I have been asked to determine the Legendre symbol $\left(\frac{14}{p}\right)$ for $p \geq 11$ and have made good progress, however, I am stuck at the very last hurdle. So far, I have found that
\begin{equation} \left(\frac{14}{p}\right) = \left(\frac{2}{p}\right)\left(\frac{7}{p}\right) \end{equation}
and we know that
\begin{equation} \left(\frac{2}{p}\right) = (-1)^{\frac{p^2 - 1}{8}} =\begin{cases} 1, & p \equiv \pm 1 \;\bmod\; 8\\ -1, & p \equiv \pm 3 \;\bmod\; 8 \end{cases} \end{equation}
and I have calculated that
\begin{equation} \left(\frac{7}{p}\right) =\begin{cases} 1, & p \equiv \pm 1, \pm 3, \pm 9 \;\bmod\; 28\\ -1, & p \equiv \pm 5, \pm 11, \pm 13 \;\bmod\; 28. \end{cases} \end{equation}
I am attempting to summarise these using the Chinese Remainder Theorem, however, I am having no luck. Using the Chinese Remainder Theorem similar to the way I used it when calculating $\left(\frac{7}{p}\right)$ would just be too many cases to consider. Am I not spotting an easy method to combine these Legendre Symbols?
This question is similar to that which was posted here, but there are no relevant answers to that post.