Use the Law of Quadratic Reciprocity and the Chinese Remainder Theorem to determine when $6$ is a quadratic residue modulo $p$.

I'm working on the following problems concerning quadratic reciprocity.

a) Use the Law of Quadratic Reciprocity and the Chinese Remainder Theorem to determine for which primes $p$, $-50$ is a quadratic residue modulo $p$.

b) Use the Law of Quadratic Reciprocity and the Chinese Remainder Theorem to determine when $6$ is a quadratic residue modulo $p$.

So for (a), you'd say that $-50$ is a quadratic residue $\text{mod}$ $p$ if there exists an integer, say $x$, such that $x^2 \equiv(-50) \text{mod } p$.

I'm not exactly sure how to proceed from here. Any pointers?

• Do you know what the Law of Quadratic Reciprocity is? – Carl Schildkraut Oct 4 '16 at 1:33
• I know the general idea of it. The general statement is that $({p}\over{q})({q}\over{p}) = (-1)^{{p-1}\over{2}{q-1}\over{2}}$ So -50 is q and I need to determine what p is? – kojak Oct 4 '16 at 1:39

We may prove in a elementary way that for every prime $p\geq 5$ $$\left(\frac{-1}{p}\right)=1\Longleftrightarrow p\equiv 1\pmod{4}$$ $$\left(\frac{-3}{p}\right)=1\Longleftrightarrow p\equiv 1\pmod{3}$$ $$\left(\frac{2}{p}\right)=1\Longleftrightarrow p\equiv \pm 1\pmod{8}$$ and since $6=(-1)\cdot(-3)\cdot(2)$ and $\left(\frac{ab}{p}\right)=\left(\frac{a}{p}\right)\cdot \left(\frac{b}{p}\right)$, by the Chinese remainder theorem we get that $6$ is a quadratic residue $\pmod{p}$ iff $\color{red}{p\!\pmod{24}\in\{1,5,19,23\}}$.
In a similar way, since $-50=(-2)\cdot 5^2$, for every prime $p\geq 7$ we have that $-50$ is a quadratic residue $\!\!\pmod{p}$ iff $-2$ is a quadratic residue, i.e. iff $\color{red}{p\pmod{8}\in\{1,3\}}$.
Hint: Try writing the conditions in terms of quadratic residues. Specifically, note that $-50=-2(5^2)$, so $-50$ being a quadratic residue $\mod p$ is equivalent to $-2$ being a quadratic residue $\mod p$ (when $p\neq 5$).