Problem

Source: Polish Mathematical Olympiad 2016 P4- Final Round

Tags: number theory, Divisibility, order of an element, auyesl



Let $k, n$ be odd positve integers greater than $1$. Prove that if there a exists natural number $a$ such that $k|2^a+1, \ n|2^a-1$, then there is no natural number $b$ satisfying $k|2^b-1, \ n|2^b+1$.