Interesting number theory question, which I feel should be reasonably straight forward, but I can't seem to crack it.
Show that 7 is a quadratic residue for any prime p of the form 28k + 1 and 28k + 3.
Now, I know that a quadratic residue $a$ (mod m) means that there exists $x$ such that $x^2\equiv a$(modm).
So if 7 is a quadratic residue mod 28k+1 or 28k+3 then $x^2\equiv 7$ (mod 28k+1 or 28k +3). But I am unsure of how to tackle this problem.
Perhaps, I could go on to say that for the first case, $x^2\equiv 6$ (mod 28k) $\equiv$ 3 (mod 14k)??
Or perhaps I am going about this wrong.
I also know the laws of quadratic reciprocity, so I need to show $(\frac{7}{28k+1})$ =1 and likewise for 28k +3
Help would be greatly appreciated. Thank you
Best Answer
Hint: If $p=28k+1$ then $\frac{7-1}{2}\cdot \frac{p-1}{2}$ is even. If $p=28k+3$ then $\frac{7-1}{2}\cdot\frac{p-1}{2}$ is odd.