# Math Help - congruence

1. ## congruence

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.

2. Since 101 is prime, Fermat's theorem applies and we must have $a^{100}\equiv 1 \mod 101$ for all $a$ relatively prime to $101$, and 0 in all other cases. Now suppose we have an $a$ with $a^{20} \equiv 3 \mod 101$. Then

$(a^{20})^5 \equiv 3^5 \mod 101$

$a^{100} \equiv 41 \mod 101$

and that is clearly impossible.