New answers tagged

0

Thanks to fgrieu's comment above and the following quote from here (PDF): Theorem: Let $p$ be a prime and let $a$ be a number not divisible by $p$. Then if $$ r \equiv s \pmod {p − 1} $$ we have $$ a^r \equiv a^s \pmod p$$ In brief, when we work $\mod p$, exponents can be taken $\mod{p − 1}$. I (think i) understood, how the first (implicit) ...



Top 50 recent answers are included