# Formalizing why the prime factorizations of some $a^n$ must have a multiplicity that is evenly divisible by $n$

I have a quick (but messy) intuition for this property(?), which goes as follows.

Given $$a^n$$, then $$a^n = a \cdot a \, \cdot \, ... \, \cdot \, a$$, where there are $$n - 1$$ multiplication operations. $$a$$ has prime factors $$x_1, x_2, ..., x_i$$, where each factor has its own power $$m$$, that is, $$a = x_1^{m_1} \cdot x_2^{m_2} \, \cdot \, ... \, \cdot \, x_i^{m_i}$$.

With that, we have: \begin{align} a^n &= a \cdot a \, \cdot \! \underbrace{...}_\text{n times} \! \cdot \, a \\ &= (x_1^{m_1} \cdot x_2^{m_2} \, \cdot \, ... \, \cdot \, x_i^{m_i}) \cdot (x_1^{m_1} \cdot x_2^{m_2} \, \cdot \, ... \, \cdot \, x_i^{m_i}) \cdot \! \underbrace{...}_\text{n times} \! \cdot \, (x_1^{m_1} \cdot x_2^{m_2} \, \cdot \, ... \, \cdot \, x_i^{m_i}) \\ &= (x_1^{m_1} \cdot x_2^{m_2} \, \cdot \, ... \, \cdot \, x_i^{m_i})^n \\ &= x_1^{m_1n} \cdot x_2^{m_2n} \, \cdot \, ... \, \cdot \, x_i^{m_in} \end{align}

Here, we can see that every prime factor of $$a^n$$, $$x_k$$ for some $$k$$, is raised to a power that has factor of $$n$$. Therefore, $$n$$ divides the multiplicity of all factors of $$a^n$$.

Would this be enough for a proof?

• Not quite, as written; what you've written is enough to show that the multiplicity of a given prime $x_i$ is at least $m_i n$, but you need to say very slightly more (using the uniqueness of prime factorization) to conclude that it is exactly $m_i n$. Oct 28, 2022 at 22:43
• "$n$ divides the multiplicity of all factors of $a^n$" This is not true. What is true is that the exponent corresponding to any prime factor dividing $a$ in the prime factorization of $a^n$ is divisible by $n$. Oct 31, 2022 at 8:14

Let $$p$$ be a prime dividing $$a$$, where $$p$$ has multiplicity $$k$$ as a factor of $$a$$. This means $$p^k | a$$ but $$p^{k+1} \nmid a$$. Then $$p^{nk} \mid a^n$$ and $$p^{nk + 1} \nmid a^n$$ (if this last claim is not obvious to you, you should prove it. It isn't too hard). That is, $$p$$ has multiplicity $$nk$$ in the factorisation of $$a^n$$. This holds for any of the prime factors of $$a$$, so the result is proved.