# Prove that for all $n\ge2\in\mathbb{Z}$ and $p$ is prime then $n^{p^p}+p^p$ is composite.

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.

• Or you could say $n^x+p<n^{px}+p^p$, which means the second parenthesis is not $1$ – Shubham Johri Feb 8 at 7:58
• Thank you for your help, I was just missing that. – user587054 Feb 8 at 8:02
• Follows from the polynomial case. – Bill Dubuque Feb 8 at 15:37

## 1 Answer

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 $$\frac{n^{p^p}+p^p}{n^{p^{p-1}}+p}\geq2.$$

• Do you mean for $p=2$? – Peter Foreman Feb 8 at 8:23
• @Peter Foreman I fixed. Thank you! – Michael Rozenberg Feb 8 at 8:24