Let $p$ be a prime. Prove that $p$ divides $ab^p−ba^p$ for all integers $a$ and $b$. Let $p$ be a prime. Prove that $p$ divides $ab^p−ba^p$ for all integers
$a$ and $b$.
 A: Hint $\ $ little Fermat $\rm\Rightarrow mod\ p\!:\ n^p \equiv n\ \Rightarrow\ f(a,b^p)\equiv f(a,b)\equiv f(a^p,b)\ $ for all $\rm\:f\in \mathbb Z[x,y]$
A: KV Raman's answer is quite right, but I'll write my answer anyway because I find it tidier.
$\mathbb{Z}_p^* = \mathbb{Z}_p\setminus \{0\}$ is a cyclic group of order $p-1$ under multiplication, so $a^{p-1} \equiv 1 \mod p$ for all integers not divisible by $p$.  If $a$ is divisible by $p$, then $a^n \equiv 0 \mod p$ for all $n$.  So we have $a^p \equiv a \mod p$ for $a\in \mathbb{Z}$.  (This is Fermat's Little Theorem.)
Now for any integers $a$ and $b$, 
\[ ab^p \equiv ab \equiv a^pb \mod p \]
and so $ab^p - ba^p \equiv 0 \mod p$, which means $p$ divides $ab^p - ba^p$.
A: $$ab^p-ba^p = ab(b^{p-1}-a^{p-1})$$
If $p|ab$, then $p|(ab^p-ba^p)$ and also if $p \nmid ab$, then gcd$(p,a)=$gcd$(p,b)=1, \Rightarrow b^{p-1} \equiv a^{p-1} \equiv 1\pmod{p}$ (by Fermat's little theorem).  
This further implies that $\displaystyle{p|(b^{p-1}-a^{p-1}) \Rightarrow p|(ab^p-ba^p)}$.  
Q.E.D.
