Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 1G / c

The classsatisfies $X^3=-1$ for every prime. Thus a solution exists for every odd prime $p$, regardless of its residue class modulo $3$.

The cube roots of minus one modulo an odd prime are controlled byFor $p\ne 3$, the quadratic factor has roots exactly when its discriminant $-3$ is a square, namely when $p\equiv 1(\mathrm{mod}3)$. In that case there are three distinct roots; when $p\equiv 2(\mathrm{mod}3)$, only $X=-1$ remains. For $p=3$,so $X=-1$ is again the unique root. Therefore