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

Find the positive integers $(2^x-1)(3^y-1)=2z^2$ have three solutionsolutions $$(1,1,1),(1,2,2),(1,5,11)$$I already know the solution of $(2^x-1)(3^y-1)=z^2$ has no solution.see See:P P.G.Walsh December 2006 [On Diophantine equations of the form] paper but there is a factor of 2$2$ that seems complicated, and I didn't know anyone had studied this before. If so, please help me with the article or link,Thanks thanks.

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

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] paper 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

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

Find the positive integers $(2^x-1)(3^y-1)=2z^2$ have three solutions $$(1,1,1),(1,2,2),(1,5,11)$$I already know $(2^x-1)(3^y-1)=z^2$ has no solution. See: P.G.Walsh December 2006 [On Diophantine equations of the form] paper 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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math110
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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] paper 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

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] paper 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

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] paper 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

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] paper 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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Martin Sleziak
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sove Solve this diophantineDiophantine equations $(2^x-1)(3^y-1)=2z^2$

Find the postivepositive 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 diophantineDiophantine equations of the form] paper 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

sove this diophantine equations $(2^x-1)(3^y-1)=2z^2$

Find the postive 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] paper 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

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

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] paper 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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math110
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