# System of exponential equations

If $x,y,z \in \mathbb{R}$ and $$\begin{cases} 2^x+3^y=5^z \\ 2^y+3^z=5^x \\ 2^z+3^x=5^y \end{cases}$$ does it imply that $x=y=z=1$?

I think so.

Let $f(x) = 5^x - 2^x - 3^x$, you have that $$x < 1 \Leftrightarrow f(x) < 0$$ $$x = 1 \Leftrightarrow f(x) = 0$$ $$x > 1 \Leftrightarrow f(x) > 0$$

Suppose that $x \geq y \geq z$:

If $x > 1$, then $5^x > 2^x + 3^x \geq 2^y + 3^z$, therefore $5^x \neq 2^x + 3^y$. This implies that $1 \geq x \geq y \geq z$.

If $z < 1$, then $5^z < 2^z + 3^z \leq 2^x + 3^y$, this implies that $1 \geq x \geq y \geq z \geq 1$ so $x = y = z = 1$.

By cyclicity of the equations you get the same result if $y \geq z \geq x$ and $z \geq x \geq y$.

Suppose $x \geq z \geq y$:

If $x > 1$ then $5^x > 2^x + 3^x \geq 2^y + 3^z$ therefore $1 \geq x \geq y \geq z$.

If $y < 1$ then $5^y < 2^y + 3^y \leq 2^z + 3^x$ therefore $1 \geq x \geq z \geq y \geq 1$ which implies that $x = y = z = 1$.

By cyclicity we have covered the other 3 orders, so the only solution should be $x = y = z = 1$.

• Thank you very much. I think it would be enough if we proved that max(x,y,z)<=1 and min(x,y,z)>=1 in the manner that you for the specific cases. – user42768 Feb 2 '15 at 16:17
• Since the equations are cyclic we can choose one variable to be the max or min and prove the two things? Thanks for the sharing. – Marco Feb 2 '15 at 17:10
• If we assume that (x,y,z) is a solution and max(x,y,z)>1, we have $$5^{(max(x,y,z))}>2^{(max(x,y,z))}+3^{(max(x,y,z))}.$$ Now on the right hand side we replace max(x,y,z) with the corresponding variables for each exponent (in a way that is similar to one of the equations), arriving at a contradiction. Then we apply the same technique for min(x,y,z). Is this wrong? – user42768 Feb 2 '15 at 17:34
• No it is not, and it is better then doing the same thing twice :) – Marco Feb 2 '15 at 18:58
• Ok. Thank you very much for your answer. – user42768 Feb 2 '15 at 19:12