Prove or disprove : $a^3\mid b^2 \Rightarrow a\mid b$

Question

I think it's true, because I can't see counterexamples.

Here's a proof that I am not sure of:

Let $p_1,p_2,\ldots, p_n$ be the prime factors of $a$ or $b$ \begin{eqnarray} a&=& p_1^{\alpha_1}\cdots p_n^{\alpha_n} \\ b&=& p_1^{\beta_1}\cdots p_n^{\beta_n} \\ a^3\mid b^2 &&\Rightarrow \frac{b^2}{a^3} \in \mathbb{Z} \\ &&\Rightarrow 2 \beta_i - 3\alpha_i \geq 0 \\ &&\Rightarrow 2 \beta_i \geq 3\alpha_i \text{ and } \beta_i \geq 0 \\ &&\Rightarrow 3 \beta_i \geq 3\alpha_i \\ &&\Rightarrow \beta_i \geq \alpha_i \\ &&\Rightarrow \beta_i - \alpha_i \geq 0 \\ &&\Rightarrow \frac{b}{a} \in \mathbb{Z}\\ &&\Rightarrow a\mid b \end{eqnarray} Is this proof correct ?

Answer 1

It's correct. Simpler: $\ a^2\mid a^3\mid b^2\Rightarrow\,(b/a)^2\! = n \in \Bbb Z\,\Rightarrow\, b/a \in \Bbb Z\,$ by RRT = Rational Root Test.

In detail: if $\,x = b/a\,$ is a root of $\,f(x) = \color{#C00}1\cdot x^2 - n\,$ then RRT implies the least denominator of $\,x\,$ must divide $\color{#c00}1\ (= $ lead coef of $f),\,$ so $\,x\in\Bbb Z,\,$ being a rational writable with denominator $= \color{#c00}1.$

Answer 2

$a^3|b^2\implies a^2|b^2\implies b^2=a^2k$. Since $a^2k$ is a square we must have $k$ is a square, Since otherwise $\sqrt{a^2k}=a\sqrt k$ which would be irrational. From here $b^2=a^2c^2\implies b=ac\implies a\mid b$