Prove that $(a,b)^2=(a^2,b^2,ab)$ I am trying to prove that $(a,b)^2=(a^2,b^2,ab)$ and was told that this follows from some very basic $\gcd$ laws. What am I not seeing?

Let $(a,b) = c$, then $a = cp, b = cq$ with $(p,q) = 1$. Thus: $(a^2,b^2,ab) = (c^2p^2, c^2q^2,c^2pq) = c^2(p^2,q^2,pq)$. We need to prove: $(p^2,q^2,pq) = 1$. We have: $(p^2,q^2,pq) = d \Rightarrow d \mid p^2, d\mid q^2 \Rightarrow d \mid (p^2,q^2) = 1$ since $(p,q) = 1 \Rightarrow d = 1$. Thus the problem is solved.
Note: They mentioned to you that use the basic facts about GCD. I think they wanted to talk about the property that: $ (p,q) = 1 \iff (p^2,q^2) = 1$.
