### samuel.shokry12's blog

By samuel.shokry12, history, 5 weeks ago, ,

Hi, I have been reading this article Discrete log, I believe it relies on trying all the possible answers a^x mod m, where x belongs to [0, m[, but that requires that a^x mod m is cyclic somehow, where a^x mod with x >= m is equal to some a^j mod m, where j < m. I have tried some manual examples, and it seems that's the case. I guess it can be proven when m is prime using Fermat's Little Theorem, but don't know how to prove it for composite m. any suggestion?

 » 5 weeks ago, # | ← Rev. 2 →   0 if $a$ and $m$ are coprime $a^x \equiv a^{x \bmod \phi (m)} \bmod m$ holdsEuler's theorem
•  » » 5 weeks ago, # ^ |   0 That helps, thanks.