If $2^n -1$ is a prime, show that $n$ is a prime.

Question

Solution:
If $n$ is composite then $n=n_{1}n_{2}$ where $n_{1}>1$ and $n_{2}>1$. Hence, $2^{n_{1}}-1 > 1$ and $2^{n_{1}} - 1|2^{n} - 1$.

Question:
Why is $2^{n_{1}} - 1|2^{n} - 1$ true?

Answer 1

Well, $2^{n}-1 = (2^{n_{1}}-1)\cdot2^{n-n_{1}} + 2^{n-n_{1}}-1 \equiv 2^{n-n_{1}}-1\mod 2^{n_{1}}-1$. You can continue to obtain $2^{n}-1\equiv 2^{n-n_{1}\cdot \frac{n}{n_{1}}}-1\equiv 0 \mod 2^{n_{1}}-1$

Answer 2

$x^p-1$ is divisible by $x-1.$ Plug in $x=2^q.$