3. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& \log _{x}(y+z)=p \\
& \log _{y}(z+x)=p \\
& \log _{z}(x+y)=p
\end{aligned}
$$

with unknowns $x, y, z$ and a non-negative integer parameter $p$.
3. The equations of the given system make sense only if the numbers $x, y, z$ are positive and different from 1. For such numbers $x, y, z$ (we will not consider any others), we obtain an equivalent system of equations by exponentiation:

$$
y+z=x^{p}, \quad z+x=y^{p}, \quad x+y=z^{p} .
$$

We will first show that in the domain $\mathrm{M}=(0,1) \cup(1, \infty)$, each solution of the system (1) consists of a triplet of the same numbers. To do this, we use the well-known fact that for $p \geqq 0$, the function $f(t)=t^{p}$ is non-decreasing (more precisely: increasing for $p>0$ and constant for $p=0$) on the set of positive numbers $t$. Suppose, on the contrary, that for some solution $(x, y, z)$, for example, $x < y$. Then, from the first equation, we have $y+z < x^{p}$, and from the second equation, we have $z+x = y^{p}$. Since $x < y$, we have $x^{p} < y^{p}$, and thus $y+z < y^{p}$. This implies $y+z - y^{p} < 0$, which contradicts the fact that $y+z = x^{p} \geq y^{p}$. Similarly, we can derive a contradiction in the case when $x > y$, and in the cases when $x \neq z$ or $y \neq z$ (the system (1) is symmetric in the unknowns $x, y, z$).

Therefore, the system (1) reduces to the equation $x+x=x^{p}$, which we need to solve in the domain $\mathrm{M}=(0,1) \cup(1, \infty)$. Since $x \neq 0$, we get the equivalent equation $2=x^{p-1}$ by dividing by $x$. This equation has no solution for $p=1$, for $p=0$ it has the unique solution $x=\frac{1}{2}$, and for natural numbers $p \geqq 2$ it has the unique solution $x=2^{\frac{1}{p-1}}$, which can be written as $x=\sqrt[p-1]{2}$ for $p \geqq 3$. (The numbers $\frac{1}{2}$ and $2^{\frac{1}{p-1}}$ clearly belong to M.)

Answer: The given system has for $p=0$ the unique solution $x=y=z=\frac{1}{2}$, for $p=1$ no solution, and for natural numbers $p \geqq 2$ the unique solution $x=y=z=2^{\frac{1}{p-1}}$.

For a complete solution, award 6 points, of which 4 points for proving the statement that each solution of the given system satisfies the condition $x=y=z$. For solving the case $p=0$ only, award 1 point.