Prove x is not congruent If $\,x \not\equiv15\pmod{17},\text{ then }x^5 \not\equiv2\pmod{17}.$
I tried to take the contrapositive:
If $\,x^5 \equiv2\pmod{17},\text{ then  }x \equiv15\pmod{17}$ and 
then I assume that $x^5=17y+2$ for some integer $y$
But I am not what to do after this step.
How do I continue? 

Proving the contrapositive is a good idea.
The key observation is that if $x$ is not divisible by $17$, then $x^{16} \equiv 1(mod\ 17)$ by Euler's theorem. If we assume $x^5 \equiv 2(mod\ 17)$, then it follows that
$$
15 \equiv -2 \equiv -2 \cdot x^{16} \equiv -2\cdot(x^5)^3\cdot x \equiv -2 \cdot 2^3 \cdot x \equiv -16x \equiv x\ (mod\ 17).
$$
