Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 6 / c / i / Solution

If the prime number $p$ divides $2^q-1$, then $p\ne 2$, because $2^q-1$ is odd. Thus the residue of $2$ is an element of the multiplicative group of a finite field ${\mathbb{Z}}_p$, and $2^q\equiv 1(\mathrm{mod}p)$.

Let $d$ be its order of a group element. The integer division writes $q=ad+r$ with $0\le r<d$. Then $1\equiv 2^q\equiv (2^d{)}^a{2}^r\equiv 2^r(\mathrm{mod}p)$, so minimality of $d$ gives $r=0$. Hence $d$ divides $q$. Since $q$ is a prime number, $d=1$ or $d=q$. The former would give $2\equiv 1(\mathrm{mod}p)$, which would mean $p$ divides $1$, an impossibility. Consequently