2004 Romanian Mathematicial Olympiad

By try and error, we can get small answers which are $0,1,2$ . (When $n=2$ , $a=b=2$ .) Now, let us use quadratic equation to show that these are the only answers.

$\begin{aligned} n^3&=a^2+b^2 \\ &=(a+b)^2-2ab \\ ab&=\frac{n^4-n^3}{2} \\ a+b&=n^2 \end{aligned}$

We can get that $a, b$ are the roots of $x^2-n^2x+\frac{n^4-n^3}{2}=0$ . The discriminant of quadratic eqn above must bigger or equal to $0$ . Hence $\begin{aligned} n^4-2(n^4-n^3)\geq 0 \\ n^3(n-2) \leq 0 \\ 0\leq n \leq 2 \end{aligned}$ Hence, the required sum is $\large 0+1+2=3$ .