Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 11G / d / Solution

Put $H=(p-1)/2$. Because $p\equiv 7(\mathrm{mod}8)$, the supplementary laws giveLet $R$ and $N$ be the quadratic residues and nonresidues, respectively, among $1,\dots ,H$, and putSince $-1$ is a nonresidue, the complete set $Q$ of least positive quadratic residues isIf $u=|N|$, its sum is

Since $2$ is a residue, multiplication by two permutes $Q$ modulo $p$. Exactly the $u$ upper-half residues $p-n$ cross $p$ when doubled. Equality of the sums before and after reduction therefore givesComparing the two formulas yieldsAs $H=(p-1)/2$,This is the lower-half quadratic-residue sum for primes congruent to seven modulo eight.

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