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:
 * $4 \divides x - y$

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.

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

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

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

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

By Square Modulo 4:
 * $\map {\nu_2} {x^{2^i} + y^{2^i} } = 1$

for $i > 0$.

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

Thus:
 * $\map {\nu_2} {x + y} = 1$

Thus:

Also see

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