Solve $x,y,z, z^x=x, z^y=y, y^y=x$

Question

I solved it and got three solutions

$(1,1,1)$ is obvious, no need to calculate it.

other two solutions are $(4,2\sqrt{2}),(4,2,-\sqrt{2})$

But actual answer is only $(1,1,1),(4,2,\sqrt{2})$.

One can see that all these three triplets are satisfying the equation.

What am I missing here.

My attempt is as follows:

$$z^x=x$$ $$x\log (z)=\log (x)$$ $$\log (z)=\frac {\log (x)}{x}$$ $$z^y=y$$ $$\log (z)=\frac {\log (y)}{y}$$

$$\frac {x}{y}=\frac {\log (x)}{\log (y)}$$ $$y^y=x$$ $$y\log (y)=\log (x)$$ $$y=\frac {\log (x)}{\log (y)}$$ $$x=y^2$$

$$y^y=x$$ $$y^y=y^2$$ $$y=2$$ $$x=2^2$$ $$x=4$$ $$z=\sqrt{2},-\sqrt{2}$$

Answer 1

$z=-\sqrt{2}$ is not permissible.

This is because $\log(-\sqrt2)$, which is used in your derivation, is undefined. This spurious solution comes about because $y$ happens to be 2.

Edit:

As pointed out below in the comments, the function $z^y$ is defined for the domain $z>0$, which is implicit. To see why, for example, $(-1)^{1/2} \ne (-1)^{2/4}$, indicating that $z^y$ could have different values for $z=-1$ and can not be viewed as a function for negative value $z$. If any negative value is allowed for the solution, it needs to be explicitly stated in the question.

Answer 2

Here is a solution without using the logarithm function. In this way, the solution $(x,y,z)=(4,2,-\sqrt{2})$ cannot be excluded.

However, it should be noted, as in Quanto's answer, that the function $z\mapsto z^y$ is generally not defined for $z\leq 0$ (or it needs to be clearly stated how it is defined), since we do not know a priori that $y$ will end up being an integer. However, at least, this solution works if we assume that $a^b$ is always defined whenever $b\in\mathbb{Z}$ (except possibly when $a=0$).

From $z^x=x$, $z^y=y$, and $y^y=x$, we have $$y^{y^2}=\left(y^y\right)^y=x^y=\left(z^x\right)^y=z^{xy}=\left(z^y\right)^x=y^x\,.$$ Hence, $$y^{y^2-x}=1\,.$$ That is, $y=1$, $y=-1$ and $y^2-x$ is an even integer, or $y^2=x$.

If $y=1$, then $x=1$, so $z=1$. If $y=-1$ and $y^2-x$ is an even integer, then $x=y^y=-1$, so $z^x=x$ means $z=-1$. Thus, if $|y|=1$, we get the solutions $$(x,y,z)=\pm(1,1,1)\,.$$

If $|y|\neq 1$, then $y^2=x=y^y$, or $y^{y-2}=1$. Thus, $y=2$ and so $x=4$. This implies $z^2=2$ (and $z^4=4$), or $z=\pm\sqrt{2}$, whence $$(x,y,z)=(4,2,\pm\sqrt{2})$$ are solutions.

P.S. I just realized that $y^{y^2-x}=1$ can also happen when $y=-1$ and $y^2-x$ is an even integer, and edited this answer accordingly.