Lifting The Exponent Lemma for p=2

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

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

Let:
 * $4 \mathrel \backslash x - y$

where $\backslash$ denotes divisibility.

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

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

Proof
Let $k = \nu_2 \left({n}\right)$.

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

By P-adic Valuation of Difference of Powers with Coprime Exponent:
 * $\nu_2 \left({x^n - y^n}\right) = \nu_2 \left({x^{2^k} - y^{2^k} }\right)$

Note that:
 * $x^{2^k} - y^{2^k} = \left({x - y}\right) \cdot \displaystyle \prod_{i \mathop = 0}^{k - 1} \left({x^{2^i} + y^{2^i} }\right)$

By Square Modulo 4:
 * $\nu_2 \left({x^{2^i} + y^{2^i} }\right) = 1$
 * for $i > 0$.

Because $4 \mid x - y$, $4 \nmid x + y$.

Thus:
 * $\nu_2 \left({x + y}\right) = 1$

Thus

Also see

 * Lifting The Exponent Lemma
 * Lifting The Exponent Lemma for Sums for p=2