Thread: [SOLVED] Solutions of Congruences

Problem: Find all the solutions of the following congruence:


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 can be rewritten in indices as



By looking at the index table, we get



Let ,

we get



Then, subtract 16 from both sides



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.

I would start by saying that . Then if you multiply both sides of the congruence by 8, it becomes , and you can read off the answer for x directly from the index table.

Opalg, I understand all of the parts until you said the answer can be read directly off the index table.

The only connection I get is this
and

Originally Posted by Paperwings

Opalg, I understand all of the parts until you said the answer can be read directly off the index table.

The only connection I get is this
and

Pay no attention to my comment. I was being stupid. I was trying to solve rather than .

But if it's any help, I notice that , so you can take y=10.

Edit ... and from the index table, that gives x=9, which works!!

Ah, thank you. I also got y = 10 and subsequently 9 although I used "guess-and-check"