If [pa ∥ m], Then [pan ∥ mn]

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:

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

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

$$\begin{align} m^n &= (p^a * \ldots)^n \\ &= (p^a)^n * (\ldots)^n \\ &= p^{an} * (\ldots)^n \end{align} $$

Therefore, \(p^{an}\) exactly divides \(m^n\).

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