# Refreshing problem

There isn’t too much going in this question.Nonetheless,I just liked it for some bloody reason.

• Let $a$ and $n$ be positive integers where $a>1$.If $a^{n}+1$ is prime,prove that $a$ is even and n is of the form $2^{m}$ where $m \in \mathbb{N}$

Lemma:Factorization of $x^{n}+1$ $x^{m}+1=(x+1)(x^{m-1}-x^{m-2}+x^{m-3}-\cdots+x^{2}-x+1)$ where $m$ is an odd positive integer.

Simply use the factorization formula to prove this.

## Proof

Since $a>1$,it is quite clear that $a^{n}+1 \neq 2$,this means that $a^{n}+1$ is odd and so $a^{n}$ is even $\Rightarrow a$ is even.

Now,to prove the second claim,assume that $n$ is not a power of 2.

This means that $\exists$ some odd prime $p$ such that $p|n$ ie $n=pk$ where k is a positive integer.

Here $p$ is an odd positive integer so by the Lemma, $a^{n}+1=(a^{\frac{n}{p}})^{p}+1=(a^{\frac{n}{p}}+1)(a^{\frac{n}{p}(p-1)})-(a^{\frac{n}{p}(p-2)})+\cdots-a+1)$

This even accounts for the case when $n$ is prime ie $p=n$

Here the factor $a^{\frac{n}{p}}+1>1$,so it contradicts the fact that $a^{n}+1$ is prime.

This completes the proof.