How to determine this solutions? (a) Determine all positive rational solutions of $x^y=y^x$                                      
(b) Determine all positive rational solutions of $x^{x+y}=(x+y)^y$
 A: a)Here is some of my work.
If $x = y$, then we get $x = y= 1$. Now WLOG, suppose $y > x$.
Let $x = \frac{a}b, y = \frac{c}d$ and suppose that both fractions are reduced.
This gives
$\left( \frac{a}b \right)^{c/d} = \left( \frac{c}d \right)^{a/b}$.
Let $\gcd(a,c) = k$ so $a = pk, c = qk, \gcd(p,q) = 1$.
This gives
$(pk)^{cb}d^{ad} = (qk)^{ad}b^{cb}$.
Which rearranges to
$k^{cb-ad} d^{ad} = q^{ad} \left( \frac{b}{p} \right)^{cb}$.
By assumption, the LHS is an integer. Furthermore, since $\gcd(p,q) = 1$, we must have $p|b$. However, since $\gcd(a,b) = 1$, $p = 1$. If we divide $q$ over, we have two cases. $q|d \implies q = 1$ or $q|k \implies k = qm$. 
In the first case, we can reduce our equation to $d^{kd}k^{kb} = b^{kb}k^{kd}$ which implies $d = b$ from taking modulo $d$ and $b$. The solution from this is just $(1,1)$.
Otherwise, $k = qm$. 
I think the only solution from this case is $(2,4)$ although I have not verified it since I must now go eat dinner. However, I have verified that these are the only solutions in integers. 
A: 
$(a)$ Determine all positive rational solutions of $x^y=y^x$

$x=\bigg(1+\dfrac1n\bigg)^n$ and $y=\bigg(1+\dfrac1n\bigg)^{n+1}$ For $n=1$ we have the only natural solution, $2^4=4^2$.
