<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 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}
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