Conditions for a^n-1 to be Prime
If
\[a^n-1\]
is primes then \[a=1\]
and \[n\]
is prime.Proof
Suppose
\[a^n-1\]
is prime. Then since\[a^n-1=(a-1)(a^{p-1}+a^{p-2}+...+a+1\]
we must have
\[a-1=1 \rightarrow a=2\]
.Also, if
\[n\]
is composite, we can write \[n=rs\]
doe some integers \[r, \; s\]
. Then\[\begin{equation} \begin{aligned} a^n-1 &= a^{rs}-1 \\ &= (a^r-1)(a^{rs-r}+a^{rs-2r}+...+a^r+1) \end{aligned} \end{equation} \]
Hence
\[a^r-1 =1 \rightarrow r=1\]
and \[n\]
is prime. 
