Problem 1. Find all pairs $(x, y)$ of integers that satisfy the equation

$$
x^{2} y+y^{2}=x^{3} \text {. }
$$

(Daniel Paleka)
First Solution. Firstly, let us consider the case $x=0$. In that case, it is trivial to see that $y=0$, and thus we get the solution $(0,0)$. From now on, we can assume $x \neq 0$.

From the trivial $x^{2} \mid x^{3}$, the equation gives $x^{2}\left|x^{2} y+y^{2} \Rightarrow x^{2}\right| y^{2}$, which means $x \mid y$.

1 point.

We use the substitution $y=k x$, where $k \in \mathbb{Z}$.

1 point.

The substitution gives us

$$
\begin{gathered}
k x^{3}+k^{2} x^{2}=x^{3} \\
k x+k^{2}=x \\
k^{2}=x(1-k)
\end{gathered}
$$

2 points.

Considering the greatest common divisor of $k^{2}$ and $1-k$, we get

$$
G C D\left(k^{2}, 1-k\right)=G C D\left(k^{2}+k(1-k), 1-k\right)=G C D(k, 1-k)=G C D(k, 1-k+k)=G C D(k, 1)=1
$$

3 points.

That leaves us with two possibilities.
a) $1-k=1 \Rightarrow k=0 \Rightarrow x=0$ which is not possible since $x \neq 0$.

1 point.
b) $1-k=-1 \Rightarrow k=2 \Rightarrow x=-4, y=-8$, which gives a solution to the original equation.

1 point.

Second Solution. We rearrange the equation into:

$$
y^{2}=x^{2}(x-y)
$$

It can easily be shown that if $y \neq 0, x-y$ must be square.

1 point.

If $y=0$, from the starting equation we infer $x=0$, and we have a solution $(x, y)=(0,0)$.

In the other case, we set $x=y+a^{2}$, where $a$ is a positive integer. Taking the square root of the equation gives:

$$
|y|=|x| a
$$

1 point.

Because $x=y+a^{2}>y$, it is impossible for $y$ to be a positive integer, because then the equation would say $y=x a>x$, which is false. That means $y<0$, and also:

$$
-y=|x| a
$$

2 points.

If $x$ is positive, we can write:

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

which rearranges into

$$
-y(a+1)=a^{3}
$$

so $a^{3}$ is divisible by $a+1$, which is not possible for positive $a$ due to $a^{3}=(a+1)\left(a^{2}-a+1\right)-1$.

2 points.

We see that $x$ cannot be zero due to $y$ being negative, so the only remaining option is that $x<0$ also. We write:

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

which can similarly be rearranged into

$$
-y(a-1)=a^{3}
$$

and this time $a^{3}$ is divisible by $a-1$.

1 point.

Analogously, we decompose $a^{3}=(a-1)\left(a^{2}+a+1\right)+1$, so $a-1$ divides 1 and the unique possibility is $a=2$.

2 points.

The choice $a=2$ gives $y=-8$ and $x=-4$, which is a solution to the original equation.

1 point.

## Notes on marking:

- Points awarded for different solutions are not additive, a student should be awarded the maximal number of points he is awarded following only one of the schemes.
- Saying that $(0,0)$ is a solution is worth $\mathbf{0}$ points. The point is awarded only if the student argues that, disregarding the solution $(0,0)$, we must only consider $x \neq 0$, or a similar proposition.
- Failing to check that $(0,0)$ is a solution shall not be punished. Failing to check that $(-4,-8)$ is a solution shall result in the deduction of 1 point only if a student did not use a chain of equivalences to discover the solution.