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Are there only two solutions for $$\sum_{k=0}^m3^k=2^n$$

Such as $3^0=2^0$ and $3^0+3^1=2^2$

Note

• If $m$ is even then $\sum_{k=0}^m3^k$ will be odd.

• $$\sum_{k=0}^m3^k=\sum_{k=0}^m\binom{m+1}{k+1}2^k=\sum_{k=0}^{m}\sum_{l=0}^{k}\binom{m+1}{k+1}\binom{k}{l}$$

Edit: generalization may be more interesting (motivation).

We can also ask as, Are there finitely many pairs of $(m, n)$ for any positive integer $(x, y)$ greater than $1$ such that $1+x+x^2+...+x^m=y^n$ Or $x^{m+1}-2\cdot y^{n}=1$

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    $\begingroup$ The general question of integer solutions to $1+x+x^2+\cdots+x^m=y^n$ has been the subject of much study and many papers. I don't have any details at my fingertips, but one place to look would be Richard Guy's book, Unsolved Problems In Number Theory, and another would be Mordell's book, Diophantine Equations. One solution is $(x,m,y,n)=(3,4,11,2)$. $\endgroup$
    – Gerry Myerson
    Commented Jun 12, 2021 at 23:53
  • $\begingroup$ @GerryMyerson Thanks for the suggestion. I'll be careful next time. $\endgroup$
    – Pruthviraj
    Commented Jun 13, 2021 at 2:02

1 Answer 1

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Yes, those are the only solutions. To see this note that $$1+3+9+27 \cdots 3^m = \frac{3^{m+1}-1}{2}.$$

So we are looking for solutions of $\frac{3^{m+1}-1}{2}=2^n$, or equivalently looking for solutions of $3^{m+1} -1 = 2^{n+1}$. But this equation has only the obvious solutions, a result which one can prove with a little modular arithmetic. (Historical note: That (1,2), (2,3), (3,4) and (8,9) are the only pairs of a consecutive power of 2 and a power of 3 was first proven by Gersonides in the 1300s. It is a fun exercise if you haven't seen it before.)

(Edit: Aside from the historical interest I'm not sure this problem should really be here. Maybe the question and answer should be moved over to Math.stackexchange.)

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    $\begingroup$ It can be hard to tell at sight whether such an equation is recreational or research-level. Occasionally even published papers that apply advanced methods to this kind of equation turn out to be replaceable with elementary arguments. There's probably little harm in keeping this Q&A in mathoverflow. $\endgroup$
    – Noam D. Elkies
    Commented Jun 12, 2021 at 17:56
  • $\begingroup$ How about the pair $(2,3)$? $\endgroup$
    – Zach Teitler
    Commented Jun 13, 2021 at 1:12
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    $\begingroup$ @ZachTeitler Fixed, thanks. $\endgroup$
    – JoshuaZ
    Commented Jun 13, 2021 at 1:13
  • $\begingroup$ elementary proofs for $ 3^s - 2^t = \pm 1 $ math.stackexchange.com/questions/3873689/… $\endgroup$
    – Will Jagy
    Commented Jun 13, 2021 at 2:24

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