We are told 2 is a primitive root mod 101. Show that there is no solution to $x^{6} \equiv 2 \bmod 101$. I know how to show that there is no solution if we are dealing with $x^{k} \equiv 2 \bmod 101$ where $k$ divides $100$.  I basically arrive at a contradiction whereby the order of $2$ is less than $100$.  But here $k$ does not divide our $p-1$.  Have not taken abstract algebra but should be solvable using elementary number theory.
 A: Since $2$ is a primitive root mod $101,$ we have:
(i). If $\gcd (x,101)=1$ then $2^n\equiv x \mod 101$ for some $n.$
(ii). If $2^m\equiv 1 \mod 101$ then $100=\phi(101)$ is a divisor of $m.$
Now $x^6\equiv 2\mod 101$ implies $\gcd (x,101)=1.$
So if $x^6\equiv 2 \mod 101,$ let $x\equiv 2^n\mod 101$ by (i). Then   $2^{6n}\equiv 2 \mod 101$ so  $2^{6n-1}\equiv 1 \mod 101.$ By (ii) this implies $100|(6n-1),$ which is impossible.
A: Since $101$ is prime, WLOG we may consider $1\le x \le 100$. Since $2$ is given to be a primitive root, we know that for each $(1\le y \le 100)$ there exists a unique $(1\le z \le 100)$, such that $2^y \equiv z \bmod 101$.
Therefore, there exists one unique $1\le k \le 100$ such that $2^k \equiv x \bmod 101$, which would imply $2^{6k} \equiv 2 \bmod 101$ for any $x^6 \equiv 2 \bmod 101$
But $2^1 \equiv 2 \bmod 101$, or since $2^{100}\equiv 1 \bmod 101$, $2^{1+100n} \equiv 2 \bmod 101$, which would then require $6k=1+100n$. Since $6k$ and $1+100n$ have different parity, this an impossibility. So the assumption that $\exists x$ such that $x^6 \equiv 2 \bmod 101$ must be wrong.
