Find all positive integers $x$ and $y$ that satisy $x^{y^2}=y^x$... I have to find all the numbers that  satisfy the condition above, or the interval of them all, or a condition to them... Is there a way without using $\log$ or $\ln$? 
 A: In the real domain, there is an explicit solution in terms of Lambert function
$$x=-\frac{y^2}{\log (y)}\, W\left(-\frac{\log (y)}{y^2}\right)$$ the derivative $$x'=\frac{y (1-2 \log (y)) }{\log ^2(y)
  }\frac {W\left(-\frac{\log (y)}{y^2}\right)^2} {1+W\left(-\frac{\log (y)}{y^2}\right)}$$ cancels when $y=\sqrt e$ and for this value $$x=-2 e \,W\left(-\frac{1}{2 e}\right)\approx 1.26107$$ and the second derivative (not reported here) test shows that this is a maximum; it also cancels for $y=0$ ($x=0$) and the second derivative  test shows that this is a minimum.
When $y$ becomes large, the asymptotics is $$x=1+\frac{\log
   (y)}{y^2}+O\left(\frac{1}{y^2}\right)$$ which makes that if $y\to \infty$ then $x\to 1$ from above.
All of the above shows that the range of $x$ is quite limited while the range of $y$ is $[0,\infty)$.
A: Is it true that if $x^{y^2}=y^x$ then $x^{y^2}=x^{2y}$
No, take $x=27$, $y=3$. 
Find all $x$, $y$ that satisfy $x^{y^2}=y^x$
Here's a sum that leads nowhere since I can't simplify it further. Probably best to use the Lambert $W$ function instead.
$$\begin{align}\frac{\ln(y)}{y^2}-\frac{\ln(x)}{x}&=0
\\
&=\sum_{i=0}^\infty\left[(y^2-x)+\sum_{j=0}^{2i}(-1)^j\left({{2i+1}\choose j}\left[xy^{2i+1-j}-y^2x^{2i+1-j}\right]\dots
\\
\dots+{2i \choose j}\left[y^2x^{2i-j}-xy^{2i-j}\right]\right)\right]\end{align}$$
A: This gives a possibly non-complete set of solutions:
If you power both sides to $y^{-2}x^{-1}$ you get to the equation
$$ x^{x^{-1}}=y^{y^{-2}}=\left(y^{y^{-1}}\right)^{y^{-1}} $$
Now you can, for instance, identify $x=y$, which gives the solution $x=y=1$.
Another identification you can do is $x=y^{y^{-1}},\,x^{-1}=y^{-1}$, which gives the same solution.
Actually the solution $x=y=-1$ is also possible, although I haven't figured out how to obtain it...
A: $$x = -{\frac {{y}^{2}}{\ln  \left( y \right) }{\rm W} \left(-{\frac {\ln 
 \left( y \right) }{{y}^{2}}}\right)}
$$
where $W$ is (a branch of) the Lambert W function.
The principal branch is real for all $y > 0$, the $-1$ branch is real for $y > 1$.
EDIT: Positive integer solutions include $(x,y) = (1,1)$, $(16,2)$ and $(27,3)$.  I think those are all, but I'm not sure.
A: There are exactly three positive integer solutions, $(x=1,y=1)$, $(x=16,y=2)$, and $(x=27,y=3)$, and here is a proof that doesn't use calculus or the Lambert function.
Observe that $(x=1,y=1)$ is a solution so that we may now assume $x,y>1$.
Fix some solution $(x\neq1,y\neq1)$. Reduce $\frac{y^2}{x}$ to $\frac{m}{n}$ in reduced terms. Then $x^m=y^n$ with $\gcd(m,n)=1$. Any maximal prime power dividing the right side has its exponent divisible by $n$. And since $m$ is relatively prime to $n$, then any maximal prime power dividing $x$ also has its exponent divisible by $n$. Hence $x^{1/n}$ is an integer. Define $z$ to be $x^{1/n}$, so that $x=z^n$ and $y=x^{m/n}=z^m$.
Now the original equation is $$z^{nz^{2m}}=z^{mz^n}$$ where we can remind everyone that $z>1$, $m,n>0$ with $\gcd(m,n)=1$. This implies $$nz^{2m}=mz^n$$
When $n>2m$, you have an integer equation $n=mz^{n-2m}$ which implies $m$ divides $n$. So for some $k>0$, you may write $n=2m+mk$ (for $k>0$), which means $2m+mk=mz^{2m+mk-2m}\implies2+k=z^{mk}$. This equation only has hope to have an integer solution for small $z$. For $z\geq4$ it's easy to show there are no solutions because the right side grows too fast with respect to $k$. That leaves $z=2,3$ to consider, and $(z=2,m=1,k=2,n=4)$ and $(z=3,m=1,k=1,n=3)$ are the only such solutions. These lead back to $(x=16,y=2)$ and $(x=27,y=3)$. 
$n=2m$ is clearly never a solution to $nz^{2m}=mz^n$.
If $n<2m$, you have an integer equation $nz^{2m-n}=m$ which implies $n$ divides $m$. So for some $k>0$ you can write $n=2m-kn$, which means $nz^{n+kn-n}=\frac{n+kn}{2}\implies2z^{kn}=1+k$. Even with the smallest considerable $z$ ($z=2$), the left side just grows too fast with respect to $k$ for there to be integer solutions.
This establishes there are only three integer solutions: $(x=1,y=1)$, $(x=16,y=2)$, and $(x=27,y=3)$.
