Proving that $x^5 = x \pmod{10}$ for every integer $x$. 
Show that  $x^5 = x \pmod{10}$ for every integer $x$.

How can I approach this? Should I use induction? I am stuck trying to get it in terms of $x+1$.
Some feedback would be appreciated.
 A: We are to prove that $x^5-x$ is divisible by 10.
Also $x^5-x=x(x^4-1)=x(x-1)(x+1)(x^2+1)$, therefore it is the product of two consecutive numbers, $x(x-1)$, hence divisible by $2$.(recall product of $r$ consecutive numbers is divisible by $r!$)
also, $n^p-n$ is divisible by a prime $p$ (Fermat's little theorem). Plug $p=5$ to show it is divisible by $5$.
since it is divisible by $2$ and $5$ hence is divisible by $10$
A: Here is an induction proof, since that's what you're trying to do. The base case is $1^5\equiv 1$, which is obvious.
Assume $n^5\equiv n$. Now we have
$$
(n+1)^5=n^5+5n^4+10n^3+10n^2+5n+1\\
\equiv n + 1+5(n^4+n) +10(n^3+n^2)\\
\equiv n+1+5n(n^3+1)\\
\equiv n+1
$$
We have $5n(n^3+1)\equiv 0$ since it's $5$ times an odd number times an even number (either $n$ or $n^3+1$ is always even).
Thus the result follows.
A: You could check this by brute force. There are only ten cases to check. For instance:
$$0^5 \equiv 0 \mod 10$$
$$1^5 \equiv 1 \mod 10$$
$$2^5 = 32 \equiv 2 \mod 10$$
$$3^5 = 243 \equiv 3 \mod 10$$
$$4^5 = 1024 \equiv 4 \mod 10$$
etc.
It's not a slick way to prove it, but it works if you can't think of anything else.
A: By Fermat's little theorem,
$$x^5\equiv x\pmod 5$$
But $x^5\equiv x\pmod 2$ is obvious by checking the even/odd case, therefore
$$x^5\equiv x\pmod{10}$$
