Assume that p and q are odd primes and $p \equiv q \pmod {28}$. Show that $(\frac{7}{p}) = (\frac{7}{q})$.

Question

Assume that p and q are odd primes and $p \equiv q \pmod {28}$. Show that $(\frac{7}{p}) = (\frac{7}{q})$.

Could anyone give me a hint for the solution please?

Answer 1

Use the quadratic reciprocity law. You have:

$$\left( \frac 7p \right) = (-1)^{\frac{7-1}{2}\frac{p-1}{2}}\left( \frac p7 \right) = (-1)^{\frac{p-1}{2}}\left( \frac p7 \right)$$ $$\left( \frac 7q \right) = (-1)^{\frac{7-1}{2}\frac{q-1}{2}}\left( \frac q7 \right) = (-1)^{\frac{q-1}{2}}\left( \frac q7 \right)$$

Now since $p \equiv q \pmod{28}$ we have that $p \equiv q \pmod 4$ and so $\frac{p-1}2$ and $\frac{q-1}{2}$ have the same parity. Thus $\left( \frac 7p \right) = \left( \frac 7q \right)$ if and only if $\left( \frac p7 \right) = \left( \frac q7 \right)$. However the last equality is trivial, since $p \equiv q \pmod 7$.