Is it possible to sum the first $n$ triangular numbers to equal a perfect cube? To state it another way, other than the obvious trivial case of n=1, are there any natural number solutions to the equation below? $$\sum_{i=1}^n T_i = \sum_{i=1}^n \frac{i(i+1)}{2} = \frac{n(n+1)(n+2)}{6} = m^3, n \neq m \in \mathbb{N}$$ If no such solution exists, how would I go about proving that it can't be done?
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OK, it does appear that my question has been asked, and answered, previously. Sorry I didn't find it when I first searched for it. – Brian J. Fink Jul 06 '19 at 16:34