natural solutions for $9m+9n=mn$ and $9m+9n=2m^2n^2$ Please help me find the natural solutions for $9m+9n=mn$ and $9m+9n=2m^2n^2$ where m and n are relatively prime.
I tried solving the first equation in the following way: $9m+9n=mn \rightarrow (9-n)m+9n=0 $
$\rightarrow m=-\frac{9n}{9-n}$
Thanks in advance.

$$mn=9n+9m \Rightarrow (m-9)(n-9)=81$$
This equation is very easy to solve, just keep in mind that even if $m,n$ are positive, $m-9,n-9$ could be negative. But there are only 6 ways of writing 81 as the product of two integers.
The second one is trickier, but if $mn >9$ then it is easy to prove that
$$2m^2n^2> 18mn > 9m+9n $$
Added Also, since $9|2m^2n^2$ it follows that $3|mn$. Combining this with $mn \leq 9$ and $m|9n, n|9m$ solves immediately the equation. 
P.S. Your approach also works, if you do Polynomial long division you will get $\frac{9n}{n-9}=9 +\frac{81}{n-9}$. Thus $n-9$ is a divisor of $81$.
P.P.S. Alternately, for the second equation, if you use $2\sqrt{mn} \leq m+n$ you get
$$18 \sqrt{mn} \leq 9(m+n)=2m^2n^2$$
Thus $$(mn)^3 \geq 81$$ which implies $mn=0 \text{ or } mn \geq 5$.
