Past exam of the short options course of the University of Oxford / 2023 / 1 / b

For $p>3$, quadratic reciprocity givesThe second Legendre symbol is $1$ when $p\equiv 1(\mathrm{mod}3)$ and $-1$ when $p\equiv 2(\mathrm{mod}3)$. Combining it with the sign determined by $p(\mathrm{mod}4)$ shows that a quadratic residue occurs precisely for

For odd $p\ne 5$, quadratic reciprocity has no sign factor because $5\equiv 1(\mathrm{mod}4)$:The nonzero squares modulo $5$ are $1,4$, so the answer for odd primes is $p\equiv 1,4(\mathrm{mod}5)$. The excluded-prime wording also allows $p=2$, where $5\equiv 1$ has a square root. Thus the full answer is

The primes $2,5$ both work, with the zero residue as a solution. For every other prime, multiplicativity of the Legendre symbol and quadratic reciprocity giveThe first factor is positive for $p\equiv 1,7(\mathrm{mod}8)$; the second is positive for $p\equiv 1,4(\mathrm{mod}5)$. They must have the same sign. The Chinese remainder theorem therefore yields