My question is two-fold. First off, let's define $v_p$ to be the valuation on $\mathbb{Q}$ defines by setting $v_p( \frac{a}{b} p^n)=n$, where $(a,p)=(b,p)=1$. How exactly does $v_p$ extend to $\mathbb{Q}_p$ (if it all)? Is this hard to show?
Secondly, let $\alpha \in \mathbb{Q}_p$ be non-zero. I was wondering how difficult it would be to understand the values of $|1+\alpha^2|_p$. We have $$|1+\alpha^2|_p\leq \max( 1, |\alpha|_p^2).$$ Is there anything more interesting that can be said?
With the definition of $p$-adic numbers as (potentially) infinite series $z=\sum_{n=N_0}^\infty a_np^n$ with each coefficient $a_n\in\{0,1,\cdots,p-1\}$, then $v(z)$ is the smallest $n$ with $a_n\ne0$. This value may be negative, but if nonnegative, we say that $z\in\Bbb Z_p$.
The valuation-behavior of the polynomial $1+z^2$ is only somewhat interesting, in that if $v(z)>0$, then $v(1+z^2)=0$, as you know. What is rather more interesting is to compare the valuation of $f(z)$ with the valuation of $z$ when $f$ is a polynomial (or even a power series) with no constant term. And this question is yet more interesting when you look at possibilities for $z$ that lie in algebraic field extensions of $\Bbb Q_p$. For then, if you keep the convention that $v(p)=1$, you’ll have all rational numbers being possibilities for $v(z)$. Then the “valuation function” of $f$ will be a concave function (concave downward) whose graph is polygonal. This valuation function is closely related to the Newton Polygon of $f$, which also is defined by means of the “additive” valuation $v$ rather than the (multiplicative) absolute value, which many people find more intuitive.