I can't find a way to prove that the following equation has only one solution :
$$ X = \frac{2^Q - 1}{2^{P+Q} - 3^P} $$
with $X,P,Q$ integers $> 0$.
One trivial solution is $X = 1, P = 1, Q = 1$.
Does anyone has an idea ? Best regards
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Sign up to join this communityI can't find a way to prove that the following equation has only one solution :
$$ X = \frac{2^Q - 1}{2^{P+Q} - 3^P} $$
with $X,P,Q$ integers $> 0$.
One trivial solution is $X = 1, P = 1, Q = 1$.
Does anyone has an idea ? Best regards
I'm more used to the formulation in the following form: $$ X(2^{P+Q} - 3^P)=2^Q-1 \\ 2^Q(2^Px -1) = 3^Px -1 $$ and then $$ 2^Q = {3^P \cdot X - 1 \over 2^P \cdot X - 1} \tag 1 $$ and Ray Steiner has proved in 1976 in the context of the Collatz-problem (using Rhin's result given in the other answer), that there is only one solution (which you've already noticed). A bit more about this (and the references) can be found in the wikipedia-article on the Collatz-problem and also in a remark in Lagarias' survey about the research in the Collatz-problem.
Footnote: The Waring-problem (which I mentioned in my comment just a minute ago) leads to a small modification; there the lhs is only required to be integer (instead of a power of 2) and even this conjecture (that there are no solutions for $P \gt 6$ ) seems to hold.
Unless $3^P$ is very close to $2^{P+Q}$, the right hand side will be smaller than 1. Hence the linear form $(P+Q)\log 2 - P\log 3$ is exceptionally small, and you should be able to obtain effective upper bounds for $P$ and $Q$ by Baker's method. Looking at the continuous fraction of $\frac{\log 2}{\log 3}$ you can probably reduce the upper bound to a range where you can check everything using a computer.