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

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 $k$.
