The formula $x^k-1=(x-1)\left(x^{k-1}+x^{k-2}+\cdots+x+1\right)$ proves that $2^n-1$ is composite whenever $n$ is composite. (For example, $2^{15}$ has the factorization $\left(2^5-1\right)\left(2^{10}+2^5+1\right)$.) If $n$ is prime, $2^n-1$ may be, and often is, prime. Such primes are called Mersenne primes. They are a popular topic because they are so easily proved to be prime (when they are prime). Use the corollary of this chapter to prove that if $n$ is prime and if $p$ is a prime factor of $2^n-1$, then $p \equiv 1 \bmod n$. Moreover, to determine whether $p$ divides $2^n-1$, one only needs to do the simple computation of $2^n$ mod $p$. Clearly $2^2-1=3,2^3-1=7$, $2^5-1=31$ are all prime. That $2^7-1=127$ is prime follows from the above ideas without computation. Use these ideas to determine whether the next few numbers $2^n-1$ for $n=11,13,17, \ldots$ are prime.