Let $p$ and $q$ be prime numbers such that $2^p$$-1$ is a multiple of $q$. Prove that $q>p$.
I began by noticing that the equation $2^p$$-1$ can be modified to fit Fermat's Little Theorem. I am pretty sure that this is when $a^{p-1}≡1($mod$p)$.
However, I am stuck here. I have no idea how to progress and solve this question. The only thing I can do is that $2^p$≡$1$ mod$q$.
If anyone can help me solve this question. I would appreciate it. Thank you very much!
We can use Fermat's Little Theorem to say that $a^{q-1} \equiv 1 \pmod q$. Now since $q$ is a prime that divides $2^p-1$, we can set $a$ equal to $2$ to obtain $2^{q-1} \equiv 1 \pmod q$.
Now I claim that the order of 2 modulo $q$ is the smallest positive integer $d$ such that $2^d \equiv 1 \pmod q$. And since $2^p \equiv 1 \pmod q$, we know that d divides p. Moreover, since $2^{q-1} \equiv 1 \pmod q$, we know that the order $d$ also divides $q-1$.
Okay, so now let's consider the two cases for $d$. Either $d=p$ or $d < p$.
If $d=p$, then $p$ divides $q-1$. And since $p$ is a prime that divides $q-1$, and $q-1<q$, this implies that $q>p$.
But if $d < p$, then $d$ is a divisor of $p$. Since $p$ is prime, this means that $d=1$. Notice now that this gives us the statement $2^1 \equiv 1 \pmod q$, which is NOT TRUE. By way of contradiction, this means that $d=p$. And then $d=p \implies p \leq q-1 \implies p < q$, so we're done.
