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 ?

  • 1
    $\begingroup$ I don't think $(a,b)=(9,10)$ is a solution. $\endgroup$ – Isaac Browne Jun 21 '17 at 17:31
  • $\begingroup$ Since there is no further solution for $a\le 10^5$, it is very unlikely that further solutions exist. $\endgroup$ – Peter Jun 21 '17 at 20:03

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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.