# Examples of Computing Jacobi Symbols

Recall from the Jacobi Symbols page that if $P, Q \in \mathbb{Z}$ and $Q$ has prime power factorization $Q = q_1^{e_1}q_2^{e_2}...q_k^{e_k}$ then the Jacobi symbol of $P$ over $Q$ is defined to be:

(1)where the terms in the product on the right are Legendre symbols. We proved some basic properties of Jacobi symbols, which are summarized below:

- 1. $\left ( \frac{PP'}{Q} \right ) = \left ( \frac{P}{Q} \right ) \left ( \frac{P'}{Q} \right )$.

- 2. $\left ( \frac{P}{QQ'} \right ) = \left ( \frac{P}{Q} \right ) \left ( \frac{P}{Q'} \right )$.

- 3. If $P \equiv P' \pmod Q$ then $\left ( \frac{P}{Q} \right ) = \left ( \frac{P'}{Q} \right )$.

We also noted the following very important result:

**If $\left ( \frac{P}{Q} \right ) = -1$ then $x^2 \equiv P \pmod Q$ has no solutions, just like with Legendre symbols. However, if $\left ( \frac{P}{Q} \right ) = 1$ then NOTHING can be determined!**

## Example 1

**Compute the Jacobi symbol $\left ( \frac{5}{12} \right )$. Does $x^2 \equiv 5 \pmod {12}$ have any solutions?**

We have that:

(2)Since $3 \equiv 3 \pmod 8$ we have that $\left ( \frac{2}{3} \right ) = -1$. Hence $\left ( \frac{5}{12} \right ) = -1$. We conclude that $x^2 \equiv 5 \pmod {12}$ has no solutions.

## Example 2

**Compute the Jacobi symbol $\left ( \frac{16}{60} \right )$. Does $x^2 \equiv 16 \pmod {60}$ have any solutions?**

We have that:

(3)We cannot conclude whether or not $x^2 \equiv 16 \pmod {60}$ has a solution by Jacobi symbols. However, it clearly does have a solution, namely $x = 4$.