# Evaluate the Legendre symbol $(\frac{14}{p})$ for $p > 2$.

Let $$p > 2$$. I try to compute the Legendre symbol $$\left(\frac{14}{p}\right)$$, but I have some difficulties. Here is my attempt so far:

$$\left(\frac{14}{p}\right) = \left(\frac{2}{p}\right)\left(\frac{7}{p}\right).$$

$$\left(\frac{2}{p}\right) = (-1)^{\frac{p^2-1}{8}} = \begin{cases} \ \ \ 1, & \text{if } p \equiv \pm 1\mod 8 \\ -1, & \text{if } p \equiv \pm 3\mod 8. \end{cases}$$

If $$p$$ is odd, $$\left(\frac{7}{p}\right) = (-1)^{\frac{7-1}{2} \frac{p-1}{2}} \left(\frac{p}{7}\right) = (-1)^{\frac{p-1}{2}}\left(\frac{p}{7}\right)$$,

where $$(-1)^{\frac{p-1}{2}} = \begin{cases} \ \ \ 1, & \text{if } p \equiv 1\mod 4 \\ -1, & \text{if } p \equiv 3\mod 4 \end{cases}$$ and $$\left(\frac{p}{7}\right) = \begin{cases} \ \ \ 1, & \text{if } p \equiv 1, 2, 4\mod 7 \\ -1, & \text{if } p \equiv 3, 5, 6\mod 7. \end{cases}$$

Applying the Chinese Remainder Theorem I obtain

$$\left(\frac{7}{p}\right) = \begin{cases} \ \ \ 1, & \text{if } p \equiv \pm 1, \pm 3, \pm 9\mod 28 \\ -1, & \text{if } p \equiv \pm 5, \pm 11, \pm 13 \mod 28 \end{cases}.$$

To multiply $$\left(\frac{2}{p}\right)$$ and $$\left(\frac{7}{p}\right)$$ I have to apply the Chinese Remainder Theorem again, but I don't see how to continue.

• If $p$ is even, then $(7|p)$ isn't a Legendre symbol. – Gerry Myerson Oct 29 '18 at 2:16
• @Will Jagy Thanks for your comment. I don't understand how to apply CRT, since $gcd(56, 8) \neq 1$. Can you maybe show this to me ? – Crystal Oct 29 '18 at 10:02
• You say $(14/p)$ for $p>2$ and then ask $(14/2)$? I don't understand... – Dietrich Burde Oct 29 '18 at 10:36
By definition we have $${\displaystyle \left({\frac {n}{2}}\right)=\left\{{\begin{matrix}0&{\mbox{if }}n\equiv 0{\pmod {2}}\\1&{\mbox{if }}n\equiv 1,7{\pmod {8}}\\-1&{\mbox{if }}n\equiv 3,5{\pmod {8}}\end{matrix}}\right.}$$ Concerning your question "I try to compute the Legendre symbol $$(\frac{14}{2})$$", the answer is $$0$$, but as Kronecker symbol, since it is not a Legendre symbol.
• Did I write $(\frac{14}{2})$ ? Sorry, then this was just a typo. – Crystal Oct 29 '18 at 18:24