6
$\begingroup$

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 $3$.

$\endgroup$
  • 1
    $\begingroup$ 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 '18 at 17:09
  • $\begingroup$ I extended the search upto $n=2\cdot 10^5$ without finding a further solution. $\endgroup$ – Peter Jun 20 '18 at 17:23
4
$\begingroup$

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".

$\endgroup$
2
$\begingroup$

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) $2^3-3^2=1.$

$\endgroup$

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.