Properties of Jacobi symbol Let $\left(\frac{a}{n}\right) $ be Jacobi symbol .
It is well known that Jacobi symbol for $a=-1$ and $a=2$ satisfies the following:
$\left(\frac{-1}{n}\right) =
\begin{cases}
1, & \text{if } n \equiv 1 \pmod{4} \\
-1, & \text{if } n \equiv 3 \pmod{4}
\end{cases}$
and , 
$\left(\frac{2}{n}\right) =
\begin{cases}
1, & \text{if } n \equiv 1,7 \pmod{8} \\
-1, & \text{if } n \equiv 3,5 \pmod{8}
\end{cases}$
Are there similar properties for other values of $a$ ?
 A: Yes, it does for all $a$ - that's arguably the point of quadratic reciprocity.
For one example,
$$\left(\frac{5}{n}\right)=\left(\frac{n}{5}\right)(-1)^{\frac{n-1}{2}\cdot\frac{5-1}{2}}=\left(\frac{n}{5}\right)=\begin{cases}
\hphantom{-}1&\text{if }n\equiv 1,4\bmod 5\\
\hphantom{-}0&\text{if }n\equiv 0\bmod 5\\
-1 & \text{if }n\equiv 2,3\bmod 5
\end{cases}$$
Even if $a$ is negative or even, you factor $-1$ and any $2$'s out and still get a breakdown in terms of residues of $n$. For example,
$$\left(\frac{-5}{n}\right)=\left(\frac{-1}{n}\right)\left(\frac{5}{n}\right)=\left(\frac{-1}{n}\right)\left(\frac{n}{5}\right)(-1)^{\frac{n-1}{2}\cdot\frac{5-1}{2}}=\left(\frac{-1}{n}\right)\left(\frac{n}{5}\right)$$
$$=\left.\begin{cases}
\hphantom{-}1&\text{if }n\equiv 1\bmod 4\\
-1&\text{if }n\equiv 3\bmod 4
\end{cases}\right\}\left.\begin{cases}
\hphantom{-}1&\text{if }n\equiv 1,4\bmod 5\\
\hphantom{-}0&\text{if }n\equiv 0\bmod 5\\
-1 & \text{if }n\equiv 2,3\bmod 5
\end{cases}\right\}$$
$$=\begin{cases}
\hphantom{-}1 & \text{if }n\equiv 1,3,7,9\bmod 20\\
\hphantom{-}0 & \text{if }n\equiv 5,15\bmod 20\\
-1 & \text{if }n\equiv 11,13,17,19\bmod 20
\end{cases}$$
