<s>[INST] System of equations: $3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y$ This was taken from a local mathematical olympiad in Romania. It is from the year 2003.
Solve for $x, y, z \in \mathbb{R}$:
$$
\left\{ 
\begin{array}{c}
3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y \\ 
3^y + 4^y + 5^y = 2^y \cdot 3^{y -1} \cdot z \\ 
3^z + 4^z + 5^z = 2^z \cdot 3^{z -1} \cdot x
\end{array}
\right. 
$$
So, the first thing I notice was $4^x = 2^{2x} = a$ so we get a system of quadratic equations. This proved to be quite hard to deal with, so I gave it up. I also tried dividing by $2^x \cdot 3^{x - 1} \cdot y$ and the others, but it was weird to deal with. Do you have any ideas?
 [/INST] We have:
$$\left( \frac{3}{6} \right)^x+\left( \frac{4}{6} \right)^x+\left( \frac{5}{6} \right)^x =\frac{y}{3} \tag{1}$$
$$\left( \frac{3}{6} \right)^y+\left( \frac{4}{6} \right)^y+\left( \frac{5}{6} \right)^y =\frac{z}{3}\tag{2}$$
$$\left( \frac{3}{6} \right)^z+\left( \frac{4}{6} \right)^z+\left( \frac{5}{6} \right)^z =\frac{x}{3}\tag{3}$$
Let's denote $f(t) =\left( \frac{3}{6} \right)^t+\left( \frac{4}{6} \right)^t+\left( \frac{5}{6} \right)^t  $, this function is decreasing in $t$ (you can prove it easily).
WLOG, suppose $x=\min{\{x,y,z\}}$. This implies that $x \le y$.
From (1) and (3), we have: $x \le y \implies f(x) \ge f(z) \implies x \le z$.
From (2) and (3), we have: $ x \le z \implies  f(z) \le f(y) \implies z \ge y$.
From (1) and (2), we have: $ y \le z \implies  f(x) \le f(y) \implies x \ge y$.
But we already suppose $x=\min{\{x,y,z\}}$ , hence, $x=y$. And from (1) and (2), we have $f(x) = f(y) \implies y = z$. So, $x=y=z$.
It suffices now to solve the equation
$$f(x)=\left( \frac{3}{6} \right)^x+\left( \frac{4}{6} \right)^x+\left( \frac{5}{6} \right)^x =\frac{x}{3}$$
We notice that $f(x)$ is decreasing function while $=\frac{x}{3}$ is an increasing function. Thus, there is at most 1 solution. And because $x=3$ is the solution, then $x=y=z=3$ is the unique solution of the system equation.
</s>