Past exam of the options II course of the University of Oxford / 2005 / C / 2 / b

Fermat's little theorem states that for a prime $p$ and an integer $a$ not divisible by $p$,Equivalently, $a^p\equiv a(\mathrm{mod}p)$ for every integer $a$, with divisibility by $p$ allowed in that latter formulation.

Since $n>1$, its factorial is even, so $(n!{)}^2+1$ is odd and its prime divisor $p$ is odd. Also $p$ cannot divide $n!$, since it would then divide1. Modulo $p$,Raise this equation to the integer power $(p-1)/2$. Fermat's little theorem givesAs $p$ is odd, $1$ and $-1$ are distinct residues. Thus the exponent is even, and

If a prime divisor $p$ of $(n!{)}^2+1$ satisfied $p\le n$, it would divide $n!$ and therefore divide the difference $(n!{)}^2+1-(n!{)}^2=1$, impossible. Hence $p>n$.

Every integer $(n!{)}^2+1>1$ has a prime divisor, and (ii) makes each of its prime divisors congruent to 1 modulo 4. Taking arbitrarily large $n$ produces such primes beyond every bound. ThereforeEquivalently, if their list were finite, choosing $n$ at least its largest member would produce a new member larger than that supposed maximum.