## Task 3 - 080923

Give all pairs $(x, y)$ of natural numbers for which $x^{3}-y^{3}=999$!
Apparently, every solution satisfies $x>y$. Therefore, there exists a positive $z \in \mathbb{N}$ with $x=y+z$. This leads to the equivalent equation $z\left(z^{2}+3 y z+3 y^{2}\right)=999$.

Thus, $z$ must be a divisor of $999=3^{3} \cdot 37$. Since $z\left(z^{2}+3 y z+3 y^{2}\right)$ is divisible by 3 if and only if $z$ is divisible by 3, and because $z \leq z^{2}+3 y z+3 y^{2}$, the only remaining possibilities are $\left(z, z^{2}+3 y z+3 y^{2}\right)=(3,333)$ and $\left(z, z^{2}+3 y z+3 y^{2}\right)=(9,111)$.

1. Case: $\left(z, z^{2}+3 y z+3 y^{2}\right)=(3,333)$

Then $9+9 z+9 y^{2}=333$, so $y^{2}+3 y=108$. The only natural solution to this equation is $y=9$. This leads to the solution $(x, y)=(12,9)$.

2. Case: $\left(z, z^{2}+3 y z+3 y^{2}\right)=(9,111)$

Then $81+27 y+3 y^{2}=111$, so $y^{2}+9 y=10$, thus $y=1$. Therefore, $(x, y)=(10,1)$ is a solution.

In total, there are exactly two solution pairs, namely $(12,9)$ and $(10,1)$.

Solution by Nuramon