Italian Mathematical Olympiad - February Round

Solution:

The answer is $\mathbf{( E )}$. Let us start by observing that the following algebraic identity holds
$$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right)$$
Setting $x=2^{a}$, $y=2^{b}$, $z=2^{c}$, the condition $x+y+z \mid x^{3}+y^{3}+z^{3}$ is equivalent to $x+y+z \mid 3 x y z$; in particular, we want to look for the triples $(a, b, c)$ such that $2^{a}+2^{b}+2^{c} \mid 3 \cdot 2^{a+b+c}$.
If we assume $a \leq b \leq c$, simplifying a $2^{a}$ we obtain that $1+2^{b-a}+2^{c-a} \mid 3 \cdot 2^{b+c}$. If it were $b>a$, on the left we would have an odd number $\geq 5$, while on the right a number whose odd part is 3, which is impossible. Hence $b=a$ holds and we must find the solutions of $2+2^{c-a} \mid 3 \cdot 2^{c+a}$. At this point we split into four cases

- If $c=a$, $2+1 \mid 3 \cdot 2^{2 a}$ is always satisfied, so we have all triples of the form $(a, a, a)$ with $0 \leq a \leq 13$ (since $2^{13}<10000<2^{14}$ ), which are 14.

- If $c=a+1$, we can rewrite the condition as $2+2 \mid 3 \cdot 2^{2 a+1}$, which holds for $a \geq 1$ and from which we obtain the triples $(a, a, a+1)$ and their permutations; hence this case has in total $3 \cdot 12$ solutions (since we also need $a+1 \leq 13$)

- If $c=a+2$, the condition becomes $2+4 \mid 3 \cdot 2^{2 a+2}$, which holds for $a \geq 0$. We thus obtain the triples $(a, a, a+2)$ and their permutations, which are valid up to $a+2 \leq 13$, hence $3 \cdot 12$ solutions in this case as well.

- If $c \geq a+3$, letting $d=c-a-1$ we have $1+2^{d} \mid 3 \cdot 2^{2 a+d}$. Since $d \geq 2$, the number on the left is odd and $\geq 5$, so there are no solutions.

The total number of ordered triples is therefore $14+36+36=86$.