Solve the following system of equations:

$$
\begin{gathered}
\log _{2}(x y+4)-1=\log _{3}\left(x^{2}+y^{2}+2\right) \\
y=x-1
\end{gathered}
$$
Let's denote the common value of the two sides of (1) briefly as $z$. Thus, the first term on the left side is $z+1$, and according to the definition of the logarithm,

$$
x y+4=2^{z+1}, \quad x^{2}+y^{2}+2=3^{z}
$$

After substituting (2), $x$ can be easily eliminated from the system:

$$
x^{2}-x=2^{z+1}-4, \quad 2 x^{2}-2 x=3^{z}-3
$$

Thus, the right side of the second equation is twice as large as that of the first:

$$
\begin{gathered}
3^{z}-3=2\left(2^{z+1}-4\right) \\
3^{z}=2^{z+2}-5
\end{gathered}
$$

Accordingly, $z$ is given by the abscissa of the common point of the curve $u=3^{z}$ and the curve $u=2^{z}$ shifted 2 units to the left and 5 units downward in the right-angled $z, u$ coordinate system (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_54f0d11b0c057e2c3c6cg-1.jpg?height=875&width=501&top_left_y=766&top_left_x=801)

We find two common points: above $z=1$ and $z=3$.

In the case of $z=1$, from (3) we have

$$
x^{2}-x=0
$$

and thus, considering (2) as well,

$$
x_{1}=0, \quad y_{1}=-1 ; \quad x_{2}=1, \quad y_{2}=0
$$

In the case of $z=3$, similarly,

$$
x_{3}=-3, \quad y_{3}=-4 ; \quad x_{4}=4, \quad y_{4}=3
$$

All four value pairs indeed satisfy the original system of equations.

Dezső Bogár (Tatabánya, Árpád Gymnasium, 3rd grade)

Remarks. 1. We obtain only formally different solutions if we stick to logarithms. Using (2), (1) takes the following form:

$$
\log _{2}\left(x^{2}-x+4\right)-1=\log _{3}\left[2\left(x^{2}-x\right)+3\right]
$$

Accordingly, we are looking for such values of the expression

$$
x^{2}-x=t
$$

for which the values of the functions

$$
\log _{2}(t+4)-1 \quad \text { and } \quad \log _{3}(2 t+3)
$$

are equal. The corresponding $x$ values can be easily obtained from (4), and $y$ from (2).

![](https://cdn.mathpix.com/cropped/2024_05_02_54f0d11b0c057e2c3c6cg-2.jpg?height=569&width=1195&top_left_y=125&top_left_x=454)

Plotting the two functions (Figure 2), we find common values at $t=0$ and $t=12$ (1 and 3, respectively), and the corresponding $x$ and $y$ values are:

| $t$ | 0 |  | 12 |  |
| :---: | :---: | :---: | :---: | :---: |
| $x$ | 0 | 1 | -3 | 4 |
| $y$ | -1 | 0 | -4 | 3 |

István Géczy (Debrecen, Fazekas M. Gymnasium, 3rd grade)

2. In neither version of the solution did we prove that the two curves have no more intersection points, as this requires a basis not provided by the school curriculum.