Lifting The Exponent Lemma/Lemma

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Lemma

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

Let $p$ be an odd prime.

Let:

$p \divides x - y$

and:

$p \nmid x y$.


Then

$\map {\nu_p} {x^p - y^p} = \map {\nu_p} {x - y} + 1$

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


Proof

Let $\map {\nu_p} {x - y} = k$.

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

We have:

\(\ds x^p - y^p\) \(=\) \(\ds (p^k m + y)^p - y^p\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 0}^p \paren {\binom p i \paren {p^k m}^{p - i} y^i} - y^p\) Binomial Theorem
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 0}^{p - 2} \paren {\binom p i \paren {p^k m}^{p - i} y^i} + \binom p {p - 1} \paren {p^k m} y^{p - 1}\) picking out the last two terms from the summation
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 0}^{p - 2} \paren {\binom p i \paren {p^k m}^{p - i} y^i} + p^{k + 1} m y^{p - 1}\)


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:

$\map {\nu_p} {x^p - y^p} = k + 1$

$\blacksquare$


Sources

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