Problem: Find all the solutions of the following congruence:

$\displaystyle 3x^5 \equiv 1 ( \mod{23} ) $

Since 5 is a primitive root of 23. I constructed an index table modulo 23 for the p.r. 5.

Number: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

Index : 22, 2, 16, 4, 1, 18, 19, 6, 10, 3, 9, 20, 14, 21, 17, 8, 7, 12, 15, 5, 13, 11

Now, since $\displaystyle 3x^5 \equiv 1 ( \pmod{23} ) $ can be rewritten in indices as

$\displaystyle ind_{5}3+5ind_{5}x \equiv ind_{5}1 \pmod{22} $

By looking at the index table, we get

$\displaystyle 16 + 5ind_{5}x \equiv 22 \pmod{22} $

Let $\displaystyle y = 5ind_{5}x$,

we get

$\displaystyle 16 + 5y \equiv 22 \pmod{22}$

Then, subtract 16 from both sides

$\displaystyle 5y \equiv 6 \pmod{22} $

From here I am stuck. I believe I am supposed to multiply 5 by some number to get 1 mod 22. From here I can solve for y and then for x.

Any help is greatly appreciated. Thank you.