I recently found this problem which asks you to find an algorithm to find all $X$ such that $X^A \equiv B \pmod{2K + 1}$.
Is there something special about the modulus being odd that allows us to solve it?
I recently found this problem which asks you to find an algorithm to find all $X$ such that $X^A \equiv B \pmod{2K + 1}$.
Is there something special about the modulus being odd that allows us to solve it?
The author of the problem once wrote an article about baby-step-giant-step algorithm (but in Chinese).
The baby-step giant-step is a meet-in-the-middle algorithm computing the discrete logarithm.
I think this article may help you.