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. 
 A: 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.
A: Notice that if
$(a+b) \equiv 2 \pmod 5$, then $(a+b)^5 \equiv 2^5 \equiv 32 \equiv 2 \pmod 5$. 
So, going from behind we have
$(a+b)^5 \equiv 2 \equiv 32 \equiv 2^5 \pmod 5$, and then $(a+b) \equiv 2 \pmod 5$.
A: *

*Use the binomial theorem to expand (a+b)^{5}.  Note that all the middle terms are multiples of 5 and so contribute zero residue, thus (a+b)^{5} == a^5+b^5 (mod 5).

*Having completed (1), now use Fermat's Little Theorem.  Thus with 5 being prime, a^{5} == a and b^{5} == b.
This works for any prime exponent, because in (1) the middle terms in the binomial expansion are all multiples of the chosen prime.
