Suppose the <prime numbers> were the finite list $p_1,\ldots,p_r$. 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 \b[there are infinitely many <prime numbers>]. The argument does not require $N$ itself to be prime.

Solution

Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 1I / i / Solution

 Solution (source code)