## Valuation of congruent elements in a local division ring

Let $$K$$ be a complete local division ring (note $$v$$ its valuation). For $$x,y\in K$$ ($$y\ne0$$), one puts $$x^y=yxy^{-1}$$. Let $$r\in\mathbb N$$. Consider $$x,y\in K$$ and $$a,b\in K^*$$ such that $$v(x-y)\ge r$$ and $$v(a-b)\ge r$$. Do we have $$v(x^a-y^b)\ge r$$? In the commutative case, it is obvious but in the non-commutative case, I can not see the answer.

## Asymptotic formula for number of square-free numbers congruent to $1$ modulo $p$
Knowing that $$\sum_{n\leq x}\mu^2(n)=\frac{6}{\pi^2}x+O(\sqrt{x})$$, prove that:
$$\sum_{n\leq x,\,\,n\equiv 1(\text{mod }p)}\mu^2(n)=\frac{6}{\pi^2(p-1)}x+O(\sqrt{x})$$
Using Dirichlet charaters, we have: \begin{align*} \sum_{n\leq x,\,\,n\equiv 1(\text{mod }p)}\mu^2(n)&=\frac{1}{\varphi(p)}\sum_{n\leq x}\sum_{\chi(\text{mod }p)}\mu^2(n)\chi(n)\ &=\frac{1}{\varphi(p)}\sum_{\chi(\text{mod }p)}\sum_{n\leq x}\mu^2(n)\chi(n)\ &=\frac{1}{\varphi(p)}\sum_{n\leq x}\mu^2(n)\chi_0(n)+\underbrace{\frac{1}{\varphi(p)}\sum_{\chi\neq \chi_0}\sum_{n\leq x}\mu^2(n)\chi(n)}_{=:\Delta(x)} \end{align*} Where $$\chi_0$$ is the principal character. I was able to prove that $$\Delta(x)=O(\sqrt{x})$$ basically by using the fact that $$\mu^2(n)=\sum_{d|n}\mu(d)$$, and that $$\left|\sum_{n\leq x}\chi(n)\right|\leq \varphi(p)$$. My problem is the main term, which doesn’t match my calculation.
By definition, $$\chi_0(n)=1$$ if $$p\not| n$$ and $$\chi_0(n)=0$$ if $$p|n$$. Therefore: \begin{align*} \frac{1}{\varphi(p)}\sum_{n\leq x}\mu^2(n)\chi_0(n)&=\frac{1}{\varphi(p)}\sum_{n\leq x}\mu^2(n)-\frac{1}{\varphi(p)}\sum_{n\leq x,\,\,p|n}\mu^2(n)\ &=\frac{1}{\varphi(p)}\sum_{n\leq x}\mu^2(n)-\frac{1}{\varphi(p)}\sum_{n\leq \frac{x}{p}}\mu^2(n)\ &=\frac{1}{\varphi(p)}\left(\frac{6}{\pi^2}x-\frac{6}{\pi^2}\frac{x}{p}\right)+O(\sqrt{x})\ &=\frac{6x}{\pi^2\varphi(p)}\left(1-\frac{1}{p}\right)+O(\sqrt{x})\ &=\frac{6}{\pi^2p}x+O(\sqrt{x}) \end{align*} What am I missing?