# 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: $9m+9n=mn \rightarrow (9-n)m+9n=0$ $\rightarrow m=-\frac{9n}{9-n}$

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$$mn=9n+9m \Rightarrow (m-9)(n-9)=81$$

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 $mn=0 \text{ or } mn \geq 5$.

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Hint: if $m$ and $n$ are relatively prime, $mn$ and $m+n$ are relatively prime.

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why is that true? @RobertIsrael – Carry on Smiling Aug 24 '12 at 22:57
@Khromonkey If they are not they have a common prime divisor. But if $p|mn$ it must divide one of them. And if $p$ also divides their sum...... (you should be able to finish the argument) – N. S. Aug 24 '12 at 23:02
@Khromonkey Suppose $(a,b)=1$. We show that $(ab,a+b)=1$. Indeed, suppose $d$ divides both $a+b$ and $ab$. Then it divides $a(a+b)-ab=a^2$ and it divides $b(a+b)-ab=b^2$. But if $(a,b)=1$, then $(a^2,b^2)=1$, so $(ab,a+b)=1$, as desired. – Pedro Tamaroff Aug 24 '12 at 23:05
@RobertIsrael ok so then m and n are not relatively prime then what? – Carry on Smiling Sep 12 '12 at 22:18
You asked for solutions where $m$ and $n$ are relatively prime. – Robert Israel Sep 12 '12 at 23:49