Let $p$ be an odd prime. Does the equation $$2^x-3p^y=5$$ only have finitely many solutions in positive integers $x$ and $y$?
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2$\begingroup$ I don't know why this was downvoted, but this is an exponential Diophantine equation (the exponents are not fixed, but are part of the problem) and as such I suspect it might be a hard problem to solve directly. But it is known that such equations would have finitely many solutions in positive integers $x$, $p$, $y$ as a consequence of the $abc$ conjecture. See page numbers 43 and 44 of Lang's paper "Old and New Conjectured Diophantine Inequalities" for the equation $Au^m + Bv^m = k$, which can be seen at projecteuclid.org/download/pdf_1/euclid.bams/1183555717. $\endgroup$– KConradCommented Sep 2, 2018 at 1:49
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2$\begingroup$ I edited your question. It had lots of typos, and it did not use TeX. $\endgroup$– GH from MOCommented Sep 2, 2018 at 1:59
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1$\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$– GH from MOCommented Sep 2, 2018 at 4:10
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1$\begingroup$ Thank you for editing my question. This is my first time to use the mathoverflow .I do not know how to accept this officially. I'm sorry. And I am trying to find how to accept it. $\endgroup$– qriloveCommented Sep 2, 2018 at 5:43
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1 Answer
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The solutions of your equation can be injected into the solutions of the $S$-unit equation over $\mathbb{Q}$, where $S=\{\infty,2,3,5,p\}$. As the latter is known to have finitely many solutions by the results of Siegel, Mahler, Lang (see Chapter 5 in Bombieri-Gubler: Heights in Diophantine geometry), your equation also has finitely many solutions.
By a result of Beukers and Schlickewei (see Theorem 5.2.1 in the above mentioned book), the number of solutions is at most $2^{72}$ for any given prime $p$. In fact the earlier work of Evertse (Inventiones, 1984) yields the better bound $3\cdot7^{7}$ for the number of solutions (apply Theorem 1 with $\lambda=1/5$, $\mu=3/5$, $S=\{\infty,2,p\}$).
answered Sep 2, 2018 at 1:57
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1$\begingroup$ The book you mention did not have Lang as a co-author. $\endgroup$– KConradCommented Sep 2, 2018 at 5:55
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1$\begingroup$ One can show that there are only two solutions with $y > 1$ (corresponding to $x=5$ and $x=9$), for arbitrary $p$. Presumably, there are infinitely many $p$ of the shape $(2^x-5)/3$. $\endgroup$ Commented Sep 3, 2018 at 22:34
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$\begingroup$ @MikeBennett: Very interesting. Can you give some details or references for your claim? $\endgroup$ Commented Sep 4, 2018 at 3:18
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1$\begingroup$ One should be able to handle the more general equation $2^x-3z^y=5$ by appealing to lower bounds for linear forms in 2-adic logs to bound $x$ and $y$ and then using Frey curves ( a la Kraus) to knock off the remaining values of $y$ (likely one would have to do the cases where $2$ or $3$ divide $y$ separately via integral points on elliptic curve arguments or something similar). The most obvious couple of Frey curves give you congruences to forms of level $30$. The details shouldn't be too bad, but I'm not sure my claim follows explicitly from anything currently in the literature. $\endgroup$ Commented Sep 4, 2018 at 15:21