We know that $\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$. Interestingly, $1^3+2^3+2^3+4^3=(1+2+2+4)^2$. Are there other non-consecutive numbers $a_1, a_2, \ldots, a_k$ such that $$\sum_{k=1}^n a_k^3 = \left(\sum_{k=1}^n a_k \right)^2?$$
$\sum_{k=1}^n a_k^3 = \left(\sum_{k=1}^n a_k \right)^2$
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algebra-precalculus
summation
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0There is 2,2,4,4 and the trivial $a_k=n$ for all $k$. – 2012-10-19