# Lifting The Exponent Lemma/Lemma

## Lemma

Let $x, y \in \Z$ be distinct integers.

Let $p$ be an odd prime.

Let:

$p \mathrel \backslash x - y$

and:

$p \nmid x y$.

Then

$\nu_p \left({x^p - y^p}\right) = \nu_p \left({x - y}\right) + 1$

where $\nu_p$ denotes $p$-adic valuation.

## Proof

Let $\nu_p \left({x - y}\right)=k$.

Then $x=p^k m + y$ where $p \nmid m$.

We have:

 $\displaystyle x^p - y^p$ $=$ $\displaystyle (p^k m + y)^p - y^p$ $\displaystyle$ $=$ $\displaystyle \sum_{i=0}^{p}\left({\binom{p}{i} \left({p^k m}\right)^{p-i} y^i }\right) - y^p$ Binomial Theorem $\displaystyle$ $=$ $\displaystyle \sum_{i=0}^{p-2}\left({\binom{p}{i} \left({p^k m}\right)^{p-i} y^i }\right) + \binom{p}{p-1} (p^k m) y^{p-1}$ Picking out the last two terms from the summation $\displaystyle$ $=$ $\displaystyle \sum_{i=0}^{p-2}\left({\binom{p}{i} \left({p^k m}\right)^{p-i} y^i }\right) + p^{k+1} m y^{p-1}$

Note that all terms in the above expression have a factor of $p$ to the order at least $k+1$.

So, $p^{k+1} \mid x^p - y^p$.

Also note that all terms in the summation have a factor of $p$ to the order at least $k+2$.

But in the term $p^{k+1} m y^{p-1}$, since $p \nmid m$ and $p \nmid y$,

we have $p^{k+2} \nmid p^{k+1} m y^{p-1}$.

So, $p^{k+2} \nmid x^p - y^p$.

So by definition of $p$-adic valuation, $\nu_p \left({x^p - y^p}\right)=k+1$.

$\blacksquare$