Solve (in real numbers) the system of equations

$$
x-y=2, \quad x y+z^{2}=-1
$$
I. solution: We have two equations to determine three unknowns, which is usually not sufficient. However, the requirement that we can only accept real numbers in the root system (in accordance with high school knowledge) makes the solution unique.

From the first equation, substituting $x=y+2-\mathrm{t}$ into the second and reducing to 0, recognizing the complete square terms,

$$
y^{2}+2 y+1+z^{2}=(y+1)^{2}+z^{2}=0
$$

The sum of two squares in the real number system can only be 0 if each is 0, meaning their bases are also 0. Thus, $y=-1, z=0$, and $x=1$. The triplet satisfies the given system.

László Simai (Kisújszállás, Móricz Zs. g. III. o. t.)

Remark. We can also reach the goal without recognizing the complete square, if we calculate the discriminant of (1) as a quadratic equation in $y$: $D=-4 z^{2}$. This is non-negative only if $z=0$.

2. By transforming the system as follows: $x+(-y)=2, x(-y)=z^{2}+1$, it is easy to see that $x$ and $-y$ are the two roots of the quadratic equation $t^{2}-2 t+\left(z^{2}+1\right)=0$. From this, starting with the conclusion $z=0$, the above solution is derived.

II. solution: The left side of the second equation decreases or remains unchanged by omitting the non-negative $z^{2}$: $x y \leqq-1$. Therefore, $x$ and $y$ have opposite signs. Specifically, $x$ is positive and $y$ is negative, because $x$ is larger, as the difference between them is positive according to the first equation. Let the absolute value of $y$ be $y_{1}$, i.e., $-y=y_{1}$. Then

$$
x+y_{1}=2, \quad x y_{1} \geqq 1
$$

Thus, the arithmetic mean of the positive numbers $x$ and $y_{1}$ is 1, so its square is also 1, and the square of their geometric mean is at least 1. However, according to the well-known inequality for the arithmetic and geometric means (or their squares) of two non-negative numbers, the latter cannot be greater, so $x y_{1}=1$, and in this case, they are equal: $x=y_{1}=-y=1, y=-1$ and from the second equation, $z=0$.

Péter Vámos (Budapest, Than K. vegyip. t. IV. o. t.)

Remarks. 1. The conditions $x>0, y<0, x y \leqq-1$, i.e., $y \leqq-1 / x$ in the Cartesian coordinate system are satisfied by the points on and inside the fourth quadrant branch of the hyperbola $y=-1 / x$ (i.e., any interior point of a chord of the branch), and the first equation by the points on the line $y=x-2$.

![](https://cdn.mathpix.com/cropped/2024_05_02_690e678f7c929afa392bg-1.jpg?height=360&width=374&top_left_y=1430&top_left_x=865)

Intuitively, the line is tangent to the hyperbola, with a single common point: $(1 ;-1)$.

István Sólyom (Budapest, Vörösmarty M. g. III. o. t.)

2. The observation in the previous remark, of course, cannot be considered a solution to the problem, but only a visual representation of the (calculated) solution.