Selected Problems from Open Contests

All the numbers described in the problem are divisible by $3$ (as their sum of digits is divisible by $3$). Note that all perfect squares leave the remainder $0$ or $1$ when divided by $3$, and therefore, for the sum of the two perfect squares to be divisible by $3$, they both have to be divisible by $3$. Now, as the numbers are both divisible by $3$, their squares are divisible by $9$ and thus the sum of the squares is divisible by $9$. However, the number described in the problem is not divisible by $9$, a contradiction.

Note that when a cube number is divided by $3$, it will leave the remainder of either $0$, $1$ or $-1$. Indeed, $(3k)^3 = 9(3k^3)$ and $(3k \pm 1)^3 = 27k^3 \pm 27k^2 + 9k \pm 1 = 9(3k^3 \pm 3k^2 + k) \pm 1$. The numbers described in the problem give the remainder $3$ when divided by $9$ (as their sum of digits gives the remainder $3$ when divided by $9$), therefore we conclude that it is impossible to express them as sums of two cubes.