Since 101 is prime, Fermat's theorem applies and we must have for all relatively prime to , and 0 in all other cases. Now suppose we have an with . Then
and that is clearly impossible.
I am trying to find the smallest positive a such that a^20 == 3 mod 101. I am kind of leaning toward there being no solutions but I don't know how to prove that. I think Fermat's Little Theorem is useful here? Any hints would be appreciated. Thanks.