Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 6 / c (Prime divisors of Mersenne numbers)

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

The multiplicative group of a finite field $p$ has group order $p-1$: its elements are exactly the nonzero residue classes. For any prime factor $p$ of $2^q-1$, i and Lagrange's theorem therefore implySince $q$ and $p-1$ are positive integers, $q\le p-1$, and henceThis applies to every prime factor, including repeated prime factors viewed as the same prime number.

Suppose there were only finitely many prime numbers. There is at least one, namely $2$, so this finite set would have a largest element $q$. The integer $2^q-1$ is greater than $1$ and therefore has a prime factor $p$: for example, its least divisor greater than $1$ must be a prime number, since any proper factor of that divisor would contradict minimality.

By ii, $p>q$, contradicting the choice of $q$. Therefore there are infinitely many prime numbers, proving Euclid's theorem by orders of group elements in finite multiplicative groups.