49th Mathematical Olympiad in Ukraine

Consider the equality: $n=8(s^6+t^6)$. We have a representation of number $n$ as a sum of cubes. For the representation of number $n$ as a sum of squares we rewrite last equality as $n=8(s^6+t^6)=2(2s^3{)}^2+2(2t^3{)}^2=(2s^3-2t^3{)}^2+(2s^3+2t^3{)}^2$.

Let us show that it is impossible to represent the number as a sum of sixth powers. Assume that $n=8(s^6+t^6)=x^6+y^6$, among all similar representations we choose that which has the minimal value of the sum $s+t+x+y$. We now prove that the numbers $s,t,x,y$ are even. Implying that $x$ and $y$ are odd, i.e. $x,y\equiv \pm 1(\mathrm{mod}4)$ then $x^6+y^6\equiv 2(\mathrm{mod}4)$ not divisible by 4, a contradiction. Therefore $x=2x_1,y=2y_1$ and we have the equality $8(s^6+t^6)=64(x_1^2+y_1^2)$. Similarly we prove that $s,t$ are even too, then $s=2s_1,t=2t_1$, and for the numbers $s_1,t_1,x_1,y_1$ holds the equality $8(s_1^6+t_1^6)=x_1^6+y_1^6$ but for all that $x_1+y_1+s_1+t_1=\frac{x+y+s+t}{2}$, which contradicts to the choice of numbers $s,t,x,y$ which have the minimal value of the sum $s+t+x+y$. It shows that such representations is impossible.