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?
2026-03-28 18:17:14.1774721834
Is there a tetrahedral number that is also a perfect cube? How would I prove or disprove?
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