As a warning, this whole thing may prove terribly useless. It will especially seem unnecessary for now, but I am hoping it helps with a later problem. Also, we’re just sort of testing out the MathJAX.

Lately, I have been looking into cases for $a,b,n\in\mathbb{N}$ where $$a\mid b^n \implies a\mid b.\tag{1}$$ While I have grown up on the definition of *divide* being $$a\mid b \iff \exists k\in\mathbb{Z}\, \text{ s.t. }\, a\cdot k = b,$$ this latest quest has led me to seek a new definition. So, I will first try to look at this definition, and then maybe we can look into some of the cases where $(1)$ is true.

First, let’s throw out when $a=1$. Since $1$ divides everything already, we will count this as trivial. We will write $a$ in terms of its prime factorization: $$a=\prod_{i=1}^{\infty}p_i^{\alpha_i}$$

Where each $p_i$ be the $i$th prime number, i.e. $p_3=5.$ To see an example, we are writing the number $12$ as $12=2^2\cdot 3^1\cdot 5^0\cdot 7^0\cdots$ where the multiplication is infinite, but at some point the $\alpha_i$ have become $0$.

Let the factorization of $b$ be written as $$b=\prod_{i=1}^{\infty}p_i^{\beta_i}.$$

Then, we have this different definition of *divide:* $$a\mid b \iff \alpha_i\leq \beta_i, \forall i$$

Now, looking back at $(1)$, let’s assume $a$ is prime. Then for some $i=k$ $\alpha_k = 1.$ Then notice that $a\mid b^n \implies 1 \leq n\beta_k.$ As we are only considering non-negative, this means that each of $n$ and $\beta_k$ must be $1$ or greater, which tells us that $\alpha_k\leq\beta_k.$ This is the definition of divides, and so $a\mid b,$ giving $(1).$ In the case of $a$ prime, this implication is already well-known. We simply used a different approach.

Indeed, even for composite $a$, if each $\alpha_i\leq 1$ we will find the same result. This gives us the “strongest” case so far, which is that $(1)$ holds for any squarefree $a.$