For all n there exists n distinct integers whose sum is a perfect square and whose product is a perfect cube

Question

Show that for all $n$ we can find $n$ distinct natural numbers whose sum is a perfect square and whose product is a perfect cube?

$$a_1 + ... + a_n = m^2 \quad\text{ and }\quad a_1a_2...a_n=k^3$$

Answer 1

For any $n\in\mathbb{N}$

$\sum _{k=1}^n k^3=\left(\frac{1}{2} n (n+1)\right)^2$

For instance $1+ 8+ 27+ 64+ 125+ 216+ 343+ 512=1296=34^2$

and $1\times 8\times 27\times 64\times 125\times 216\times 343\times 512=65\,548\,320\,768\,000=40320^3$