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I want to find solutions of $3^n-1=2m^2$ other than $(n, m)=(0, 0)$, $(n, m)=(1, \pm1)$, $(n, m)=(2, \pm2)$ and $(n, m)=(5, \pm11)$.

There are no other "small" solutions ($n<1000$). For even $n$, we can let $3^{n/2}=k$ and solve Pell equation $k^2-2m^2=1$, and similarly for odd $n$, let $3^{(n-1)/2}=k$ and the equation becomes Pell-like equation $3k^2-2m^2=1$. The complete set of solution is known in both cases, but I cannot prove none of large $k$ can be power of $3$.

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    $\begingroup$ No further solutions upto $n=10^5$. It is possible that I read somewhere that there are no further solutions, but I am not sure and neither I have an idea how to prove it. $\endgroup$
    – Peter
    Jun 20, 2018 at 17:09
  • $\begingroup$ I extended the search upto $n=2\cdot 10^5$ without finding a further solution. $\endgroup$
    – Peter
    Jun 20, 2018 at 17:23

2 Answers 2

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According to Dickson's "History of the Theory of Numbers" vol. 2, pg. 694,

"E. Fauquembergue [158] proved that $1+3+3^2+ · · · +3^n=y^2$ only when n=0, 1, 4, by using the powers of $a+b\sqrt{-2}$ to treat $3^{n+1}=1+2y^2$."

The reference is to "Mathesis, (2), 4, 1894, 169-170".

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This is only to say one can disregard the case of $n \ge 4$ even, using Catalan's conjecture/theorem.

For even $n=2k\ge 4,$ if $3^n-1-2m^2$ then $(3^k-1)(3^k+1)=2m^2,$ where the factors on the left have gcd $2.$ This means one factor is $2a^2$ and the other is $b^2,$ where (unneeded) $\gcd(a.b)=1.$ So we get either $3^k-1=b^2$ or $3^k+1=b^2,$ where now $k \ge 2.$

We have then a difference of nontrivial powers equal to $1.$ Now apply Catalan's conjecture (theorem). We'll have either $3^k-b^2=1$ or $b^2-3^k=1,$ and neither fits the only solution (Catalan) $2^3-3^2=1.$

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