A solution to $y^5 \equiv 2\pmod{251} $ 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 $\gcd(5,250) \neq 1$.
 A: Let $x$ be a primitive root modulo $251$, so that every non-zero residue modulo $251$ is a power of $x$ and if $x^m=1$ mod $251$ then $m$ is divisible by $250$. Write $2=x^k$ for some integer $k$ and use the fact that $x^{50k}=2^{50}=1$ mod $251$ to obtain that $k$ is divisible by $5$. Thus $2=x^{5l}=(x^l)^5$ mod $251$ for some integer $l$.
A: A probabilistic root finding algorithm for finite fields can do the job. The following is described in the book 
Rudolf Lidl & Harrald Niederreiter, Finite Fields, Cambridge University Press, 1997, 168pp
all the numbers of $F_{251}$ except $0$ are roots of $x^{250}-1$ and therefore  roots of either $x^{125}-1$ or $x^{125}+1$. to find a root from a polynomial $f(x)=x^5-2$, take a randomly selected number from ${1,...,250}$. I will take the number $1$. Then calculate 
$$gcd(f(x-1),x^{125}-1)=x^3-79*x^2+4*x-89$$
$$gcd(f(x-1),x^{125}+1)=x^2+74*x+79$$
Repeat the process until you have only linear factors. For these polynomials you get the factors $$(x-108)*(x-91)*(x+120)$$
for the second you get the factors
$$(x-44)*(x+118)$$
Therefore $$f(x-1)=(x-108)*(x-91)*(x+120)*(x-44)*(x+118)$$. Now substitute $x+1$ for $x$ to get $$f(x)=(x-107)*(x-90)*(x-43)*(x+119)*(x+121)$$ and the zeroes $$197,90,43,132,130$$ for $$x^5-2=0$$
A: One can find the roots of your polynomal by checking every number from ${1,...,250}$  and use some tricks like those @Steve showed to save some labour. But calculating the roots of a polynomial $\mod p$ for a large prime is not a trivial task and I think you will not have success with your approach. But there are efficient algorithms. Two of them based on facts from the theory of finite groups and finite fields are described in
Eric Bach; Jeffrey Shallit: Algorithmic Number Theory, Volume 1: Efficient Algorithms, Chapter 7 
A: For these special numbers it is possible to calculate the roots using pencil and paper. I give you hints because I assume you want to do the calculations by yourselve.  
Observe that there is a relation between the base $2$, a 5th power and the modulus $251$:
$$ 3^5+2^3=251 $$
Use this to construct a solution of 
$$x^5 \equiv 2 \mod 251$$
How many 5th roots of $2$ are there?  Construct the others from the one you have found.
