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there are some integer pairs $(m,n)$ that satisfy $$\frac{m^2+mn+n^2}{m+2n}=\frac{13}{3}.$$ Find the value of $m + 2n.$

I tried expressing the numerator in terms of $m+2n$ but it resulted in nothing remotely useful

hints or suggestions would be appreciated aswell as solutions.

from the 2019 SAIMC https://chiuchang.org/imc/wp-content/uploads/sites/2/2019/08/2019-IWYMIC-answer.x17381.pdf

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    $\begingroup$ Well, trial and error yields a pair $(m,n)$ that works pretty quickly...but of course there could be others. $\endgroup$
    – lulu
    Aug 14 '19 at 19:29
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We have $$3m^2+(3n-13)m+3n^2-26n=0,$$ which gives $$(3n-13)^2-12(3n^2-26n)\geq0$$ or $$27n^2-234n-169\leq0,$$ which gives not so many cases: $$n\in\{0,1,2,3,4,5,6,7,8,9\}.$$ Can you end it now?

I got only $n=2$ gives $m=5$ and $n=7$ gives $m=-5.$

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I can rewrite the equation $\frac{m^2+mn+n^2}{m+2n}=\frac{13}{3}$ in the form: $3m^2+3mn+3n^2=13(m+2n)$ and so $3m^2+m(3n-13)-26n+3n^2=0$. Using the quadratic formula: $$m=\frac{-(3n-13)\pm\sqrt{(3n-13)^2-4\cdot3\cdot(3n^2-26n)}}{6}$$

The polynomial $(3n-13)^2-4\cdot3\cdot(3n^2-26n)$ is positive only if $\frac{13}{3}-\frac{\sqrt{26}}{3\sqrt3}\leq x \leq \frac{13}{3}+\frac{\sqrt{26}}{3\sqrt3}$. The only cases when the polynomial is a square are $n=2$ $\land$ $m=5$ or $n=7$ $\land$ $m=-5$.

Let $A_1=m_1+2n_1=5+2\cdot 2=9$ and $A_2=m_2+2n_2=-3$.

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  • $\begingroup$ $m+2n$ has to be divisible by $3$. $\endgroup$
    – Anurag A
    Aug 14 '19 at 20:18
  • $\begingroup$ Sorry, my mistake. Corrected. $\endgroup$
    – Matteo
    Aug 14 '19 at 20:19
  • $\begingroup$ can you please explain what you did with the $\frac{13}{3}-\frac{\sqrt{26}}{3\sqrt3}\le x \le \frac{13}{3}+\frac{\sqrt{26}}{3\sqrt3}$ I don't understand the reasoning behind it $\endgroup$
    – Tyrone
    Aug 15 '19 at 14:33
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    $\begingroup$ @Tyrone: simply put: $(3n-13)^2-4\cdot3\cdot(3n^2-26n)\geq0$. $\endgroup$
    – Matteo
    Aug 15 '19 at 15:04

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