<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] Observe that your system is equivalent to $$\begin{cases}f(x)=y\\ f(y)=z\\f(z)=x\end{cases}$$ Where $$f(x):=3\cdot \left[\left(\frac{3}{6}\right)^x+\left(\frac{4}{6}\right)^x+\left(\frac{5}{6}\right)^x\right]$$ Let $g(x)=f(x)-x$. Adding up the three equations yields $\sum g(x)=0$, so at least one of $g(x), g(y), g(z)$ has to be $\leqslant 0$. It's easy to see that $g(x)$ is strictly decreasing and has its only root at $x=3$.
Case 1. $g(x)\leqslant 0\iff x\geqslant 3$. Since $f(x)$ is also strictly decreasing, this implies that $y=f(x)\leqslant f(3)=3$. Thus $y\leqslant 3\implies z=f(y)\geqslant f(3)=3$. Hence, $z\geqslant 3\implies x=f(z)\leqslant f(3)=3$. So $x\leqslant 3$. But this is absurd unless $x=y=z=3$.
Cases 2 and 3 are analogous.
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