[Number Theory]
Problem: Find all triples $(x, y, z)$ of positive integers such that $x \leq y \leq z$ and

$$
x^{3}\left(y^{3}+z^{3}\right)=2012(x y z+2) \text {. }
$$

Solution: First note that $x$ divides $2012 \cdot 2=2^{3} \cdot 503$. If $503 \mid x$ then the right-hand side of the equation is divisible by $503^{3}$, and it follows that $503^{2} \mid x y z+2$. This is false as $503 \mid x$. Hence $x=2^{m}$ with $m \in\{0,1,2,3\}$. If $m \geq 2$ then $2^{6} \mid 2012(x y z+2)$. However the highest powers of 2 dividing 2012 and $x y z+2=2^{m} y z+2$ are $2^{2}$ and $2^{1}$ respectively. So $x=1$ or $x=2$, yielding the two equations

$$
y^{3}+z^{3}=2012(y z+2), \quad \text { and } \quad y^{3}+z^{3}=503(y z+1)
$$

In both cases the prime $503=3 \cdot 167+2$ divides $y^{3}+z^{3}$. We claim that $503 \mid y+z$. This is clear if $503 \mid y$, so let $503 \nmid y$ and $503 \nmid z$. Then $y^{502} \equiv z^{502}(\bmod 503)$ by FERMAT's little theorem. On the other hand $y^{3} \equiv-z^{3}(\bmod 503)$ implies $y^{3 \cdot 167} \equiv-z^{3 \cdot 167}(\bmod 503)$, i. e. $y^{501} \equiv-z^{501}(\bmod 503)$. It follows that $y \equiv-z(\bmod 503)$ as claimed.

Therefore $y+z=503 k$ with $k \geq 1$. In view of $y^{3}+z^{3}=(y+z)\left((y-z)^{2}+y z\right)$ the two equations take the form

$$
k(y-z)^{2}+(k-4) y z=8 \tag{1}
$$
$$
k(y-z)^{2}+(k-1) y z=1 \tag{2}
$$

In (1) we have $(k-4) y z \leq 8$, which implies $k \leq 4$. Indeed if $k>4$ then $1 \leq(k-4) y z \leq 8$, so that $y \leq 8$ and $z \leq 8$. This is impossible as $y+z=503 k \geq 503$. Note next that $y^{3}+z^{3}$ is even in the first equation. Hence $y+z=503 k$ is even too, meaning that $k$ is even. Thus $k=2$ or $k=4$. Clearly (1) has no integer solutions for $k=4$. If $k=2$ then (1) takes the form $(y+z)^{2}-5 y z=4$. Since $y+z=503 k=503 \cdot 2$, this leads to $5 y z=503^{2} \cdot 2^{2}-4$. However $503^{2} \cdot 2^{2}-4$ is not a multiple of 5 . Therefore (1) has no integer solutions.

Equation (2) implies $0 \leq(k-1) y z \leq 1$, so that $k=1$ or $k=2$. Also $0 \leq k(y-z)^{2} \leq 1$, hence $k=2$ only if $y=z$. However then $y=z=1$, which is false in view of $y+z \geq 503$. Therefore $k=1$ and (2) takes the form $(y-z)^{2}=1$, yielding $z-y=|y-z|=1$. Combined with $k=1$ and $y+z=503 k$, this leads to $y=251, z=252$.

In summary the triple $(2,251,252)$ is the only solution.

Final answer: $(2,251,252)$