# Zeros of a sequence in $\overline{\mathbb F_q(T)}$

Let $$\beta$$ be an element of $$\overline{\mathbb F_q(T)}\setminus\overline{\mathbb F_q}$$. Is it true that the sequence $$(\beta^{q^n}-T)_n$$ admits infinitely many zeros, that is there exist infinitely many distinct places $$\mathcal P_1,\cdots,\mathcal P_n\cdots,$$ of $$\mathbb F_q(T)(\beta)$$ such that for every $$n\in\mathbb N$$, there exists a $$k\in\mathbb N$$ such that $$v_{\mathcal P_n}(\beta^{q^k}-T)>0$$, where $$v_{\mathcal P_n}$$ is the valuation of $$K$$ associated to the place $$\mathcal P_n$$. That is an analogue of the following number theory problem: Let $$\alpha,\beta\ge2$$ be an integer. Does the sequence $$(\alpha^n-\beta)_n$$ admit infinitely many divisors?

An alternate way to phrase your question is: consider $$X = \mathbb P^1\times \mathbb P^1$$ with a given curve $$C \subset X$$ corresponding to $$\beta$$. We are interested in the geometric points of intersection of $$I_N = C\cap (\phi^n)^*\Delta$$ where $$\Delta$$ is the diagonal and $$\phi: X \to X$$ is the partial Frobenius $$(x,y) \to (x,y^q)$$. This goes back to a paper of Chai and Oort ("Hypersymmetric abelian varieties") where they were thinking of the question in the modular curve context. The arguement is pretty easy so I'll sketch it:
The basic idea is to compute intersection numbers in two different ways and show that, for these to match up, there have to be more and more points of intersection as $$n \to \infty$$. First, we can easily compute the global intersection number. Since $$\beta$$ is not constant, it corresponds to a class $$(A,B) \in Pic(X) \cong \mathbb Z^2$$ with $$A \neq 0$$ while the class of $$(\phi^n)^*\Delta$$ is $$(1,q^n)$$. Therefore the global intersection number is $$q^nA + B \to \infty$$ as $$n \to \infty$$.
On the other hand, for any point $$c \in C$$, we can show that the local intersection number is bounded. If $$C$$ is locally cut out by an equation $$f(x,y) = 0$$, then the local intersection number is computed by the vanishing order of $$f(x,x^{q^n}) = 0$$ but we know that $$f$$ has a term of the form $$x^k$$ for some $$k = k(c) \geq 1$$ since it is not divisible by $$y$$. Therefore, the local intersection number is bounded by $$k$$ for $$n \gg 0$$.