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How many pairs (a, b) of positive integers are there such that $a\ge b$ and $$2\left(\sqrt\frac{15}{a}+\sqrt\frac{15}{b}\right) $$ is an integer

From the 2013 IMC (http://imc-official.chiuchang.org/files/problem/2013-IWYMIC-Individual.pdf)

I genuinely have no idea how to solve this, hints would be appreciated

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Clearly the only solutions are when the radicals evaluate to rational numbers. Hence WLOG we can substitute $a=15m^2$ and $b=15n^2$ with $m\ge n$ giving the expression $$2\left(\frac1m+\frac1n\right)=\frac{2m+2n}{mn}$$ So, in order for the resulting expression to be an integer, we need $$2m+2n\equiv0\mod{mn}$$ which is equivalent to the two simultaneous equations $$2n\equiv0\mod{m}$$ $$2m\equiv0\mod{n}$$ The first equation implies that $n=m/2$ or $n=m$ as $m\ge n\gt0$ which agrees with the second equation. In the first case ($n=m/2$), the entire original expression simplifies to $$\frac{6}{m}$$ Hence the only possible values of $m$ and $n$ are $(m,n)=(2,1),(6,3)$ in order for the expression to be an integer. In the second case ($n=m$) we get $$\frac{4}{m}$$ in which case the only solutions are $(m,n)=(1,1),(2,2),(4,4)$. All of these solutions for $m$ and $n$ correspond to exactly one solution for $a$ and $b$ hence there are $5$ solutions in which the expression results in an integer.

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  • $\begingroup$ "Clearly the only solutions are when the radicals evaluate to rational numbers" or if $(\sqrt{\frac 1a} + \sqrt{\frac1 b}) = q\sqrt{15}$ for a rational $q$. Can you claim that is only possible if $\sqrt{\frac{15}a}$ and $\sqrt{\frac{15}b}$ are both rational? $\endgroup$ – fleablood Jul 31 at 21:13
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    $\begingroup$ @fleablood $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ either both rational or irrational as they sum to a rational (actually, an integer divided by $2$). However, as asked and shown in The sum of two irrational square roots, in particular the answer by Mathmo123, shows that the sum of two irrational square roots is always irrational. Thus, this means here that $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ must both be rational. $\endgroup$ – John Omielan Jul 31 at 21:29
  • $\begingroup$ Right... but I think that needs to be pointed out. $\endgroup$ – fleablood Jul 31 at 21:37
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From

$$a=15m^2,\\b=15n^2$$ we draw $$2\left(\frac1m+\frac1n\right)=k$$ or $$2(m+n)=kmn$$

where $1\le k\le4$.

By brute force, $$(m,n,k)=(4,4,1),(3,6,1),(6,3,1),(2,2,2),(1,2,3),(2,1,3),(1,1,4).$$

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    $\begingroup$ You should mention that you (correctly) use that the sum of square roots of rationals is rationals iff each of the summands is rational $\endgroup$ – Hagen von Eitzen Jul 31 at 20:27

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