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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 $\gcd(5,250) \neq 1$.

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    There is not just one answer but 5 answers!2012-03-07
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    You were almost there. The non-zero elements are a cyclic group of order 250, right? Since $2$ has order $50$, you're simply asking if an $x$ satisfying $50x \equiv 0 \pmod {250}$ is a multiple of $5$. (this is essentially what steve's answer is doing)2012-03-07

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