Past exam of the options II course of the University of Oxford / 2007 / C / C2

Wilson's theorem states that for a prime $p$, $(p-1)!\equiv -1(\mathrm{mod}p)$. For odd $p$, pair each nonzero residue with its multiplicative inverse. Only the self-inverse residues remain unpaired; $x^2=1$ in a field gives $x=\pm 1$. Their product is $-1$, proving the result. For $p=2$ the assertion is immediate.

If composite $n\ge 6$ is not the square of a prime, choose a prime divisor $p$ such that $p$ and $n/p$ are distinct. Both occur among the factors of $(n-1)!$, so their product $n$ divides it. In the remaining case $n=p^2$, $p\ge 3$, and the two distinct factors $p,2p$ both occur in the factorial and have product divisible by $p^2$. Thus the factorial is zero modulo every such composite $n$.

For prime $n$, Wilson's congruence makes the numerator sine squared equal ${\sin }^2(\pi /n)$, so its ratio is one. For composite $n\ge 6$, the numerator vanishes. The only exceptional composite in the sum is four; its ratio is ${\sin }^2(3\pi /2)/{\sin }^2(\pi /4)=2$. Therefore for $x\ge 4$ the sum counts the primes plus two, and subtracting two gives exactly the required prime count. All statements concern the integers $2\le n\le \lfloor x\rfloor$.