I need to show that the following equation has a solution. (I am not asked for the answer, which I know by Mathematica to be $y=43$. )
$y^5 \equiv 2 \pmod{251}. $
I know that the order of 2 is 50, so $2^{50} \equiv 1$. Could we raise both sides of the equation to the power of 50, which would give the trivial result of $y^{250} \equiv 1$? My first approach was to consider $(y^5)^k=y^{5k}=yy^{5k-1}$ and then finding the value of $k$ such that $5k \equiv 1 (\bmod 250)$, however this doesn't work as $\gcd(5,250) \neq 1$.