# Lifting The Exponent Lemma for p=2

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## Theorem

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

Let $n \ge 1$ be a natural number.

Let:

$4 \divides x - y$

where $\divides$ denotes divisibility.

Then

$\map {\nu_2} {x^n - y^n} = \map {\nu_2} {x - y} + \map {\nu_2} n$

where $\nu_2$ denotes $2$-adic valuation.

### Corollary

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

Let $n \geq1$ be an even natural number.

Then

$\nu_2 \left({x^n - y^n}\right) = \nu_2 \left({x + y}\right) + \nu_2 \left({x - y}\right) + \nu_2 \left({n}\right) - 1$

where $\nu_2$ denotes $2$-adic valuation.

## Proof

Let $k = \map {\nu_2} n$.

Then $n = 2^k m$ with $2 \nmid m$.

$\map {\nu_2} {x^n - y^n} = \map {\nu_2} {x^{2^k} - y^{2^k} }$

Note that:

$x^{2^k} - y^{2^k} = \paren {x - y} \cdot \ds \prod_{i \mathop = 0}^{k - 1} \paren {x^{2^i} + y^{2^i} }$
$\map {\nu_2} {x^{2^i} + y^{2^i} } = 1$

for $i > 0$.

Because $4 \divides \paren {x - y}$, $4 \nmid \paren {x + y}$.

Thus:

$\map {\nu_2} {x + y} = 1$

Thus:

 $\ds \map {\nu_2} {x^{2^k} - y^{2^k} }$ $=$ $\ds \map {\nu_2} {x - y} + \sum_{i \mathop = 0}^{k - 1} \map {\nu_2} {x^{2^i} + y^{2^i} }$ p-adic Valuation is Additive $\ds$ $=$ $\ds \map {\nu_2} {x - y} + k$

$\blacksquare$