special equation $z^y$ - $y^z$ = $x^y$ I have prepared an equation by myself $z^y$ - $y^z$ = $x^y$. This equation has infinitely $(0, n, n)$ and $(n, 0, n)$ solutions for some positive integer $n$. Can this equation can be treated as Diophantine equation or not? How we can find surface of this curve? Is there any member, who have an idea to discuss...plz.
Thanks in advance.
 A: Over $\mathbb{N}$ or $\mathbb{Z}$, there is a third "trivial" solution set: $\{(n-1,1,n)\}$, as well as some "sporadic" solutions: $(1,2,3),~(-1,3,2),~(0,2,4)$. One should examine the existence of solutions with $z\ne y>1$ over $\mathbb{Z}$ or with $y\ne1$ more generally, but I haven't done that yet.
Over $\mathbb{R}$ or $\mathbb{C}$, the equation is a surface with infinitely many solutions. Over $\mathbb{R}$,
$$
x = \left(z^y-y^z\right)^{1/y}
$$
always has a solution when $z^y-y^z > 0$, i.e.
for $(y,z)$ in
$$
R_1=
\left\{
\matrix{
(y,z) \quad: & \\\\
0 ~ < ~ z ~ < & \infty \\
0 ~\le~ y ~\le&
\min
 \left( h_1^{-1}
  \left( h
   \left( \max
    \left( e,~ z
    \right)
   \right)
  \right) ,~ z
 \right)
}\right\}
\quad
\text{or}
\quad
R_3=
\left\{
\matrix{
(y,z) & : \\\\
e                                     &\le& y & < & \infty \\
h_1^{-1}\left(\tfrac{\ln y}{y}\right) &\le& z &\le& y
}\right\}
$$
where $h_1^{-1}(y)\le e$ and $h_2^{-1}(y)\ge e$ are the inverse functions
of the restrictions $h_1(z),~h_2(z)$ of $y=h(z)=\frac{\ln z}{z}$
to $I_0\cup I_1$ for $I_0=(0,1],~I_1=(1,e]$ and $I_2=[e,\infty)$,
depicted below right in blue and green respectively.
These are the regions $1$ & $3$ on the graph below left
that include the horizontal line through $(e,e)$.


If we further restrict $h_1$ to $I_1$, so that $ I_1
\stackrel{h_1}{\longrightarrow} I_0
\stackrel{h_2}{\longleftarrow } I_2 $,
then the nonlinear part of the left graph above
separating regions $1$ & $2$ from regions $3$ & $4$,
a self-inverse function on $(1,\infty)=I_1\cup I_2$,
$$
g(z)=\left\{\matrix{
h_2^{-1}\circ h_1=h_2^{-1}\left(\tfrac{\ln z}{z}\right) \quad&1<z\le e \\
h_1^{-1}\circ h_2=h_1^{-1}\left(\tfrac{\ln z}{z}\right) \quad&e\le z<\infty \\
}\right.
$$
is tangent to $yz=e^2$ at $(e,e)$
but with a smaller growth (decay) rate as $y\to1$ ($\infty$).
Also, for $y=g(z)$,
$$
0>g'(z)=\left\{\matrix{
\frac{y^2}{z^2} \cdot
\frac{1-\ln z}{1-\ln y}
=\frac{\ln(\frac{z}{e})/(\frac{z}{e})^2}{\ln(\frac{y}{e})/(\frac{y}{e})^2}
&& y,~z\ne e \\\\
-1 && y=z=e
}\right.
$$
while
$G(z)=\frac{e^2}{z}$ has slope
$G'(z)=-\frac{e^2}{z^2}$
so that, still with $y=g(z)$,
$$
\frac{g'(z)}{G'(z)} = -
\left(\frac{y}{e}\right)^2
\frac{\ln(z/e)}{\ln(y/e)}
\implies
-\frac{G'(z)}{g'(z)}\ln\frac{z}{e}
=\frac{\ln(y/e)}{(y/e)^2} \le \frac1y
$$
using the well-known inequality $\ln t\le\frac{t}{e}$
(which can be proved graphically) with $t=\frac{y}{e}$.
All of the above follows from noting that for $y,~z>0,~x=0~\iff$
$$ z^y = y^z
\quad\iff\quad  y\,\ln z = z\,\ln y
\quad\iff\quad  \frac{ln y}{y} = \frac{ln z}{z} $$
$$   \iff\qquad h(y)=h(z)
\quad\iff\quad  W(-\ln y)=W(-\ln z) $$
and analyzing the function $h(z)=\frac{ln z}{z}$
on its maximal real intervals of invertibility as above.
Note also that $y=g(z) \iff y=z=e$ or $y\ne z$ and
$W(-\ln y)=W(-\ln z)$, where $W$ is the Lambert W function.
Some other facts about $h(z)$ are that its reciprocal,
$\frac{z}{\log z}$ (where $\log$ has base $e$),
is the analytic asymptote of the prime counting function $\pi(z)$,
which is also related to the logarithmic integral function.
Some further facts are that it has derivative is $h'(z)=\frac{1-\ln z}{z^2}$,
it satisfies the differential equation
$(zh)'=zh'+h=z^{-1}$ or $h'+z^{-1}y=z^{-2}$,
and it has antiderivative
$$
H(z)=\int_1^zh(s)ds=\left[\frac12(\ln s)^2\right]_1^z=\frac12(\ln z)^2.
$$
In $\mathbb{R}^3$, there are also solutions whenever
$y=\frac{p}{q}\in\mathbb{Q}$ is expressed in lowest terms
and $p$ is odd, for then, we can take the $p^{\text{th}}$
root of $z^y-y^z$ even if it is negative:
$$
x = \left(z^{p/q}-\left(\tfrac{p}{q}\right)^z\right)^{q/p}
$$
This will have a unique real solution $x$
iff one of the several conditions is met...
