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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.

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  • $\begingroup$ You'll need to work modulo $56$. $\endgroup$ – Lord Shark the Unknown Dec 5 '18 at 10:30
  • $\begingroup$ @LordSharktheUnknown any chance you could expand on the method of converting my working to modulo 56? $\endgroup$ – YGrade Dec 5 '18 at 10:50
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    $\begingroup$ $x \equiv 1 \pmod {28}$ means $x \equiv 1 \pmod {56}$ or $x \equiv 29 \pmod {56}$. Similiarly $x \equiv -3 \pmod 8$ means $x \equiv 5 \pmod {56}$ or $x \equiv 13 \pmod {56}$, or $\ldots$, or $x \equiv 29 \pmod {56}$, or $\ldots$, or $x \equiv 53 \pmod {56}$. You now know the value of both partial Legendre symbols for $p \equiv 29 \pmod {56}$, and can finally conculde $\left(\frac{14}p\right)$ for $p \equiv 29 \pmod {56}$. Continue this for all the other residue classes. A simple (though slightly long) table will help, I don't think using CRT will be less work in this case. $\endgroup$ – Ingix Dec 5 '18 at 11:24

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