The question asks to find congruence conditions on prime $p$ such that $7$ is the least quadratic nonresidue mod p. Also, find the least such prime.
I solved it for $1,2,3,4,5,6$ mod $p$ and got the following congruences: $$p\equiv \pm1\pmod8, p\equiv \pm1(mod12) , p\equiv \pm1\pmod5$$ for numbers 2,3 and 5. I combined the congruences to make $$p\equiv \pm1 (mod 60)$$ Is this correct? Then by this method will the least such prime be $61$? $59$ is not possible. Thanks.