:
$$n^4+4=n^4+4n^2+4-4n^2=(n^2+2n+2)(n^2-2n+2),$$ which is composite for all $n\geq2.$
But for $p>2$ we see that
$$n^{p^p}+p^p=\left(n^{p^{p-1}}\right)^p+p^p$$ is divisible by $n^{p^{p-1}}+p$ because $p$ is odd now.
Also, we see that $$\frac{n^{p^p}+p^p}{n^{p^{p-1}}+p}\geq2.$$