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

Question

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

Answer 1

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.

Answer 2

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