Is there $p$ such that $a^p\,$mod$\,r=0$ and $r$ is a prime number and $1\le a<r$.
I am believing that $a^p\,$mod$\,r=0$ will be not equal to $0$ any value of $p$ for all values of $a$ from $1$ to $r-1$.
Is it correct? please provide any counter example if not.
If $a^p\equiv 0\mod r$, it means that $r$ divides $a^p$ (or that $a^p=0$ which is not the case).
As $r$ is prime, then $r$ must divide $a$, but $a<r$ so $r$ cannot divide $a$. This contradiction proves that the premise is false that there is $p$ such as $a^p\equiv 0\mod r$ for some $a$ with $1\le a<r$.
