Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 18H / iii / Solution

For $p\nmid n$, the reduction of the cyclotomic polynomial ${\Phi }_n$ has precisely the primitive $n$th roots as its roots. If $p\mid {\Phi }_n(x)$, the residue class of $x$ has multiplicative order $n$ in ${\mathbb{F}}_p^{\times }$, so Lagrange's theorem gives $n\mid p-1$. Conversely, if $n\mid p-1$, cyclicity of the multiplicative group supplies an element of order $n$, which is a root of ${\Phi }_n$; any integer representative supplies $x$. HenceThree explicit examples are ${\Phi }_3(X)=X^2+X+1$ with $x=2,p=7$; ${\Phi }_4(X)=X^2+1$ with $x=2,p=5$; and ${\Phi }_6(X)=X^2-X+1$ with $x=3,p=7$. Each prime is coprime to the relevant $n$ and has the stated congruence.