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 $n=1$?

  • 1
    $\begingroup$ 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). $\endgroup$ – Xarles Feb 17 at 15:41
  • 4
    $\begingroup$ 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. $\endgroup$ – 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.


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