# Proof of a special case of Pillai’s conjecture

In the last post we conjectured that given fixed integers $A, B, C, D, E>0$ with $BD>1$, the equation

$(*)\qquad\qquad\qquad\qquad\quad A\cdot B^x-C\cdot D^y=E$

has only finitely many solutions $(x,y)$ in integers. Let us prove this conjecture.

If $(*)$ has no solution we are done. Otherwise suppose that $(x_0,y_0)$ is a solution. Assume the contrary, and let $S$ denote the infinite set of solutions $(x,y)$. For any $(x,y)\in S$, we have

$AB^x-AB^{x_0}=CD^y-CD^{y_0}$.

So, if $(*)$ has infinitely many solutions, then writing

$\displaystyle u_x=AB^x-AB^{x_0},\quad v_y=CD^y-CD^{y_0}$

gives that $u_x=v_y$ for infinitely many $x,y\in\mathbb N_0$. So by Theorem 1.1 in this paper (or this one),

$B^r=D^s$

for some positive integers $r,s$. So $\min\{B,D\}\mid\max\{B,D\}$ (exercise). But then $\min\{B,D\}^{\min\{x,y\}}$ divides $E$ for all $(x,y)\in S$, implying $\min\{x,y\}$ is bounded over all $(x,y)\in S$. But then, by $(*)$, $\max\{x,y\}$ must also be bounded over all $(x,y)\in S$. Thus $S$ is finite, a contradiction.