What are the positive rational solutions of $x^{(x+y)} = (x+y)^y$? I saw this problem in the Problem-Solving through Problems book by Larson (# 3.3.25b).
I got to here:
$$x \log(x) = y\log\left(1+ \frac yx\right)$$
But I can't seem to find a way to reduce this further.
 A: Here's a partial solution.  The first part is for general $x,y$ and is incomplete.  The second part is a complete solution for integer $x,y$.
Part 1: Set $z=x+y$. Then you may rearrange as $$x^zz^x=z^z$$
Now, these are fractions, so set $x=\frac{a}{b}$, $z=\frac{c}{d}$.  Assume without loss that $a,b$ are relatively prime, and so are $c,d$.  We get 
$$\left(\frac{a}{b}\right)^z\left(\frac{c}{d}\right)^x=\left(\frac{c}{d}\right)^z$$
or 
$$a^z c^xd^z=c^zb^zd^x$$
To clear denominators raise both sides to the $bd$ power, to get
$$a^{bc}c^{ad}d^{bc}=c^{bc}b^{bc}d^{ad}$$
Becase $z>x$ we have $bc>ad$.  Set $k=bc-ad>0$ for convenience. We divide and get
$$a^{bc}d^{k}=b^{bc}c^{k}$$
Finally, we have an expression with positive integers.  Because $a,b$ are relatively prime, $a^{bc}|c^{k}$ and $b^{bc}|d^k$.  Because $c,d$ are relatively prime, $c^{k}|a^{bc}$ and $d^k|b^{bc}$. Hence $a^{bc}=c^k$ and $b^{bc}=d^k$.
Now, rearrange $a^{bc}=c^{bc-ad}$ to get $c^{ad}=(c/a)^{bc}$. Since $c$ is an integer, $a|c$.  Similarly, rearrange $b^{bc}=d^{bc-ad}$ to get $d^{ad}=(d/b)^{bc}$, to conclude $b|d$.  These are helpful, but it doesn't give explicitly all rational solutions.
Part 2: If we assume integer solutions (not just rational), then $c=d=1$.  Because $a|c$ we may write $c=aq$.  Now $a^{bc}=c^k$ becomes $$a^{aq}=(aq)^{aq-a}$$
If $a=1$, then $q=c=1$, which is impossible since $y>x$.  Otherwise we take logs, getting $$aq \ln a = a(q-1) (\ln a + \ln q)$$ This rearranges to $$\frac{q}{q-1}\ln a=\ln a + \ln q$$  Divide by $\ln a$ (since $a>1$), subtract 1, and add fractions to get $\frac{1}{q-1}=\frac{\ln q}{\ln a}$.  Cross-multiply and exponentiate to get the final solution $$a=q^{q-1}$$
This works for any natural $q$; we take $a=q^{q-1}, c=aq=q^q, b=c-a=a(q-1)$.  In particular, this finds Oleg's solutions (from $q=2,3$), but also $(64,192)$ (corresponding to $q=4$), etc.
