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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.

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    $\begingroup$ Or you could say $n^x+p<n^{px}+p^p$, which means the second parenthesis is not $1$ $\endgroup$ – Shubham Johri Feb 8 at 7:58
  • $\begingroup$ Thank you for your help, I was just missing that. $\endgroup$ – user587054 Feb 8 at 8:02
  • $\begingroup$ Follows from the polynomial case. $\endgroup$ – Bill Dubuque Feb 8 at 15:37
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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 $$\frac{n^{p^p}+p^p}{n^{p^{p-1}}+p}\geq2.$$

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

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