$x = (3^n - 2^n) / (2^{m+n} - 3^n)$
$n$, $m$, and $x$ must be positive integers $\geq 1$
$x$ must also be odd
The trivial solution is
$n=1, m=1, x=1$
Are there any other solutions? If not, is there a way to prove there are no other solutions?
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Rearranging to $(2^{m+n} - 3^n)x = 3^n-2^n$ and testing a couple of low values of $n$ makes it seem improbable that there are other solutions, but I can't rule it out just because $n\leq 8$ doesn't work. – Arthur Mar 14 '24 at 23:27
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(This also makes it obvious that $x$ must be odd, even without the problem specifying it; the right-hand side is odd, so the left-hand side must also be odd.) – Arthur Mar 14 '24 at 23:33
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Some initial ideas: if $n=m$ then $f(x)=\frac{3^x-2^x}{4^x-3^x}$ is decreasing, so $x(n)<1$ for $n>1$. If $m>n$, $2^{n+m}-3^n>4^n-3^n$, so again $x(n)<1$ for $n>1$. If $m<n$, then $3^n-2^n$ is divisible by $2^m$, but $2^{n-m}$ is an an integer while $(\frac{3}{2})^{n-m}$ isn't so x isn't an integer. – Vit Mar 15 '24 at 00:07
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3Does this answer your question? How would I go about proving the following statement? - found using an Approach0 search. Note there are also other duplicate, or nearly duplicate, questions, e.g., If $ab > 1$, can $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ be an integer? and ... – John Omielan Mar 15 '24 at 00:13
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2(cont.) Collatz Conjecture: Does it follow that since there is no $1$-cycle, if $ab > 1$, then $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ cannot be an integer?. – John Omielan Mar 15 '24 at 00:16