<s>[INST] Find the values of $p$ such that $\left( \frac{7}{p} \right )= 1$ (Legendre Symbol) 
Show that if $p$ is an odd prime coprime to $7$, then $\left( \frac{7}{p} \right) = 1$ if and only if $p \equiv \pm 1, \pm 3,$ or $\pm 9 \pmod{28}$. HINT: If $p$ is an odd prime, determine which values can $p$ take $\mod28$, and consider each of these values in turn. Note that if we know $p \mod 28$ then we know  $p \mod 4$, and hence we know whether $\frac{p-1}{2}$ is odd or even.

Here, $\left( \frac{a}{b} \right)$ is the Legendre symbol. 
The bit I don't understand in the hint is, what do they mean by consider the values that $p$ can take $\mod 28$. Do they mean the values that would make $p$ a quadratic residue $\mod 28$, i.e all the $x$ values satisfying $x^2 \equiv \mod 28$, because then isn't this just $1,4,9,16,25$?
What do they mean the to "consider each of these values in turn"?
 [/INST] Here is a solution that avoids the casework suggested by Zev Chonoles. Note that the computation of $\left(\tfrac{-7}{p}\right)$ is much easier. Once we know this, the computation of $\left(\tfrac{7}{p}\right)$ falls to CRT. By the Quadratic Reciprocity law, $$\left(\frac{-7}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{7}{p}\right)=\left(\frac{p}{7}\right)(-1)^{(p-1)/2}(-1)^{(p-1)(7-1)/4}=\left(\frac{p}{7}\right)(-1)^{2(p-1)}=\left(\frac{p}{7}\right)=p^3.$$ Thus, in order for $\left(\tfrac{-7}{p}\right)=1$, we want $p^3\equiv 1\pmod{7}$. A difference of cubes factorization yields $(p-1)(p^2+p+1)\equiv 0\pmod{7}$, so $p\equiv 1\pmod{7}$ or $p^2+p+1\equiv 0\pmod{7}$. In order to make the latter case easy to work with, note that $$p^2+p+1\equiv p^2+p-6=(p+3)(p-2)\equiv 0\pmod{7},$$ so $p\equiv -3,2\pmod{7}$. Summarizing, $\left(\frac{-7}{p}\right)=1\iff p\equiv 1,2,4\pmod{7}$. Now, using that $\left(\frac{7}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{-7}{p}\right)$ along with a straightforward application of CRT should yield the required result.
EDIT: I'm being silly - no need to solve the cubic, just note that the quadratic residues modulo $7$ are $1$, $2$, and $4$.
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