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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$\begingroup$ This is very near related to an (open) conjecture of a detail in the Waring-problem of sums of like powers. I cannot at the moment show the exact relation/transformation, but you might search for keyword "waring" in MSE where I've given answers datailing this problem, sometimes to questins of the user Fred Kline there. I can come back to this possibly in the evening (west European time) $\endgroup$– Gottfried HelmsCommented Jun 25, 2015 at 11:52
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$\begingroup$ Thank you everybody for your answers. I've read a little about Ray Steiner proof, and as commented by Lagarias "it is surprising that such heavy weapons are needed for such a tiny result". I was thinking ther was an easier proof, something like Gersonides proof for 3^n-2^n=1 $\endgroup$– BenLazCommented Jun 25, 2015 at 13:05
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2 Answers
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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.
answered Jun 25, 2015 at 12:08
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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.
answered Jun 25, 2015 at 11:16
Jan-Christoph Schlage-PuchtaJan-Christoph Schlage-Puchta
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2$\begingroup$ Georges Rhin has a paper where he computes an explicit irrationality measure for $\log 2/\log 3$. He shows that $|a\log 2+b\log 3+c|\le \max(|a|,|b|,|c|)^{-13.3}$. That should make a very small search indeed. $\endgroup$ Commented Jun 25, 2015 at 11:46
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$\begingroup$ @Anthony : didn't the argument of Rhin actually mean, that the approximation of the sum of logarithms to the nearest integer is actually bad instead of good so instead of "$\le$" in your formula you needed "$\ge$" ? (I don't have access to Rhin's paper, it seems just logical to me; also I found in a paper of J. Simons/B.de Weger the argument $ (K+l) \log2 - K \log3 \gt e^{-13.3(0.46057+ \log K)} \qquad $ with the reference to Rhin's result. ) $\endgroup$ Commented Jul 17, 2015 at 20:33
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$\begingroup$ @Gottfried: Yes exactly. Thanks for the correction. $\endgroup$ Commented Jul 17, 2015 at 20:50