If [pa ∥ m] And [pb ∥ n], Then [pmin(a,b) ∥ m+n]

According to the fundamental theorem of arithmetic, \(m\) can be represented as:

$$ m = {p_1}^{k_1} * {p_2}^{k_2} * \cdots $$

But we are only interested in the \(p\) in question:

$$ m = p^a * \ldots $$

As for \(n\):

$$ n = p^b * \ldots $$

We can represent \(m+n\) as:

$$\begin{align} m+n &= (p^a * \ldots) + (p^b * \ldots) \\ &= p ^{\min \{ a, b \}} * ((\ldots) + ( \ldots)) \end{align} $$

Therefore, \(p^{\min(a, b)}\) exactly divides \(m+n\).

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