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The charm of elliptic curves is that given one or two integer points, one can find others by the group law. However the easy to guess points from the title just pump me around trough a cyclic group of order six, so they are no help for the question whether there are any others.

My interest in this curve comes from the question if there are any triangular numbers that are also cubes, beyond the number 1. So we want to solve $\frac{m(m-1)}{2} = n^3$ and the substitution $y = 2m -1, \hspace{.1cm} x = 2n$ gives the above curve. This means that I am mostly interested in integer points $(x, y)$ for which $x$ is even and $y$ is odd, while both are positive; as expected there is only one of these among the easy to guess points (corresponding to $\frac{m(m-1)}{2} = n^3 = 1)$ but are there any others?

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  • $\begingroup$ Now that we are talking links, do you by any chance know of a place online where the result that Mordell curves have only finitely many points is being proved? So far I could only find proofs of much stronger theorems like Siegel's version talking about integral points on all sufficiently complicated curves. $\endgroup$ – Vincent Jul 24 '15 at 4:31
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    $\begingroup$ Fermat claimed to have proven that no triangular number can be a third, fourth, or fifth power. The general case (i.e., that no triangular number can be a perfect $n$th power for any $n > 2$) was solved recently, but required pretty heavy mathematical machinery. $\endgroup$ – Kieren MacMillan Mar 9 '16 at 3:35
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Those are the only integer points; in fact, they are the only rational points. For a lovely descent proof due to Euler, see Conrad’s paper.

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