Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 1I (Number Theory)

Suppose the prime numbers were the finite list $p_1,\dots ,p_r$. The integer $N=p_1\cdots p_r+1>1$ has a prime factor, by repeatedly factoring any composite factor until a prime is reached. But division by any $p_j$ leaves remainder one, so that prime factor is absent from the list. This contradiction proves there are infinitely many prime numbers. The argument does not require $N$ itself to be prime.

For any positive integer $m$, put $N=(m+1)!$. Each of $N+2,N+3,\dots ,N+m+1$ is composite, since $N+j$ is divisible by $j$ and larger than $j$. Let $p$ be the largest prime number at most $N+1$, and let $q$ be the smallest prime number at least $N+m+2$. Such $q$ exists by part (i), and no prime lies strictly between $p$ and $q$. Therefore these are consecutive prime numbers withSince $m$ is arbitrary, gaps between consecutive prime numbers are unbounded.