3x^5=1 (mod 29) and 3^x=2 (mod 29) find all solutions for both congruences.
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Originally Posted by mathisthebestpuzzle 3x^5=1 (mod 29) Write, (since is an inverse of mod ). Now find the index of for a given primitive root of . Can you finish up the problem?
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!
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