Properties of $v_p(a-b) =n, p$ adic order when $\gcd(a,p)=\gcd(b,p)=1$ about $a^p-b^p$.

Question

$v_p(a-b) =n, p \not = 2$ adic order when $\gcd(a,p)=\gcd(b,p)=1$
I need to show for every $n \in \Bbb N$ that: $$v_p(a^p-b^p)=n+1$$

I know how to proof the case for $p=2$ when $n \ge 2 $ but for this one i can`t find way to show $x^p-y^p$ such for $p=2$: $a^2 -b^2=(a+b)(a-b)$

Answer 1

We are given that $a=b+up^n$ with $u$ not a multiple of $p$. Then $$ a^p-b^p=\sum_{k=1}^p{p\choose k}b^{p-k}u^kp^{nk}$$ Using the fact that $p\mid {p\choose k}$ for $0<k<p$, we see: The summand for $k=1$ is divisible precisely by $p^{n+1}$, the summand for $k=p$ is divisible by $p^{np}>p^{n+1}$, and all in-between summands are divisible at least by $p^{2n+1}$.

Answer 2

Since $p\mid a-b$ we have $a\equiv _p b$, so

$$a^p-b^p = \underbrace{(a-b)}_{m\cdot p^n}(\underbrace{a^{p-1}+a^{p-2}b+...+b^{p-1}}_{\equiv p\cdot a^{p-1}}) = m\cdot p^n\cdot p \cdot k$$

where $a^{p-1}+a^{p-2}b+...+b^{p-1} =p\cdot k$ and obviously $p\not{|}\; mk$