Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
2  
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

2 Answers 2

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

share|improve this answer
    
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
1  
Thank you for the clarification! –  EuYu Aug 24 '12 at 23:49
up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

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.