## Task 5 - 341245

Determine all pairs $(x ; y)$ of non-negative integers $x, y$ for which the following holds:

$$
x^{3}+8 x^{2}-6 x+8=y^{3}
$$
Due to

$$
8 x^{2}-6 x+8=8\left(x-\frac{3}{8}\right)^{2}+\frac{55}{8}
$$

it follows initially that

$$
y^{3} \geq x^{3}+\frac{55}{8}
$$

which, in conjunction with $x, y \in \mathbb{N}$ and the fact that due to

$$
y^{3}-x^{3}=(y-x)\left(y^{2}+x y+x^{2}\right)=8 x^{2}-6 x+8
$$

$x$ and $y$ must have the same parity, immediately implies

$$
y \geq x+2
$$

However, the equality sign must hold here, because if $y=x+k$ for some $k \geq 3$, then due to

$$
\forall x \in \mathbb{N}: y^{3}=x^{3}+3 k x^{2}+3 k^{2}+k^{3}>x^{3}+8 x^{2}-6 x+8
$$

equality could no longer occur, as one can see immediately by comparing the coefficients of the two polynomial functions in $x$.

The rest is simple, as using the substitution $y=x+2$ transforms the original equation into a simple equation in $x$ alone, namely

$$
(x+2)^{3}-x^{3}-8 x^{2}+6 x-8=2 x(9-x)=0
$$

from which we can immediately read off the only two solutions

$$
(x, y) \in\{(0,2),(9,11)\}
$$

of the problem.