showing a number theory mod 5 problem How can i solve this problem ?

Given that $(a + b)^5 \equiv 2 \pmod5$, show that $(a+b) \equiv 2 \pmod5$. 

I have used if-then logic but I am stuck. 

if you put $x = a + b$ it's a little simpler. We just want to show that if $x^5 \equiv 2 \pmod 5$ then $x \equiv 2 \pmod 5$.
To do that we can just try all 5 possible numbers of $x$:


*

*$x = 0$ then $0^5 \equiv 0 \pmod 5$

*$x = 1$ then $1^5 \equiv 1 \pmod 5$

*$x = 2$ then $2^5 \equiv 32 \equiv 2 \pmod 5$

*$x = 3$ then $3^5 \equiv 243 \equiv 3 \pmod 5$

*$x = 4$ then $4^5 \equiv (-1)^5 \equiv -1 \equiv 4 \pmod 5$


This proves the statement.

Furthermore you might notice that $x^5 \equiv x \pmod 5$, this is Fermats little theorem so if you have this theorem that gives a quicker proof of the statement.
