Determine the odd primes $p$ for which $−28$ is a quadratic residue $\bmod p$.

Question

1. Determine the odd primes $p$ for which $−28$ is a quadratic residue $\bmod p$. (I've done something though I'm unsure if I'm working towards the correct answer. I have no solution for this.)

When $Q_p$ is the set of quadratic residues of modulo $p$.

Using the fact that if $-28\in Q_p$ the Legendre symbol is $\left(\frac{-28}{p}\right) = 1$. I've gathered that $\left(\frac{-28}{p}\right) = \left(\frac{-7}{p}\right)\left(\frac{2}{p}\right)\left(\frac{2}{p}\right) = \left(\frac{-7}{p}\right)$. By considering primes less than $7$ we see that $-28$ is not quadratic residue for any.

Hence $\left(\frac{-28}{p}\right) = \left(\frac{p-7}{p}\right)$, by Euler's criterion $\left(\frac{p-7}{p}\right)\equiv (p-7)^{(p-1)/2}\bmod p = 1$.

How / can I get any closer to determining a pattern for $p$ other than $1 = (p-7)^{(p-1)/2}\mod p$? Thank you.

Instead I could use the law of quadratic reciprocity? Though I'm not sure how.

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Edit: Or $\left(\frac{-7}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{7}{p}\right) = (-1)^{(p-1)/2}\left(\frac{7}{p}\right)$.

So there are two cases: $\left(\frac{7}{p}\right) = 1$ and $p\equiv 1 \mod 4$

or $\left(\frac{7}{p}\right)=-1$ and $p\equiv 3\mod 4$. How should I find a conjugacy for $\left(\frac{7}{p}\right)$? Then I'm assuming use the chinese remainder theorem to combine.

Answer 1

As you have shown, $(\frac{-7}{p})=1$.

If $p \equiv 1 \ (\mathrm{mod} \ 4)$, then by quadratic reciprocity we obtain

$(\frac{-7}{p})=(\frac{-1}{p})(\frac{7}{p})=(\frac{p}{7})=1$.

Hence $p \equiv 1 \ (\mathrm{mod} \ 7)$ or $p \equiv 2 \ (\mathrm{mod} \ 7)$ or $p \equiv 4 \ (\mathrm{mod} \ 7)$. By Chinese remainder theorem we obtain

$p \equiv 1 \ (\mathrm{mod} \ 28)$ or $p \equiv 9 \ (\mathrm{mod} \ 28)$ or $p \equiv 25 \ (\mathrm{mod} \ 28)$.

Similarly, if $p \equiv 3 \ (\mathrm{mod} \ 4)$, then

$(\frac{-7}{p})=(\frac{-1}{p})(\frac{7}{p})=(-1)(-(\frac{p}{7}))=(\frac{p}{7})=1$.

So by Chinese remainder theorem we obtain $p \equiv 15 \ (\mathrm{mod} \ 28)$ or $p \equiv 23 \ (\mathrm{mod} \ 28)$ or $p \equiv 11 \ (\mathrm{mod} \ 28)$.