Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 4 down vote accepted

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

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

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.