According to the fundamental theorem of arithmetic:
$$ \begin{align} a &= {p_1}^{x_1} {p_2}^{x_2} {p_3}^{x_3} \cdots {p_n}^{x_n} \\ b &= {p_1}^{y_1} {p_2}^{y_2} {p_3}^{y_3} \cdots {p_n}^{y_n} \\ (a,b) &= {p_1}^{\min(x_1,y_1)} {p_2}^{\min(x_2,y_2)} {p_3}^{\min(x_3,y_3)}* \cdots{p_n}^{\min(x_n,y_n)} \\ [a,b] &= {p_1}^{\max(x_1,y_1)} {p_2}^{\max(x_2,y_2)} {p_3}^{\max(x_3,y_3)}* \cdots{p_n}^{\max(x_n,y_n)} \end{align} $$
Let \(p\) be a prime. Define \(s\) and \(t\) by \(p^s ∥ a\) and \(p^t ∥ b\). Suppose \(s \le t\). Since \(p\) divides \(p^s ∥ a\) and \(p^s | b\), then \(p^s | a + b\). If \(p\) divides \(a\) exactly \(s\) times, then it can't divide \(a+b\) more than \(s\) times, so \(p^s ∥ a+b\).
We know that \(p^{\max(s,t)} ∥[a,b]\) and \(p^{\min(s,t)} ∥ (a,b)\). If \(\max(s,t)=t\), then \(p^t ∥[a,b]\). If \(p^s ∥ a+b\) and \(p^t ∥[a,b]\), then \(p^{\min(s,t)}∥(a+b,[a,b])\).