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.

Find $m + n$ if $m$ and $n$ are natural numbers such that: $$\frac {m+n} {m^2+mn+n^2} = \frac {4} {49}\;.$$

My reasoning:

Say: $$m+n = 4k$$ $$m^2+mn+n^2 = 49k$$ It follows:$$(m+n)^2 = (4k)^2 = 16k^2 \Rightarrow m^2+mn+n^2 + mn = 16k^2 \Rightarrow mn = 16k^2 - 49k$$

Since: $$mn\gt0 \Rightarrow 16k^2 - 49k\gt0 \Rightarrow k\gt3$$

Then no more progress.

share|improve this question
    
I figured out how to complete my own answer only after the three answers below had already been submitted, so I didn't know if I should have presented it as a comment, as a brand new answer, or like an edit. –  Sawyier Jan 30 '13 at 0:47
3  
I would post it as answer. You are explicitly allowed to answer your own question and accept it. Your answer is different from the rest. –  Ross Millikan Jan 30 '13 at 1:27
add comment

5 Answers

up vote 7 down vote accepted

Observe that $k$ must be a non-zero integer.

We know that $m, n$ are the roots of the quadratic equation

$$X^2 - 4kX + (16k^2 - 49k)$$

The roots, from the quadratic equation, are

$$ \frac { 4k \pm \sqrt{(4k)^2 - 4(16k^2 - 49k) }} {2} = 2k \pm \sqrt{ 49k - 12k^2}$$

The expression in the square root must be a perfect square.

Try $k = 1$, $49 k - 12k^2 = 37$ is not a perfect square.

Try $k = 2$, $49k - 12k^2 = 50$ is not a perfect square.

Try $k=3$, $49k-12k^2 = 39$ is not a perfect square.

Try $k=4$, $49k-12k^2 = 4$ is a perfect square. This leads to roots 6, 10, which have sum 16.

For $k\geq 5$, $49k - 12k^2 < 0$ has no solution.

For $k \leq -1$, $49k - 12k^2 < 0$ has no solution.

share|improve this answer
    
The roots would be $10$ and $6$ actually, which have sum of 16. You can also check that by substituting $k=4$ in $m+n = 4k$. But your reasoning was the important part and it was great, thx! It actually helped me complete mine. –  Sawyier Jan 30 '13 at 0:32
    
@Sawyier yes thanks for pointing that out. I knew it was 6 and 10, but somehow typed it wrongly. Vietas + Quadratic formula is nice for such problems. –  Calvin Lin Jan 30 '13 at 1:03
add comment

A simple approach is to use the symmetry and choose $m \ge n$, then note that $m$ cannot be very large. If it is $13$ or more the fraction will be too small: $\frac {m+n}{m^2+mn+n^2} \lt \frac {m+n}{m^2+mn} = \frac 1m$. That doesn't leave many choices. Now the quickest is to make a spreadsheet with $m$ across the top, $n$ down the side, compute $\frac {m+n}{m^2+mn+n^2} -\frac 4{49}$, and scan by eye for zeros. Alternately, you can make it a quadratic $4m^2+4mn+4n^2-49m-49n=0$ and insert the $n$'s and solve for $m$. See when you get naturals.

share|improve this answer
add comment

$4m^2+4mn+4n^2=49m+49n$

$4m^2+4mn+n^2+3n^2-49m-49n=0$

$(2m+n)^2+3n^2-(49/2)(2m+n)-(49/2)n=0$

$16(2m+n)^2-392(2m+n)+48n^2-392n=0$

$(4(2m+n)-49)^2+48n^2-392n=2401$

$3(4(2m+n)-49)^2+144n^2-1176n=7203$

$3(4(2m+n)-49)^2+(12n-49)^2=9604$

$3(8m+4n-49)^2+(12n-49)^2=9604$.

$3x^2+y^2=98^2$

Now the last equation has only finitely many integer solutions, with a finite procedure for finding them, then for each solution you can check whether $8m+4n-49=x,12n-49=y$ has integer solutions, and that should do it.

share|improve this answer
    
Those were some pretty awesome tricks you used there :) –  Sawyier Jan 30 '13 at 0:22
4  
@Sawyier It's a special case of Lagrange's method for normalizing general bivariate quadratic Diophantine equations. –  Math Gems Jan 30 '13 at 2:16
add comment

Finishing the original reasoning started in the question:

$$(\frac {m+n}{2})^2 -mn \ge0 \Rightarrow (\frac {4k}{2})^2 - (16k^2 -49k) \ge0 $$ $$ 49k - 12k^2 \ge 0 \Rightarrow k\le4$$

If: $$3\lt k \le 4 \Rightarrow k=4$$

Finally, substituting $k=4$ in $m+n = 4k$, we get $m+n=16$.

share|improve this answer
add comment

It is useful to write the stipulation of the problem in a few different ways: $$\frac{49}{4}=\frac{(m+n)^2-mn}{m+n}=m+n-\frac{mn}{m+n},$$ $$\frac{49}{4}=\frac{(m+n)n+m^2}{m+n}=n+\frac{m^2}{m+n},$$ $$\frac{49}{4}=\frac{(m+n)m+n^2}{m+n}=m+\frac{n^2}{m+n}.$$ From the first, we glean that $m+n>12$ because $49/4=12+1/4$ and $m,n>0$. From the second and third, we see that $n\le 12$ and $m\le 12$.

Another thing to notice is that $mn/(m+n)$ must reduce to a fraction with a denominator of $4$. This means that $$4mn=c(m+n)$$ for some $c$ relatively prime to $4$. This implies that $4$ divides $m+n$, so there are only three cases to consider: $m+n=16$, $m+n=20$, $m+n=24$. From the first, we have $$12+\frac{1}{4}=16-\frac{m(16-m)}{16}$$ which easily gives the solution $\{m,n\}=\{6,10\}$. The other two cases result in quadratic equations with no real solutions.

share|improve this answer
add comment

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.