Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 1H / Solution

For an odd prime $p$, the Legendre symbol $(a/p)$ is zero if $p\mid a$, one if $a$ is a nonzero square modulo $p$, and minus one otherwise. By Gauss lemma, $(2/p)=(-1{)}^N$, where $N$ counts those least positive residues of $2,4,\dots ,p-1$ which exceed $p/2$. They correspond to integers $j>p/4$ among $1\le j\le (p-1)/2$, henceFor $p\equiv 1,3,5,7(\mathrm{mod}8)$, its parity is respectively $0,1,1,0$, the same as that of $(p^2-1)/8$. ThereforeUse multiplicativity and quadratic reciprocity for the requested composite numerator. Since $91=7\cdot 13$,The first reciprocity sign is negative because both primes are $3(\mathrm{mod}4)$; the subsequent signs are positive because one prime is $1(\mathrm{mod}4)$. Thus $(91/167)=-1$.