Primes of form Power Less One

Theorem
Let $m, n \in \N_{>0}$ be natural numbers.

Let $m^n - 1$ be prime.

Then $m = 2$ and $n$ is prime.

Proof
First we note that by Integer Less One divides Power Less One:
 * $\left({m - 1}\right) \mathrel \backslash \left({m^n - 1}\right)$

where $\backslash$ denotes divisibility.

Thus $m^n - 1$ is composite for all $m \in \Z: m > 2$.

Let $m = 2$, and consider $2^n - 1$.

Suppose $n$ is composite.

Then $n = r s$ where $r, s \in \Z_{> 1}$.

Then by the corollary to Integer Less One divides Power Less One:


 * $\left({2^r - 1}\right) \mathrel \backslash \left({2^{r s} - 1}\right)$

Thus if $n$ is composite, then so is $2^n - 1$.

So $2^n - 1$ can be prime only when $n$ is prime.

Also see
Primes of the form $2^n - 1$ are called Mersenne primes.

They are particularly interesting because there is a convenient algorithm (the Lucas-Lehmer Test) which can determine the primality of such a number with high computational efficiency. Therefore the largest primes known are Mersenne.

Historical Note
The proof that if $n$ is composite, then so is $2^n - 1$, is historically attributed to, who gave it in 1603.