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In the equation $3^x=2y^2-1$, $x$, $y$ are natural numbers. I found $x=1$ or $2$ (mod $4$), and $y^2=1$ or $4$ (mod $120$) but I even don't know if the number of solutions is infinite. Is there a way to find the solution of this indeterminate equation?

eandpiandi
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  • "there was a mistake. I fixed the equation sry" What happens to YiFan's answer below, now offtopic? Bad practice... – Did Dec 14 '18 at 07:38
  • @Did Having seen the original state of the equation, it was probably more a typographical or less important error. (That is to say, it probably doesn't affect YiFan's answer in any huge way.) – PrincessEev Dec 14 '18 at 07:41
  • It unfortunately does, because the argument given in YiFan's answer no longer works. There is a solution to $2y^2 \equiv -1 \pmod 3$ so reducing modulo 3 does not lead to a contradiction (in fact there is a solution $\pmod {3^k}$ for all $k$ by Hensel's lemma). – Tob Ernack Dec 14 '18 at 07:44
  • @EeveeTrainer It does, massively. – Did Dec 14 '18 at 07:47
  • Ah, I see... My bad then. – PrincessEev Dec 14 '18 at 07:49
  • Based on this Meta post (https://math.meta.stackexchange.com/questions/9017/typo-in-question-statment) it seems it would be more appropriate for the question to be reverted to its original state, then, for the question/answer to actually be appropriately related, and the appropriate question posted as a separate question. – PrincessEev Dec 14 '18 at 07:51
  • ok i reverted it to original state – eandpiandi Dec 14 '18 at 07:57
  • You can still post the original question in another post. I'm sure some of us will be glad to help. – YiFan Tey Dec 14 '18 at 08:33
  • The version of the question with $3^x = 2y^2 + 1$ has a few solutions: $(0, 0), (1, \pm 1), (2, \pm 2)$ and $(5, \pm 11)$. I was not able to prove that these are the only ones so far. – Tob Ernack Dec 14 '18 at 09:48

2 Answers2

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Take both sides of the equation modulo $3$. If $x\geq 1$ then $2y^2\equiv 1$ modulo $3$, but if $y\equiv 0$, then $2y^2\equiv 0$, and if $y\equiv\pm 1$, $2y^2\equiv 2(\pm1)^2\equiv2$. So there are no solutions in this case. We must have $x=0$, thus $y=1$.

YiFan Tey
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You can even solve this question using Mathematical Induction. However, answer is also correct and easy.

And its solution is x=0 and y=1.