Solve the equation $x^{x+y}=y^{y-x}$ in positive integers. 
Let $x,y$ are positive integers. Solve the equation
  $$x^{x+y}=y^{y-x}$$

smartplot(x^(x+y)=y^(y-x));

I used http://www.wolframalpha.com
 A: Let $x,y$ be integers, then $x^{x+y}$ is integer, and so must be $y^{y-x}$. Therefore, $y \geq x$.
Now from fundamental theorem of arithemtic it's evident that $x = n^a$ and $y = n^b$ for some positive integers $n, a, b$, where $b > a$. If $n = 1$ we get solution $x=y=1$. Otherwise, rewrite original equation in terms of $n,a,b$ and take $\log_n$ of both sides. You'll get
$$
a(n^a + n^b) = b(n^b - n^a) \quad\iff\quad
a+b = (a-b)n^{a-b}.
$$
Let $c = b-a$, then $a+b = c + 2a$ and
$$
2a = c(n^c - 1).
$$
So you can choose arbitrary positive integers $c$ and $n$, where $c(n-1)$ is even, and get a solution
$$
x = n^{c(n^c-1)/2}, \quad y = n^{c(n^c-1)/2 + c}.
$$
UPD: As user90369 mentioned in comments, if you denote $n^c = m$, the solution will take simpler form
$$
x = m^{(m-1)/2}, \quad y = m^{(m+1)/2},
$$
where $m = 2k+1$ or $(2k)^2$ for some integer $k$.
A: Here I will show that $y= x^{a/b}$ where a and be are two consecutive odd or even numbers:
Clearly x and y have common factors, lets suppose $y= k x$ we may write:
$y = x ^ {\frac{x+y}{x-y}}= x^{\frac{(k+1)x}{(k-1)x}}=x^{\frac{k+1}{k-1}}$ 
$k+1 $ and $k -1$ are two consecutive odd or even numbers:
Examples:
$k=3 ⇒ a=k+1 =4 , b =k -1= 2 ⇒ y =x^2$; solution
x=3 and y=9 is a result  $
$ k=4 ⇒ a= 4+1=5, b=4-1=3 ⇒y=x^{5/3}$;solution
x=8 and y=32 is a result$
Hence equation $x^{x+y} =y^{y-x}$ is reduced to a simple form like $ y =x^{a/b}$ where a and b are two consecutive odd or even numbers. Clearly a and b can have certain values as showed in examples.
A: Let $y=mx$, with no assumption on $m$, aside from it being rational.  The equation becomes
$$x^{(m+1)x}=(mx)^{(m-1)x}$$
We can now take the $x$th root, obtaining $x^{m+1}=(mx)^{m-1}$, which solves to
$$x=m^{(m-1)/2}\quad\text{and}\quad y=mx=m^{(m+1)/2}$$
as in G. Strukov's answer, except we don't yet know that $m$ is an integer.
Now if $m=p/q$ (with $p$ and $q$ relatively prime) has a denominator of any sort, then $x^2$ is the $q$th root of $(p/q)^{p-q}$, which, since $\gcd(q,p-q)=1$, is irrational unless $p$ and $q$ are individually $q$th powers. But $b^q\gt b$ for all $q$ if $b\ge3$, and the possibility $q=2^2$ is easily ruled out.  So for $x$ and $y$ to be integers, $m$ must also be an integer.  
The rest is straightforward: Any odd integer $m$ gives integers $x$ and $y$, but only squares of even integers give integers. I.e., the solutions are of the form
$$(x,y)=((2k+1)^k,(2k+1)^{k+1})\quad\text{and}\quad(x,y)=((2k)^{4k^2-1},(2k)^{4k^2+1})$$
for $k\in\mathbb{N}$.
