2
$\begingroup$

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 $x=y=z=1$?

$\endgroup$
1
$\begingroup$

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 $x = y = z = 1$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user42768 Feb 2 '15 at 16:17
  • $\begingroup$ 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. $\endgroup$ – Marco Feb 2 '15 at 17:10
  • $\begingroup$ 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? $\endgroup$ – user42768 Feb 2 '15 at 17:34
  • $\begingroup$ No it is not, and it is better then doing the same thing twice :) $\endgroup$ – Marco Feb 2 '15 at 18:58
  • $\begingroup$ Ok. Thank you very much for your answer. $\endgroup$ – user42768 Feb 2 '15 at 19:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.