the fermat catalan-conjecture is mentioned. In the german version, I read that in the case $$\frac{1}{m}+\frac{1}{n}+\frac{1}{k}=1$$ satisfied by the triples $(2/4/4)$ , $(2/3/6)$ , $(3/3/3)$ , the equation $$a^m+b^n=c^k$$ with coprime positive integers $a,b,c$ has finite many solutions. In the cases $(2/4/4)$ and $(3/3/3)$ no solution exists.

What about the cae $(2/3/6)$ ? Which solutions do exist in this case ?


This answer is heavily based on ABCs of Number theory by Noam Elkies. You can read the whole article at https://dash.harvard.edu/bitstream/handle/1/2793857/Elkies%20-%20ABCs%20of%20Number%20Theory.pdf?sequence=2 .

Your case of $(2, 3, 6)$ is equivalent to the rational points of Elliptic curve $Y^2=X^3 \pm 1$. They have rank $0$ (https://en.wikipedia.org/wiki/Rank_of_an_elliptic_curve), so there are finitely many rational points. One can check that the only nonzero rational point is $(X,Y)=(2,3)$, and the solutions of your equation set is trivial multiples of $2^3+1^6=3^2$.

  • $\begingroup$ Does that mean that we know that $(2/1/3)$ is the only solution in the case $(2/3/6)$ including permutations , or do we need the abc-conjecture ? $\endgroup$ – Peter Apr 22 '18 at 9:35
  • $\begingroup$ Yes, it is known that no other solutions (which cannot be deduced from $(1/2/3)$) exist. $\endgroup$ – didgogns Apr 22 '18 at 9:38
  • $\begingroup$ And if we permut (for example $a^2+b^6=c^3$) , there is no solution , right ? $\endgroup$ – Peter Apr 22 '18 at 9:39
  • $\begingroup$ Yes, that's right. $\endgroup$ – didgogns Apr 22 '18 at 9:53
  • $\begingroup$ So, with coprime integers $a,b,c$ and the exponents $2,3,6$ in any order, we only have the solution arising in the Catalan-conjecture. Thank you very much! $\endgroup$ – Peter Apr 22 '18 at 9:55

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