What is the min amount of $n$ so that the following conclusion be true? $a^n | b^3 \to a^3 | b^2$

Question

What is the min amount of $n$ so that the following conclusion be true?

$a^n | b^3 \to a^3 | b^2$

We know If $a^n|b^3$ then there exists a $k \in \mathbb{Z}$ so that $b^3=k.a^n$ so :

\begin{align} b^3 =k.a^n\to b^2 =a^n.k.b^{-1}\to b^2 =a^3 . a^{n-3}.k.b^{-1} \end{align} So If we want $a^3 |b^2$ to be true, $a^{n-3}.k.b^{-1}$ must belong to $\mathbb{Z}$. We have to check for at least which $n$ this is true? for $n=4$ I found a counter example, but I couldn't find a clear way to find the min of $n$.

Answer 1

Recall if $\,q\in\Bbb Q,\, k\in\Bbb N_{>0}\ \ \color{#c00}{\rm then}\ \ q^k\in\Bbb Z\!\color{#c00}{\iff}\! q\in\Bbb Z,\,$ by the Rational Root Test. Using this we can power the quotients to put them over a common denominator, then the rest is straightforward.

Hence $\ \dfrac{b^{\large 3}}{a^{\large n}}\in \Bbb Z\,\Rightarrow\, \dfrac{b^{\large 2}}{a^{\large 3}}\in \Bbb Z\ $ is $\rm\color{#c00}{equivalent}$ to below (with $\,k=3,\,k=n)$

$\qquad\ \ \dfrac{b^{\large \color{#0a0}9}}{a^{\large 3n}}\in \Bbb Z\,\Rightarrow\, \dfrac{b^{\large \color{#0a0}{2n}}}{a^{\large 3n}}\in \Bbb Z,\,$ clearly true if $\,\color{#0a0}{9\le 2n},\,$ else false (e.g. let $\,a\!=\!c^{\large 3},\,b\!=\!c^{\large n},\,c>1)$

Answer 2

You already found counterexample for $n = 4$. So it suffices to show that $n = 5$ works.

Suppose we have $a^5 \mid b^3$. Taking squares gives $a^{10} \mid b^6$.

In particular, it implies $a^9 \mid b^6$, which in turn implies $a^3 \mid b^2$.