number theory -prove there is a solution to $x^5\equiv k \pmod{p}$ $p$ is a prime such that $p-1$ is not divisible by $5$.
$k$ is an integer number.
Prove there is a solution to $x^5 \equiv k \pmod{p}$.
I try to post another question but it's tell me:
This post does not meet our quality standards.
Why?
P is an odd prime.
Now by Euler-Fermat: 
$x^2=1(mod3)$    
$x^{(p-1)}=1(modp)$  
Show there is $k<2(p-1)$ such that 
$x^k=1(mod3p)$ .
Thanks.
 A: If $5$ does not divide $p-1$, then $\gcd(5, p-1) = 1$. Bézout tells you there are $a, b \in \Bbb{Z}$ such that $5 a +  (p-1) b = 1$.
Now by Euler-Fermat $5^{p-1} \equiv 1 \pmod{p}$. It follows that
$$k = k^{1} = k^{a \cdot 5 + (p-1) \cdot b} \equiv (k^{a})^{5} \pmod{p},$$
so $x = k^{a}$ is your solution.
A: Another approach, not using the hint:
Since $U_p^*=(\mathbb Z/p\mathbb Z)^*$ is of order =$p-1$, and as $\gcd(5,p-1)=1$, we know that there is no element of order =$5$. hence there is only one trivial solution to $x^5=1$ in $U_p^*$, i.e. $x=1$. So the homomorphism $x\to x^5$ has trivial kernel, i.e. it is a monomorphism. Since $U_p^*$ is a finite set, it is also an epimorphism as well. So, for any $k$, there is some $x$ such that $x^5=k$, as required to be shown.  
P.S. The notation is not standard, but adapted just for the sake of convenience.
Barring mistakes, and regards then.  
A: It is case $\rm\:(k,n) = (5,p\!-\!1) = 1\:$ of the  Easy k'th Power Criterion that I mentioned recently.
Suppose $\rm\, g^n = 1.\,$ Then exponents on $\rm\,g\,$ can be interpreted $\rm\ mod\ n\!:\ i \equiv m\:$ $\Rightarrow$ $\rm\,g^i = g^{m}.\  $ So it is 
clear that $\rm\,g^i\,$ is a $\rm\,k$'th power if $\rm\ mod\ n\!:\, k\mid i,\ $ i.e. $\rm\ i\equiv jk,\,$ so $\rm\,g^i = g^{jk} = (g^j)^k.\,$ By $\rm\color{#C00}{Bezout}$
$$\rm k\,|\, i\ \ (mod\ n)\!\iff\! \exists\,j\!:\ jk\equiv i\:\ (mod\ n)\!\iff\! \exists\, j,m\!:\ jk \!+\! mn = i\color{#C00}{\!\iff\!} (k,n)\,|\, i$$
Hence we have conceptually derived a proof of the following 
Theorem $\rm\ \ \ g^n = 1,\,\ (k,n)\mid i\:\Rightarrow\: g^i\,$ is a $\rm\,k$'th power $\ \ $ [Easy $\rm\,k$'th Power Criterion]
Proof $\rm\ \  By\ Bezout,\,\ (k,n)\mid i\:\Rightarrow\:k\mid i\ \ (mod\ n)\: \Rightarrow\:i\equiv jk\ \ (mod\ n)\:\Rightarrow\: g^i = g^{jk} = (g^j)^k$
Note $\,\ $ That $\rm\ \ k\,|\, i\:\ (mod\ n)\!\iff\! (k,n)\,|\: i\ \, $ frequently proves conceptually handy, $ $ e.g. $ $ see here. $\ $ The reason behind this will become clearer when one studies cyclic groups and (principal) ideals.
