Prove that if $(x+y)^5$ is congruent to $2$ (mod $5$) then $(x+y)$ is congruent to $2$ (mod $5$)? I need a proof of the above.  I understand that it is just an application of Fermat's Little Theorem, but how do I prove this without the theorem?
 A: You can just run through all the possible cases. Since you are not doing anything with $x$ or $y$ independently of the other, you can just assign $z = x + y$. And then there are only five cases, so it shouldn't take long to go through all the possibilities.


*

*$0^5 \equiv 0 \pmod 5$

*$1^5 \equiv 1 \pmod 5$

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

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

*$4^5 = 1024 \equiv 4 \pmod 5$
Another option is to do these calculations in base 5. But since $10 = 2 \times 5$, that doesn't offer much help.
A: Use that if $n\equiv 0,1,2,3,4\mod 5$ then $n^5\equiv 0,1,2,3,4\mod 5$
A: Hint
$$x^5-x =x(x^2-1)(x^2+1) \equiv x(x^2-1)(x^2+1-5)=(x-2)(x-1)x(x+1)(x+2) \pmod{5}$$
Now use the fact that among 5 consecutive integers one is divisible by $5$.
A: $w^5 \equiv w \pmod 5$.  Always.
Just calculate them.  There are only five cases:
Cases 1-5:  $w \equiv 0,1,2,3,4\pmod 5$ so $w^5 \equiv 0,1, 32, 243, 1024 \equiv 0,1,2,3,4 \pmod 5$.
So $(x+y)^5 \equiv x+y\pmod 5$ and if $(x+y)^5\equiv 2\pmod 5$ then $ x+y \equiv 2 \pmod 5$ is a peculiar specific application of it.
Of course we don't have to calculate them all.  As $5$ is prime Fermat's Little Theorem say explicitely that $w^5 \equiv w\pmod 5$ and so their is nothing to prove at all.
