Let their least common multiple = $N$. If $p^{2k-1} \parallel a (k \in \mathbb{N})$ for some prime $p$. Since they are coprime together, $p$ cannot divide both $b, c$. If $p \not| \ bc$, $p^{2k-1} \parallel N$, so it's a contradiction ($N$ is a perfect square) Without Loss of Generality; let $p \ |\ b$. $N$ should be a perfect square, so $V_p(b) \geq {2k-1}$ and $2\ |\ V_p(b)$. Thus, $2 \not| \ V_p(a)+V_p(b)=V_p(abc)$ which leads to contradiction. (By the assumption, $abc$ is a perfect square) Therefore if $p\ |\ a$, $2 \ |\ V_p(a)$. So $a$ is a perfect square. In the same way, we can claim that $a$, $b$ and $c$ are perfect squares.