Let $a$ and $n$ be positive integers.Prove that $a^n-1$ is prime only if $a=2$ and $n=p$ is prime.

Proof:

\begin{equation}

a^n-1^n=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)

\end{equation}

So

\begin{equation}

a=2

\end{equation}

So

\begin{equation}

a^n-1=2^n-1

\end{equation}

If $n$ is not prime,then $n=p_1p_2,p_1,p_2>1$.Then

\begin{equation}

2^n-1=2^{p_1p_2}-1=(2^{p_1})^{p_2}-1=(2^{p_1}-1)\Delta

\end{equation}

$\Delta>1$.So $2^n-1$ become a composite number,which leads to absurdity.So $n$ is a prime.