Analytic p-adic functions that take an algebraic value Suppose it exists $r\in\mathbb R$ such that the non constant p-adic function $f(z)=\sum_{n\ge0}a_nz^n$ ($a_n\in\mathbb C_p$) is defined on $\mathcal D=\{z\in\mathbb C_p\mid v_p(z)>r\}$. Does it exist $\alpha\in\overline{\mathbb Q}\cap f(\mathcal D)$? If the answer is yes, does it exist $\alpha\in{\overline{\mathbb Q}}\cap f(\mathcal D\cap\mathbb Q_p)$?
 A: For the first, yes. Without loss of generality by shifting we may assume $a_1 \neq 0$.
For $\alpha\in \mathbb Q$, let $x_0=0$ and $x_{n+1} = x_n + \frac{\alpha-f(x)}{a_1}$. To check that $x_n$ converges as $n$ goes to infinity to a root of $\alpha-f(x)$, it suffices to check that $v_p ( \alpha - f(x_{n+1})) \geq v_p ( \alpha - f(x_n)) + 1$.
To do this, it suffices to have $v_p(x_n) \geq s$ and $v_p( x_{n+1}- x_n) \geq s$ for some $s \in \mathbb R$ such that  $v_p (a_n)+n s> v_p (a_1) + s + 1$ for all $n > 1$, as then the contributions of $a_2$ and higher to $\alpha - f(x_{n+1})$ will be dominated by the contribution of $a_1$.
This is easy to ensure by first choosing such an  $s < \frac{ v_p (a_n) - v_p(a_1) - 1}{ n-1}$ for all $n>1$ (checking that this series is bounded below) and then choosing $alpha$ such that $v_p ( \alpha- a_0) > a_1 + s$, so that $v_p(x_1-x_0)>s$ and thus inductively $v_p(x_{n+1} -x_n) >s$ for all $n$.
For the second, no. Just take $f(z) = z + b$ where $b\notin \mathbb Q_p + \overline{\mathbb Q}$. Since $\mathbb C_p/\mathbb Q_p$ is uncountable, such $b$ exists.
