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"?
 A: Hint: Apply quadratic reciprocity to $\bigl(\frac{7}{p}\bigr)$. You will see that the only relevant things affecting the outcome are what $p$ is modulo 4, and what $p$ is modulo 7. By the Chinese remainder theorem, that is the same information as what $p$ is modulo 28.
A: Let $p \neq 7$ be an odd prime.
Suppose first that $p \equiv 1 \pmod{4}$.  Then by Quadratic Reciprocity $(\frac{7}{p}) = (\frac{p}{7})$, so $(\frac{7}{p}) = 1$ iff $p$ is a square modulo $7$, i.e., iff $p \equiv 1,2,4 \pmod{7}$.  We need to consolidate this mod $7$ information with our assumption that $p \equiv 1 \pmod{4}$: since $4$ and $7$ are relatively prime, this is accomplished by the Chinese Remainder Theorem, and the answer will be a set of congruence classes modulo $28$.  
Next you have to do the case that $p \equiv 3 \pmod{4}$, so you have to apply the case of Quadratic Reciprocity in which your two odd primes are both $3 \pmod{4}$: there is an extra minus sign.  Again you can use the Chinese Remainder Theorem to compile this into a list of congruence classes modulo $28$.   
A: 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 $4$.
A: Here is the answer with minor details.
by Quadratic Reciprocity rule, $$\left( \frac{7}{p} \right)=(-1)^{\frac{7-1}{2} \cdot \frac{p-1}{2}} \left(\frac{p}{7} \right)=(-1)^{3 \cdot \frac{p-1}{2}} \left(\frac{p}{7} \right)=(-1)^{ \frac{p-1}{2}} \left(\frac{p}{7} \right).$$
Now we evaluate values of each factor on RHS:
\begin{align} (-1)^{\frac{p-1}{2}}&=\left\{ \begin{aligned}1~\text{if}~p \equiv 1 ~(\text{mod}~ 4) \\ -1~\text{if}~p \equiv 3~(\text{mod}~4) \end{aligned} \right. \\ \text{and by the previous answers, we have }~\left(\frac{p}{7}\right)&=\left\{ \begin{aligned}1~\text{if}~p \equiv 1,2,4 ~(\text{mod}~ 7) \\ -1~\text{if}~p \equiv 3,5,6~(\text{mod}~7) \end{aligned} \right. \end{align}
Now we can use Chinese Remainder Theorem, and combine the factors to obtain the final result:
\begin{align} \left(\frac{7}{p}\right)=\left\{ \begin{aligned}&1~\text{if}~p \equiv \pm 1, \pm 3, \pm 9 ~(\text{mod}~ 28) \\ -&1~\text{if}~p \equiv \pm 5, \pm 11, \pm 13~(\text{mod}~28) \end{aligned} \right. \end{align}
