The aim of the following example is to illustrate how to use primitive roots and the index of an integer to solve seemingly complicated congruences.

For this example, let $p=13$ and let us attempt to solve $$7x^{10}\equiv 5 \mod 13.$$ Since every prime has a primitive root, we can easily find one. In particular, the number $2$ is a primitive root for $p=13$ . Indeed, the powers of $2$ are the following modulo $13$ :

Now we can use the properties of index to solve the equation $7x^{10}\equiv 5 \mod 13$ . By taking indices on both sides we obtain $$\ind (7x^{10})\equiv \ind 7 +10\ind x \equiv \ind 5 \mod 12$$ and so $\ind x \equiv \ind 5 - \ind 7 \equiv 9-11\equiv 10 \mod 12$ . The equivalence $10\ind x \equiv 10 \mod 12$ implies $5\ind x\equiv 5 \mod 6$ and hence $\ind x \equiv 1 \mod 6$ . Lifting this solution to modulo $12$ we obtain $\ind x \equiv 1$ or $7$ modulo $12$ , and by the table of indices, $x\equiv 2$ or $11$ modulo $13$ are the unique solutions of the congruence.