Since HCF$(m,n)=12$ we can write $m=12a$, $n=12b$ where $a$ and $b$
are coprime. The lowest common multiple of $m$ and $n$ must be
$12ab$.

Since $12ab=4\times 3ab$, $3ab$ must be a square number, and since $a$, $b$ are coprime exactly one of them is a multiple of $3$. Without loss of generality suppose that $a$ is divisible by $3$ and write $a=3c$.

$3ab=9cb$ is a square number, so $cb$ is a square number, which implies that both $b$ and $c$ are square numbers, say $c=d^2$ and $b=e^2$. Therefore $m=12a=36d^2$ and $n=12e^2$.

Then $\frac{m}{4}=9d^2$ and $\frac{n}{3}=4e^2$ are square numbers, whereas $\frac{m}{3}=12d^2$, $\frac{n}{4}=3e^2$ and $mn=36\times 12d^2e^2$ are not.

View the current weekly problem

Since $12ab=4\times 3ab$, $3ab$ must be a square number, and since $a$, $b$ are coprime exactly one of them is a multiple of $3$. Without loss of generality suppose that $a$ is divisible by $3$ and write $a=3c$.

$3ab=9cb$ is a square number, so $cb$ is a square number, which implies that both $b$ and $c$ are square numbers, say $c=d^2$ and $b=e^2$. Therefore $m=12a=36d^2$ and $n=12e^2$.

Then $\frac{m}{4}=9d^2$ and $\frac{n}{3}=4e^2$ are square numbers, whereas $\frac{m}{3}=12d^2$, $\frac{n}{4}=3e^2$ and $mn=36\times 12d^2e^2$ are not.

*This problem is taken from the UKMT Mathematical Challenges.*

View the current weekly problem