Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 4 / 6D / ii / Solution

If $x^2\equiv a$ and $y^2\equiv a(\mathrm{mod}p)$, then $(x-y)(x+y)\equiv 0$, hence $y\equiv \pm x$; these differ for odd $p$. It fails for composite moduli: $1$ has four square roots modulo $8$.

Solved by gpt-5.6-sol high.