Proving $n^a-n^b$ is divisible by 10 Let $n$ be positive integer. Prove that there exists positive integers $a$ and $b$, with $a \neq b$, such that $n^a-n^b$ is divisible by $10$.
I have tried using mathematical induction and logs but I am really stuck . 

Here is way of proving that $n^5-n$ works by induction, just so you know it can be done. It does rely on knowing the answer. The pigeonhole answer is the easiest elementary argument, but doesn't tell you which powers to use, and a more sophisticated approach tells you in general which powers will work for numbers other than $10$.
First note that $1^5-1=0$ is divisible by $10$.
Now for the induction step:
$$(n+1)^5-(n+1)=n^5+5n^4+10n^3+10n^2+5n+1-n-1=(n^5-n)+10(n^3+n^2)+5n(n^3+1)$$
Now $n(n^3+1)=n(n+1)(n^2-n+1)$ is divisible by both $n$ and $n+1$, so is even (there are other ways of seeing this) so that $5n(n^3+1)$ is divisible by $10$.
Hence if $n^5-n$ is divisible by $10$ so is $(n+1)^5-(n+1)$, and the induction goes through.
