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

Write $\chi (a)=(\frac{a}{p})$. When $p\equiv 1(\mathrm{mod}4)$, the first supplementary law for quadratic reciprocity gives $\chi (-1)=1$, so $\chi (p-a)=\chi (a)$. Pairing $a$ with $p-a$ givesThe sum of the nontrivial character $\chi$ over all nonzero residue classes is zero. Because its two half-sums are equal when $\chi (-1)=1$,Hence

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