# find all natural numbers satisfying $9(x+y)=[x,y]^{(x,y)}$

Let $(x,y)=$ gcd $(x,y)$ and $[x,y]=$lcm$(x,y)$ Find all numbers $x,y$ such that $9(x+y)=[x,y]^{(x,y)}$.

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something seems to be missing. Why do we need to define $(x,y)$ and $[x,y]$? –  N. S. Aug 24 '12 at 23:19
Hmm something is probably still wrong, since the two $9$'s cancel eachother.... –  N. S. Aug 24 '12 at 23:23
Is there anything interesting about this question? Any reason why anyone would want to know an answer? –  Gerry Myerson Aug 24 '12 at 23:51

HINT write $x= ad$ and $y=bd$ where $(a,b)=1$

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I really don't see how this helps much. Would you care to elaborate a bit? –  EuYu Aug 24 '12 at 23:41
It was posted some questions before math.stackexchange.com/questions/186519/… –  clark Aug 24 '12 at 23:42
Thank you for the clarification! –  EuYu Aug 24 '12 at 23:49

I think I have a solution but im not sure. let $x=\prod\limits_{k=1}^p p_k^{x_k}$ and let $y=\prod\limits_{k=1}^p p_k^{y_k}$. Let $m_k = x_i-y_i$if $x_i-y_i\geq 0, 0$ if $x_i-y_i<0$ let $n_k = y_i-x_i$if $y_i-x_i\geq 0, 0$ if $y_i-x_i<0$. Let $m=\prod\limits_{k=1}^p p_k^{m_k}$ and $n=\prod\limits_{k=1}^p p_k^{n_k}$.

The following properties hold: (x,y)(mn)=[x,y], (x,y)(m+n)=x+y.and m and n are prime numbers.$m,n>0$ Therefore we can rewrite the equation as

$9([(x,y)(m+n)]=[(x,y)(mn)]^{(x,y)}$ also: $m+n< mn+2$. Therefore $9(m+n)< 9mn+18<27mn$ and $mn\leq (mn)^w$ where w is a natural number

Now I see the equation for the first 4 values of $(n,m)$. However, we know that number 4 is impossible. And any number greater than that is also impossible.

$9(m+n)=mn$

$9(m+n)=2(mn)^2$

$9(m+n)=9(mn)^3$

$9(m+n)=64(mn)^4$

Therefore if it happens that none of the first equations can be solved, no numbers x and y satisfy the problem.

we will prove $9(m+n)=9(mn)^3$ has no solutions. $9(m+n)=9(mn)^3 \rightarrow m+n=(mn)^3$ If $m,n>1$ then $2\leq m+n\leq mn$ Therefore $(mn)^3>m+n.$ if m= 1 then $9n+9=9(n)^3 \rightarrow n+1=n^3$ which has no solutions. Therefore $m+n=(mn)^3$ has no solutions.

The solution to the other two problems I copy from N.S textually from natural solutions for $9m+9n=mn$ and $9m+9n=2m^2n^2$ $$mn=9n+9m \Rightarrow (m-9)(n-9)=81$$

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 $mn=0$ or $mn \geq 5$.

Therefore no numbers x, y satsfy the condition.

Is this proof correct?? thank you.

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