When is $\frac{a}{n} = n^a$ true? Title says it all really, I am trying to figure out if theres a situation where 
$\frac{a}{n} = n^a$ is true or if this is impossible.
This is not realy from somewhere, just for the sake of curiosity.

$\frac an = n^a \implies$
$a = n^{a+1} \implies $
$1 = \log_a a = \log_a n^{a+1} = (a +1 ) \log_a n \implies$
$\frac 1{a+1} = \log_a n \implies$
$n = a^{\frac 1{a+1}}$
So for all positive $a$, $n$ is solvable.  As $0/0$ (as well as $0^0$) are undefined $n = a = 0$ is not a solution (despite $0 = 0^1$).  
For $a < 0$ it gets a little more complicated.
If $n > 0$ then $a/n < 0$ while $n^a > 0$ so $n < 0$.
But $n <0$ implies $a/n > 0$ so $(-|n|)^a > 0$ so $a = -p/q$ must be that $p$ is even.  (Remember: negative numbers to irrational powers are not defined.) And as $\gcd (p,q) =1$, $q$ must be odd.
Then $n = a^{1/(a+ 1)} = a^{q/(q-p)}$.
Example: $a = -2/3; n = (-2/3)^{3}$. And $a/n = \frac 1{(2/3)^2} = (-3/2)^2 = ((-3/2)^3)^{2/3} = ((-2/3)^3)^{-2/3} = n^a$
Or $a = -2; n=-1/2$ and $a/n = 4 = (-2)^2 = (-1/2)^{-2}$.
Or $a = -4/3; n = (-4/3)^{-1/3} = -\sqrt[3]{3/4}$.  
etc.
