# Find the primes for which 7 is a quadratic residue

From what I understand about quadratic reciprocity. I have the following so far:

$$7 \equiv 3 \pmod{4}$$ Case A: If $p\equiv3\pmod4$, then 7 is a quadratic residue modulo $p$ iff $p$ is a quadratic nonresidue modulo 7.

Case B: Otherwise, $\left(\frac 7p\right)=\left(\frac p7\right)$; i.e. 7 is a quadratic residue modulo $p$ iff $p$ is a quadratic residue modulo 7.

I don't how to proceed with the actual calculations though. Any help would be appreciated!

• Which calculation? – Bill Dubuque Mar 20 '15 at 3:36
• Sorry, I didn't mean calculations. I mean how do I use this information to find the primes? – Shad0wNinja Mar 20 '15 at 3:40
• You have congruence on $\,p\,$ mod $\,4\,$ and mod $\,7\,$ so $\,p\,$ is determined mod $28\ \$ – Bill Dubuque Mar 20 '15 at 3:44
• Could you elaborate please? – Shad0wNinja Mar 20 '15 at 3:49
• e.g. if $\,p\equiv 1\pmod{4}\,$ then $\,p\,$ is a square mod $\,7\,$ so $\,p\equiv 1,2,4\pmod 7\,$ so $\,p\equiv 1,9,-3\pmod{28}.\$ Note $\,p\equiv a\pmod 4,\,$ $\,p\equiv b\pmod 7$ $\iff p\equiv 8b-7a\pmod{28}\ \$ – Bill Dubuque Mar 20 '15 at 3:53