A short way and a short way

Algebra Level 5

$\large{ \begin{cases} z^x = x \\ z^y = y \\ y^y = x \end{cases} }$

Find all triplets of positive numbers $(x,y,z)$ satisfying the system of equations above. Submit your answer as $\displaystyle \sum (x_k^2 + y_k^2 + z_k^2 )$ .

The answer is 25.

1 solution

Aditya Dhawan
Feb 21, 2016

${ z }^{ x }=x\Rightarrow z={ x }^{ \frac { 1 }{ x } }(1)\\ { z }^{ y }=y\Rightarrow z={ y }^{ \frac { 1 }{ y } }(2)\\ \therefore \quad { y }^{ \frac { 1 }{ y } }={ x }^{ \frac { 1 }{ x } }\Rightarrow x={ y }^{ \frac { x }{ y } }(3)\\ But\quad x={ y }^{ y }\quad (4)\\ Thus\quad x={ y }^{ 2 }\quad (5)\\ Subsituting\quad in\quad (3)\\ We\quad get\quad y=2\quad and\quad y=1\\ Thus\quad for\quad y=2,\quad z=\sqrt { 2 } and\quad x=4\\ For\quad y=1,\quad x=y=z=1\\ Thus\quad required\quad answer=\\ { 1 }^{ 2 }+{ 1 }^{ 2 }{ +1 }^{ 2 }{ +\sqrt { 2 } }^{ 2 }+{ 2 }^{ 2 }{ +4 }^{ 2 }=\boxed { 25 }$

Isn't (-1, -1, -1) also a solution?

Kush Singhal - 5 years, 3 months ago

Note that the problem only admits positive values.

Aditya Dhawan - 5 years, 3 months ago

Nice solution, @Aditya Dhawan

Mehul Arora - 5 years, 3 months ago

The answer should be 3. Note that the first equation is $z^x=y$ and not $z^x=x$ as mentioned in the solution. I guess there has been a typo in the problem statement.

Nitish Joshi - 5 years, 3 months ago

I think the problem has been edited. I am certain that the original version had the first equation as $z^x=x$ and not $z^x=y$ . The latter has real solutions as (1,1,1) and (-1,-1,-1) out of which only the first one is admissible. Thus according the question in its current form, the answer should indeed be 3.

Aditya Dhawan - 5 years, 3 months ago

Dear sir, look, please, at the statement of the problem. The solution you present corresponds to another one. I am sorry.

Сергей Кротов - 5 years, 3 months ago

A short way and a short way

The answer is 25.

1 solution

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