5. Find all real numbers $x$ and $y$ that satisfy the equation

$$
(x+y)^{2}=(x+3)(y-3)
$$

Solve the problem independently. You have 150 minutes for solving.

The use of notes, literature, or a pocket calculator is not allowed.

Mathematical Competition for High School Students in Slovenia

## Invitational Competition

March 28, 2003

## Problems for 4th Year
III/5. If we simplify the equation, we get $x^{2}+x(y+3)+\left(y^{2}-3 y+9\right)=0$. For this quadratic equation to have any real solution, its discriminant $D=(y+3)^{2}-4\left(y^{2}-3 y+9\right)=$ $-3(y-3)^{2}$ must be non-negative. Therefore, $y$ can only be $3$, which gives us $x=-3$. The original equation is satisfied only by the real numbers $x=-3$ and $y=3$.

Reformulating the equation $\mathbf{v}$ into a quadratic equation in $x$ (or $y$): 2 points. Writing the condition of non-negativity of the discriminant $-3(y-3) \geq 0$ (or equivalently $\mathbf{v} x$): $\mathbf{3}$ points. Stating the solutions $x=-3, y=3$: $1+1$ point.

If the contestant only states the solutions $x=-3, y=3$ and does NOT prove that they are the only ones: 2 points.