3. Determine for which real numbers $p$ the system of equations

$$
\begin{aligned}
& x^{2}+1=(p+1) x+p y-z, \\
& y^{2}+1=(p+1) y+p z-x, \\
& z^{2}+1=(p+1) z+p x-y
\end{aligned}
$$

with unknowns $x, y, z$ has exactly one solution in the set of real numbers.

The school - written part of the first round of category A takes place

## on Tuesday, December 4, 2001

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. These details will be communicated to the students before the start of the competition.
3. Let's notice that the equations of the given system differ from each other only by a cyclic permutation of the variables $x, y$, and $z$. Therefore, if the system has a solution $(x, y, z) = (x_0, y_0, z_0)$, then the triples $(x, y, z) = (y_0, z_0, x_0)$ and $(x, y, z) = (z_0, x_0, y_0)$ are also solutions. If the solution of the system (for a fixed $p$) is unique, the mentioned triples must be identical, so it must hold that $x_0 = y_0 = z_0$. The triple $(x_0, x_0, x_0)$ is clearly a solution of the given system if and only if the number $x = x_0$ is a solution of the equation $x^2 + 1 = 2px$. For each sought $p$, the last equation must have a unique solution, so its discriminant $D = 4p^2 - 4$ must be zero. This implies that necessarily $p = \pm 1$.

Now we will show that for $p = 1$, $x = y = z = 1$ is indeed the unique solution of the original system of three equations, and the same holds for $p = -1$ with the solution $x = y = z = -1$. By comparing the sum of the left-hand sides with the sum of the right-hand sides of the system, we find that any solution $(x, y, z)$ of the system also satisfies the equation

$$
x^2 + y^2 + z^2 + 3 = 2p(x + y + z),
$$

from which we can derive

$$
(x - p)^2 + (y - p)^2 + (z - p)^2 = 3(p^2 - 1).
$$

For both values $p = \pm 1$, it holds that $p^2 - 1 = 0$, so then the sum of the non-negative numbers $(x - p)^2, (y - p)^2$, and $(z - p)^2$ equals zero. This is possible only if $x = y = z = p$.

Answer: The sought values of $p$ are two: $p = 1$ and $p = -1$.

Alternative Solution. Similarly to the first solution, by adding the three given equations, we obtain equation (1). From this equation, it follows that if the system has at least one solution $(x, y, z)$ in the domain of real numbers for a given $p$, then the inequality $p^2 \geq 1$, or $|p| \geq 1$, must hold. However, if $|p| > 1$, we can easily write down two different solutions of the examined system, namely the triples $(x_1, x_1, x_1)$ and $(x_2, x_2, x_2)$, where $x_{1,2}$ are the roots of the equation $x^2 + 1 = 2px$ (its discriminant is positive due to the assumption $|p| > 1$). Therefore, we only need to evaluate the values $p = \pm 1$, for which, however, from equation (1) it immediately follows: if the original system has any solution at all, it is the triple $(x, y, z) = (p, p, p)$. A simple substitution check shows that this is indeed a solution (for $p = 1$ as well as for $p = -1$).

For a complete solution, 6 points are awarded, of which 2 points for excluding all values $|p| > 1$ and 2 points for verifying the numbers $p = \pm 1$. (For example, for justifying why $|p| = 1$ must necessarily hold, 4 points will be awarded.)