## Task 10/90

Which triples $(x ; y ; z)$ of prime numbers satisfy the equation $x^{3}-y^{3}-z^{3}=6 y(y+2)$?
Since the right side of the equation is certainly positive, the left side must also be positive. Therefore, $x>y ; z \geq 2$. Since the right side of the equation is certainly even, the left side must also be even. Therefore, either $y=2, z \geq 3$ or $y \geq 3, z=2$. Let's first consider $y=2, z \geq 3$. Then the equation, after rearrangement, takes the form

$$
x^{3}-z^{3}=(x-z)\left(x^{2}+x z+z^{2}\right)=56=7 \cdot 8
$$

From the uniqueness of prime factorization, it follows that either

$$
x^{2}+x z+z^{2}=7, \quad x-z=8 \quad \text { or } \quad x^{2}+x z+z^{2}=8, \quad x-z=7
$$

However, these systems of equations have no rational solutions, let alone prime numbers $x ; z$ as solutions. Therefore, $y \geq 3, z=2$. This makes the equation take the form

$$
x^{3}=y^{3}+6 y^{2}+12 y+8=y^{3}+3 y^{2} \cdot 2+3 y \cdot 2^{2}+2^{3}=(y+2)^{3}
$$

From this, it immediately follows that $x=y+2$. Therefore, all triples $(x ; y=x-2 ; z=2)$ are solutions if $x$ and $y=x-2$ are twin primes (with $y$ being the smaller of the two primes).