$3^{4a+1} = 2b^2+1$ diophantine equation, $(a,b) \in \mathbb{N}^2$

Question

I would like to solve the following diophantine equation : $$3^{4a+1} = 2b^2+1$$

Yet I don't know how to proceed.

What I found so far is just that $$(a, b) = (0,1)\ ,\ (1,11)$$ seem to be the only solutions...

Any ideas ?

Answer 1

This question was formulated by me on the forum in Russian here-1 and here-2 in connection with the generalization of the problem of identifying non-standard objects - counterfeit coins (see here-3). The upper bound of the length of such an algorithm corresponds to the Hamming boundary for ternary codes and has the form:

$1+ 2C_n^1+2^2C_n^2+…+2^t C_n^t=3^m,$

where $t$ is the maximum number of positions in which errors are corrected by the code. For $t = 2$ this equality becomes

$1+2n^2=3^m.$

It is known that this equation has solutions: $(n, m) = (1,1), (2,2), (11,5)$, of which only one $n = 11, m = 5$ has a value in coding theory and problem [here-3]. In this case, for the parameters $n = 11, m = 5, t = 2$, a well-known perfect ternary Golay code (Wirtakallio-Golay code) and the perfect weighting algorithms, identifying up to 2 counterfeit coins ftom 11, are constructed. On the forum, an outstanding mathematician (Russia, nickname falcao) wrote "For the case t = 2, the solution of the equation $1 + 2n ^ 2 = 3 ^ m$ was described by Nagel in 1923. True, I did not find the text of the article or proofs of the evidence on the Web." Therefore, he gave his brilliant proof of the non-existence of other (different from the above) solutions of this equation.