Solve this Diophantine equations $(2^x-1)(3^y-1)=2z^2$

Asked Apr 01 '18 at 13:46

Active Apr 03 '18 at 15:36

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Find the positive integers $(2^x-1)(3^y-1)=2z^2$ have three solution $$(1,1,1),(1,2,2),(1,5,11)$$

I already know the solution of $(2^x-1)(3^y-1)=z^2$ has no solution.see:P.G.Walsh December 2006 On Diophantine equations of the form but there is a factor of 2 that seems complicated(I also searched a lot of these articles, and I didn't find any questions about the coefficient of 2,3,4 on the right.) and I didn't know anyone had studied this before. If so, please help me with the article or link,Thanks Before I have post this problem on MO

edited Apr 01 '18 at 14:56

asked Apr 01 '18 at 13:46

math110

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For $1\le x,y\le 10^4$ , no further solution exists. – Peter Apr 01 '18 at 14:00
It looks like the comments over on MO should guide you to a proof. – B. Mehta Apr 09 '18 at 01:48
This seems to be answered in the MO post. – Mason Oct 02 '19 at 19:14

Solve this Diophantine equations $(2^x-1)(3^y-1)=2z^2$

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