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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.

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Your equation could be interesting over any of $\mathbb{Z},\mathbb{N},\mathbb{Q}$ or $\mathbb{C}$. –  bgins Apr 7 '12 at 10:41
    
@bgins!well, please explain over any of Z, N, Q and C –  gandhi Apr 7 '12 at 10:49

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up vote 2 down vote accepted

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 $(e,e)$.

x=0 log(y)/y

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...

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! it is really wonder. I am so haapy with your presentation and I will include this in my notes. Once again, thank you so much for your presentation. If you don't mind can I have your mail ID or send your id to editor126 at gmail dot com –  gandhi Apr 8 '12 at 4:48
    
Bgins! Sir, where these equations will come? I mean what kind of equations are these? is there any specific name? –  gandhi Apr 8 '12 at 4:49
    
You should do some research to see if they have been studied. I'm doing my own thinking about it, which may or may not be original (seems unlikely to be). So far I have added some references to related functions. If I have time I will go further over $\mathbb{Q}$ next, and then perhaps address $\mathbb{Z}$. –  bgins Apr 8 '12 at 23:40

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