# Diophantine equation in Laurent polynomials

(This is a modified repost of a question from MSE; since it came out of research, I thought it might be appropriate to post it here.)

Consider the equation $$\begin{equation*} P(x, x^{-1})^m + Q(x, x^{-1})^n + 1 = 0 \end{equation*}$$ where $$m, n$$ are positive integers and $$P, Q$$ are single-variable Laurent polynomials with coefficients in $$\mathbb{C}$$. I would like to know for which $$m, n$$ there exist nontrivial solutions to this equation, i.e., with neither $$P$$ nor $$Q$$ constant. We may restrict our attention to $$2\leq m\leq n$$ (we impose the first inequality because solutions are trivial to obtain if either $$m, n$$ is 1, and we impose the second inequality to avoid redundancy). A method for generating solutions, or for classifying all solutions, would be great. References related to this equation, or to how the corresponding Diophantine equation in integers might give intuition for this problem, would also be great.

As an example, a nontrivial solution when $$(m, n) = (2, 2)$$ is $$\begin{equation*} P(x, x^{-1}) = \frac{x - x^{-1}}{2}, \quad Q(x, x^{-1}) = \frac{i(x + x^{-1})}{2}. \end{equation*}$$ I am particularly interested in the cases $$(m, n) = (2, 2), (2, 3), (2, 4), (2, 5)$$. Comments on the related equation $$P(x, x^{-1})^2 + Q(x, x^{-1})^3 + Q(x, x^{-1}) = 0$$ would also be appreciated.

Write $$P(x,x^{-1})=x^{-i}A(x)$$ and $$Q(x,x^{-1})=x^{-j}B(x)$$ with polynomials $$A$$ and $$B$$ satisfying $$A(0)\ne0$$ and $$B(0)\ne0$$. WLOG, I'll assume that $$nj\ge mi$$. Then your equation becomes $$x^{nj-mi}A(x)^m + B(x)^n + x^{nj} = 0.$$ Since $$B(0)\ne0$$, this in fact forces $$nj=mi$$, which yields $$A(x)^m + B(x)^n + x^{nj} = 0.$$ Now the elementary "$$abc$$ inequality for polynomials" implies that $$\max\bigl\{ m\deg A, n\deg B, nj \bigr\} \le \deg(A) + \deg(B) + 1.$$ (I think that this is right, but the $$+1$$ might be $$+2$$.) Anyway, it shows that $$m$$ and $$n$$ can't be very large unless $$A$$ and $$B$$ have very small degree.