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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.

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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
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

5 Answers 5

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.

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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

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.

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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.

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Those were some pretty awesome tricks you used there :) –  Sawyier Jan 30 '13 at 0:22
@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

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 $m+n=16$.

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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.

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