Find the positive integers $(2^x1)(3^y1)=2z^2$ have three solution $$(1,1,1),(1,2,2),(1,5,11)$$I already know the solution of $(2^x1)(3^y1)=z^2$ 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, and I didn't know anyone had studied this before. If so, please help me with the article or link,Thanks
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1$\begingroup$ Can you show that x has to be one? Gerhard "Try It. It's Not Hard" Paseman, 2018.03.31. $\endgroup$ – Gerhard Paseman Apr 1 '18 at 5:33

1$\begingroup$ And after you show that $x=1$, as Gerhard suggests, you can take the three cases $y=3w$, $y=3w+1$, and $y=3w+2$, which reduces the problem to finding the integer points on three fairly standard elliptic curves. $\endgroup$ – Joe Silverman Apr 1 '18 at 12:26

1$\begingroup$ Actually, the argument in the linkedto reference goes through essentially line by line, with nary an elliptic curve. $\endgroup$ – Igor Rivin Apr 1 '18 at 18:14

6$\begingroup$ The Diophantine equation $2z^2+1 = 3^y$ that appears for $x=1$ has already been studied: $2z^2+1 = 4{z \choose 2} + 2z + 1$ is the volume of a radius2 ball in ternary Hamming space ${\bf F}_3^z$, so $2z^2+1 = 3^y$ is a necessary condition for the existence of a perfect 2errorcorrecting ternary code in that space. It is known that $z=1,2,11$ are the only solutions, each corresponding to a code (trivial codes for $z=1$ and $2$, and the ternary Golay code for $z=11$). $\endgroup$ – Noam D. Elkies Apr 11 '18 at 2:49

1$\begingroup$ Following the hint of @NoamD.Elkies , I found a 1958 paper of CY Lee "Some properties of nonbinary errorcorrecting codes" ieeexplore.ieee.org/document/1057446 which attributes the solution of $2z^2+1=3^y$ to T. Nagell. Since it is hard to find, I have uploaded a copy of Nagell's 1923 paper "Sur l'impossibilité de quelques equations à deux indéterminées" here dropbox.com/s/69iuu00ux30dclu/nagell1923.pdf?dl=0 (this dropbox link is not really permanent, so rehosting would be appreciated). Perhaps someone more expert than me could write an answer? $\endgroup$ – j.c. Apr 16 '18 at 19:49