If $p$ is a prime number, where $p>3$, show that $6p \vert ab^p - a^pb$ .

Question

I can't show that $3\vert ab^p-a^pb$

$a^p\equiv a \pmod p \Rightarrow ab^p-a^pb\equiv ab-ab\equiv 0\pmod p$

For $p>3$ $(p,6)=1$

So we have to show that $2|ab^p-a^pb$ and $3|ab^p-a^pb$, where I showed that it is divisible by 2.

If a and b are even or if one of them is even $\Rightarrow 2|ab^p-a^pb$

If they are odd $a=2k+1, b=2l+1$ we have

$(2k+1)(2l+1)^p-(2k+1)^p(2l+1)=(2k+1)(2l+1)((2l+1)^{p-1}-(2k+1)^{p-1})$

$=(2k+1)(2l+1)(2l-2k)((2l+1)^{p-2}+...+(2k+1)^{p-2})$

Where $2l-2k$ is divisible by $2$.

But how to show it for $3$.

Answer 1

You have $ab^p - a^pb = ab(b^{p-1} - a^{p-1})$. So work by cases. If either of $a$ or $b$ is a multiple of $3$, you're done. Otherwise, since $p-1$ is even, both $a^{p-1}$ and $b^{p-1}$ are congruent to $1 \pmod{3}$. So their difference is $0 \pmod{3}$.