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.
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.
Since 101 is prime, Fermat's theorem applies and we must have $\displaystyle a^{100}\equiv 1 \mod 101$ for all $\displaystyle a$ relatively prime to $\displaystyle 101$, and 0 in all other cases. Now suppose we have an $\displaystyle a$ with $\displaystyle a^{20} \equiv 3 \mod 101$. Then
$\displaystyle (a^{20})^5 \equiv 3^5 \mod 101$
$\displaystyle a^{100} \equiv 41 \mod 101$
and that is clearly impossible.