This is a problem from my workbook(not homework), and I can tell that it is true simply upon observation(They share no factors[other than one] and they, when multiplied have all squares as their unique prime factorization, hence they must have all squares as prime factors, which makes $u,v$ squares.)
I am not completely practiced at proofs, so I just want to check that this is rigorous, an if it isn't please advise me as to how I can make it rigorous, enough of my ramblings:
If $\gcd(u,v)=1$ and $uv$ is a square, then $u$ and $v$ are squares.
Proof
If $\gcd(u,v) = 1$ then $u=p_1^{a_1}p_2^{a_2} \dots p_k^{a_k}$, $v=q_1^{b_1}q_2^{b_2} \dots q_l^{b_l}$ where $\forall p,q \;\;, p \ne q$
$uv =p_1^{a_1}q_1^{b_1}p_2^{a_2}q_2^{b_2} \dots p_k^{a_k}q_l^{b_l}$, but $uv$ is a square, hence $u = p_1^{2c_1}p_2^{2c_2} \dots p_k^{2c_k} \;\; , v=q_1^{2d_1}q_2^{2d_2} \dots q_l^{2d_l}$ and thus $u,v$ are clearly both squares.
Thank you for your time, and please try to understand the pain I endured coding those prime factors in $\LaTeX$ on a tablet pc.