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.


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




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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 $x^{m+1}-2\cdot y^{n}=1$



Are there only two solutions for $1+3+9+...+3^m=2^n$
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.)


