Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 4 / 5F / c / Solution

Put $z=2\cos (\pi /n)$. The recurrencedefines monic integer polynomials and gives $S_k(z)=2\cos (k\pi /n)$. Hence $z$ is an algebraic integer. If $\cos (\pi /n)$ is rational, then the rational algebraic integer $z$ is an integer. Since $-2\le z<2$, this leaves the values corresponding toThis is the rational cosine of an integral submultiple of pi result. For every $n\ge 4$, the value lies strictly between $1/2$ and $1$, so it is irrational.