The $2$-adic order $\nu_2(x)$ of a nonzero integer $x$ is the highest power $k$ of $2$ such that $ 2^k $ divides $x$.
Consider these two propositions for a positive integer $n$ and odd numbers $x$ and $y$:
(a) If $4$ divides $x-y$, then $\nu_2(x^n-y^n)=\nu_2(x-y) + \nu_2(n).$
(b) If $n$ is even, then $\nu_2(x^n-y^n)=\nu_2(x-y)+\nu_2(x+y)+\nu_2(n)-1.$
Egnlish Wikipedia says that (b) is a stronger one. But I think (a) and (b) are equivalent because, if we assume (a), we may prove (b) as follows:
Proof (??) of (a) $\to$ (b). Write $n=2m$. Then we have $$ \nu_2(x^n-y^n) = \nu_2\left ((x^2)^m-(y^2)^m \right) = \nu_2(x^2-y^2)+\nu_2(m). $$ But $\nu_2(ab)=\nu_2(a)+\nu_2(b)$, we conclude that $$ \nu_2(x^n-y^n) = \nu_2(x-y) + \nu_2(x+y) + \nu_2(n) -1. $$
Is this correct?