Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 11I / a / Solution

For an odd prime $p$, the Legendre symbol isEuler criterion states that

The Gauss lemma says that, for $p\nmid a$, if $m$ of the least positive residues oflie in $(p/2,p)$, thenTo prove it, replace every residue above $p/2$ by its negative. The resulting absolute residues are distinct up to sign and therefore form a permutation ofMultiplying the congruences givesCancel the nonzero factorial and apply Euler's criterion.

For $a=2$, the residues are $2,4,\dots ,p-1$. The number above $p/2$ iswhose parity gives the second supplementary law for quadratic reciprocityConsequently, for odd primes,The congruence is also soluble for $p=2$.