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.
• You seem to have forgotten about $7$ – user14972 Nov 10 '14 at 11:22
• You're right. Is this correct that $p\equiv -1 (mod 4)$ for 7? – Paradox 101 Nov 10 '14 at 11:28
The lowest common denominator of $8$,$12$ and $5$ is 120, so we are looking $\pmod{120}$. There are two independent conditions, that is $\pm1\pmod{24}$ and $\pm1\pmod5$, which give rise to four possible conditions: $p\equiv\pm1,\pm49\pmod{120}$