If $p$ is prime and $p\equiv 3 \pmod 5$, show that for every $a$, $x^5\equiv a \pmod p$ is solvable. I tried all sort of things. I know it is supposed to be easy but I can't seem to be thinking anymore. I could really use even the most basic lead here. 
I tried working with primitive roots and quadratic residue but nothing seems to do any process. I find no point in elaborating a certain way for I have few ways started, none of which leads anywhere. 

If $a\equiv 0\pmod{p}$, the result is obvious. So suppose that $a$ is not divisible by $p$. 
Note that $p-1$ and $5$ are relatively prime. So there exist integers $x$ and $y$ such that $5x+(p-1)y=1$. By choosing the signs of $x$ and $y$ suitably, we conclude that there are positive integers $s$ and $t$ such that $5s=(p-1)t+1$. Thus 
$$a^{(p-1)t+1}=a^{5s}=(a^s)^5.\tag{1}$$
But
$$a^{(p-1)t +1}=a\cdot (a^{p-1})^t \equiv a\pmod{p}\tag{2}.$$
From (1) and (2) we conclude that $(a^s)^5\equiv a\pmod{p}$.
