# Does this equation have more than one integer solution?

Consider the following diophantine equation $$n = (3^x - 2^x)/(2^y - 3^x),$$ where $$x$$ and $$y$$ are positive integers and $$2^y > 3^x$$.

Does $$n$$ have any other integer solutions besides the case when $$x=1$$ and $$y=2$$, which give $$n=1$$?

• It is not a solution because of the last inequality, but $x=2$ and $y=3$, so $n=-5$, it is an integer solution of the equation (which I am reluctant to say is Diophantine). – Xarles Feb 17 at 15:41
• Something that may help: $n + 1 = \frac{2^y - 2^x}{2^y - 3^x}$, so there is such an $n$ iff $2^y - 3^x | 2^y - 2^x$. The denominator is odd, so there is such an $n$ iff $2^y - 3^x | 2^{y - x} - 1$. Notice that this implies that $2^{y - 1} < 3^x < 2^y$, so for each $y$, there's only one $x$ to check. – user44191 Feb 19 at 5:42

This follows quickly from the observations of user44191. Check each $$1 \leq x \leq 66$$ and note that, for $$x \geq 67$$, we have $$y < 1.6x$$. Applying lower bounds for linear forms in two complex logarithms (as in, say, Theorem 5.2 of de Weger's thesis), we have that $$2^y - 3^x \geq 3^{0.9x},$$ since $$3^x > 10^{15}$$. From the fact that $$2^y-3^x \mid 2^{y-x}-1$$, it follows that $$2^{0.6x}-1 \geq 3^{0.9x},$$ a contradiction.