Problem 1318

III/4. 1. method. Let $x+y=z$, so $x=z-y$. Then $y=z(2z-2y+3y) = z(2z+y)$, from which we can express

$$y=-\frac{2z^2}{z-1}=-\frac{2z^2-2+2}{z-1}=-2(z+1)-\frac{2}{z-1}$$

Since $y$ is an integer, $z-1$ must divide 2, so $z-1$ is equal to $2, 1, -1$ or -2. We get in turn $z=3, z=2, z=0$ and $z=-1$, from which we can calculate that the pairs $(x, y)$ are equal to $(12,-9), (10,-8), (0,0)$ and $(-2,1)$.

Introduction of $x+y=z$

1 point

Writing $y=-\frac{2z^2}{z-1}$ or equivalent 1 point

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(If the contestant writes pairs $(x, y)$ that satisfy the equation but does not justify that there are no other pairs, award a maximum of 4 points.)

2. method. We observe that the number $x+y$ must divide $y$. Therefore, we can write $y=k(x+y)$ for some integer $k$. If $k=0$, it follows that $y=0$ and from the original equation also $x=0$. Otherwise, we can express $x=\frac{y}{k}-y$ and substitute into the equation. We get

$$y=\frac{y}{k}\left(\frac{2y}{k}-2y+3y\right)=\frac{y^2}{k}\left(1+\frac{2}{k}\right)$$

The case $y=0$ has already been considered, so let $y \neq 0$. Then we can divide both sides by $y$ and express

$$y=\frac{k^2}{k+2}=\frac{k^2+2k-2k}{k+2}=\frac{k(k+2)-2k-4+4}{k+2}=k+\frac{-2(k+2)+4}{k+2}=k-2+\frac{4}{k+2}$$

From this, it follows that $k+2$ is a divisor of 4. We consider six cases, as $k+2$ can be $\pm 1, \pm 2$ and $\pm 4$. For $k+2=1$, we get $y=1$ and $x=-2$, for $k+2=2$ we get $k=0$, which we have already considered separately. If $k+2=4$, then $y=3$ and $x=-\frac{3}{2}$, which is not an integer.

The case $k+2=-1$ gives $y=-9$ and $x=12$, and $k+2=-2$ gives $y=-8$ and $x=10$. The remaining case is $k+2=-4$, from which we get $y=-9$ and $x=\frac{21}{2}$, which is also not an integer.

Observation that $x+y$ divides $y$

1 point

Writing $y=\frac{k^2}{k+2}$ or equivalent 1 point

Conclusion that $k+2$ divides 4

1 point

Solutions $(x, y) \in \{(12,-9),(10,-8),(0,0),(-2,1)\}$

1 point

(If the contestant writes pairs $(x, y)$ that satisfy the equation but does not justify that there are no other pairs, award a maximum of 4 points.)

3. method. The equation can be rewritten as

$$2x^2 + 5xy + y(3y-1) = 0$$

For the quadratic equation in $x$ to have an integer solution, the discriminant must be a perfect square. Thus, $D=y^2+8y=a^2$, from which it follows that $(y+4)^2-16=a^2$ or $(y+4-a)(y+4+a)=16$. The numbers $y+4-a$ and $y+4+a$ are of the same parity, so both are even. We can assume that the number $a$ is non-negative, so $y+4+a \geq y+4-a$. We consider the following cases.

If $y+4-a=2$ and $y+4+a=8$, then $y=1, a=3$, and the quadratic equation has one integer root $x=-2$. If $y+4-a=4=y+4+a$, then $a=y=0$ and $x=0$. From $y+4-a=-4=y+4+a$ we get $a=0, y=-8$ and $x=10$. The remaining case is $y+4-a=-8$, $y+4+a=-2$, where $y=-9, a=3$ and $x=12$.

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(If the contestant writes pairs $(x, y)$ that satisfy the equation but does not justify that there are no other pairs, award a maximum of 4 points.)