Determine if there are $2018$ different positive integers such that the sum of their squares is a perfect cube and the sum of their cubes is a perfect square.
Solution
Let $ c=2018$
Let $ A=\sum ^{c}_{j=1} j^{2}$ and $ $$ B=\sum ^{c}_{j=1} j^{3}$.
Let $ x_{k} =kA^{4} B^{3}$ for each $ k\in \mathbb{N}$.
Then $ \sum ^{c}_{k=1} x^{2}_{k} =A^{8} B^{6}\sum ^{c}_{k=1} k^{2} =\left( A^{3} B^{2}\right)^{3}$
and $ $$ \sum ^{c}_{k=1} x^{3}_{k} =A^{12} B^{9}\sum ^{c}_{k=1} k^{3} =\left( A^{6} B^{5}\right)^{2}$.
Therefore, there exist $2018$ different positive integers satisfying the condition of the problem.