System of equations with $3^x+4^x+5^x$

Question

Solve over reals:

\begin{align} 3^x + 4^x + 5^x &= 2^{x} 3^{x-1} y \\ 3^y + 4^y + 5^y &= 2^{y} 3^{y-1} z \\ 3^z + 4^z + 5^z &= 2^{z} 3^{z-1} x \end{align}

I searched for it, but can't find anything similar. I don't really know how to begin. Maybe divide by $2^{x}3^{x-1}$ and then what? Can somebody give me a starting idea? Thank you.

Answer 1

Yes, your idea works.

Indeed, rewrite our system in the following form: $$\frac{3(3^x+4^x+5^x)}{6^x}=y,$$ $$\frac{3(3^y+4^y+5^y)}{6^y}=z$$ and $$\frac{3(3^z+4^z+5^z)}{6^z}=x.$$ Now, let $y>3$.

Thus, $$\frac{3(3^x+4^x+5^x)}{6^x}>3$$ or $$\left(\frac{3}{6}\right)^x+\left(\frac{4}{6}\right)^x+\left(\frac{5}{6}\right)^x>1$$ and since $f(x)=\left(\frac{3}{6}\right)^x+\left(\frac{4}{6}\right)^x+\left(\frac{5}{6}\right)^x$ decreases and $\left(\frac{3}{6}\right)^3+\left(\frac{4}{6}\right)^3+\left(\frac{5}{6}\right)^3=1,$ we obtain $$x<3,$$ which gives $$\left(\frac{3}{6}\right)^z+\left(\frac{4}{6}\right)^z+\left(\frac{5}{6}\right)^z<1$$ and $$z>3,$$ which gives $$\left(\frac{3}{6}\right)^y+\left(\frac{4}{6}\right)^y+\left(\frac{5}{6}\right)^y>1$$ and $$y<3,$$ which is a contradiction.

Similarly, we'll get a contradiction for $y<3.$

But, for $y=3$ from the first equation we obtain $x=3$ and from the third we'll get $z=3,$ which gives the answer: $$\{(3,3,3)\}.$$

Answer 2

One of the answers is $x = 3 , y = 3 , z = 3$

Answer 3

Define $$ f(x) = \frac{3^x+4^x+5^x}{2^x3^{x-1}} $$ The system is: $f(x) = y, f(y) = z, f(z)= x$.

The function $f(x)$ decreasing, so the function $f(f(f(x)))$ is decreasing. Thus, there is at most one solution for $f(f(f(x))) = x$. As Ilya noted, $3$ is that solution.