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Find all positive integers $a, b$ such that $3^a = 2b^2 + 1$.

This means that $\displaystyle\frac {3^a − 1} 2$ is a perfect square. Then checking for a few small values I get a few solutions as: $(a, b) \in \{(0, 0), (1, 1), (2, 2), (5, 11)\}$

How to proceed to find out all the possible solutions? Please help.

High Regards, Shamik Banerjee

player3236
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Shamik Banerjee
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    See https://math.stackexchange.com/questions/124565/constructive-proof-need-to-know-the-solutions-of-the-equations or https://math.stackexchange.com/questions/1612956/find-all-natural-numbers-x-y-such-that-3x-2y21 or https://math.stackexchange.com/questions/873147/finding-non-negative-integers-m-such-that-1-sqrt-2m-has-real-part or the 18 times the question has been asked on art of problem solving website. Also, https://math.stackexchange.com/questions/2331436/34a1-2b21-diophantine-equation-a-b-in-mathbbn2 – Gerry Myerson Nov 14 '20 at 11:46
  • Thank you very much. – Shamik Banerjee Nov 15 '20 at 13:15

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