# solve congruences

• February 17th 2009, 07:42 PM
mathisthebestpuzzle
solve congruences
3x^5=1 (mod 29)

and

3^x=2 (mod 29)

find all solutions for both congruences.
• February 17th 2009, 10:32 PM
ThePerfectHacker
Quote:

Originally Posted by mathisthebestpuzzle
3x^5=1 (mod 29)

Write, $x^5 \equiv 10(\bmod 29)$ (since $10$ is an inverse of $3$ mod $29$).
Now find the index of $10$ for a given primitive root of $29$.
Can you finish up the problem?
• February 17th 2009, 11:00 PM
mathisthebestpuzzle
hacker i thank you so much, but i had already solved that one since i posted it.

i am stuck on the second one though...any help would be wonderful!