elementary number theory - Show that if a and b are positive integers, and $a^3 | b^2$, then a | b

Show that if a and b are positive integers, and $a^3 | b^2$, then a | b.

If p is a prime divisor of a, and $p^r$ is the highest power of p dividing a. Then $p^{3r} | a^3$, and so $p^{3r} | b^2$. If $p^s$ is the highest power of p dividing b, then 3r ≤ 2s. And so we have, r ≤ (2/3)s < s.

But how come this implies that $p^r | b$, and eventually a | b.

Answer

Since $a^2\mid a^3$, we have $a^2\mid b^2$. Then let $c=b/a$. We want to show $c\in \mathbb{Z}$. Note that $c$ satisfies $x^2-\frac{b^2}{a^2}$. This is monic in $\mathbb{Z}[x]$, and since any rational solution to a monic in $\mathbb{Z}[x]$ is actually an integer, we get that $c\in\mathbb{Z}$.

Thm: If $f(x)=x^n+a_{n-1}x^{n-1}+...+a_0$ is such that $a_i\in \mathbb{Z}$, and we have that for some $c\in \mathbb{Q}$ $f(c)=0$, then $c\in \mathbb{Z}$

Proof: Write $c=b/a$ with $c$ a reduced fraction (meaning $a$ and $b$ are coprime). Then we have:

$$(b/a)^n+a_{n-1}(b/a)^{n-1}+...+a_0=0$$ Multiply everything by $a^n$. You get that $b^n=aD$ for $D\in \mathbb{Z}$. Hence, $a\mid b^n$. If $a$ is not $\pm 1$, then there is a prime $p$ such that $p\mid a$. This means that $p\mid b^n$ which then implies that $p\mid b$. This contradicts the fact that $a$ and $b$ are coprime. Hence we get that $a=\pm 1$, so $c\in \mathbb{Z}$ just as we wanted.