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Answer: two of them


$m$ and $n$ are multiples of $12$, so say $m=12a$ and $n=12b$

$12$ the highest common factor $\therefore$ $a$ and $b$ don't have any common factors

Lowest common multiple of $12a$ and $12b$ is $12ab$ (since $a,b$ have no common factors)

$12ab$ is a square number so it is a product of squares

$12ab=4\times3\times a \times b$

Say $3a$ is a square, so $a=3p^2$ for some number $p$ (so that $3a=9p^2=(3p)^2$)
and $b=q^2$

So $m$ and $n$ are  $36p^2$       and       $12q^2$
$\Rightarrow \frac m3$ and $\frac n3$ are $12p^2$ (not square), $4q^2$ (square)
     $ \frac m4$ and $\frac n4$ are $9p^2$ (square),        $3q^2$ (not square)
     $mn = 12\times36\times p^2\times q^2$ (not square)


This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.