elementary number theory - If $b^2$ is the largest square divisor of $n$ and $a^2 mid n$, then $a mid b$.

I am trying to prove this:

$n$, $a$ and $b$ are positive integers. If $b^2$ is the largest square
divisor of $n$ and $a^2 \mid n$, then $a \mid b$.

I want to prove this by contradiction, and I don't want to go via the fundamental theorem of arithmetic to do the contradiction. Can I prove this purely from properties of divisibility and GCD?

Since $b^2$ is the largest square divisor of $n$,

$$a^2 \le b^2 \implies a \le b.$$

Let us assume that $a \nmid b$. Now, I want to arrive at the fact that there is a positive integer $c$ such that $c > b$ and $c^2 \mid n$. This will be a contradiction to the assumption that $b^2$ is the largest square divisor of $n$.

How can I do this?

Answer

Hint $\ $ Suppose $\rm\:A^2\:|\:B^2C,\:$ and $\rm\:C\:$ is squarefree. Cancel $\rm\:(A,B)^2\:$ to get $\rm\:a^2\:|\:b^2 C,\ (a,b)=1.\:$ Hence $\rm\:a^2\:|\:C\:$ by Euclid's Lemma. But $\rm\:C\:$ is squarefree, so $\rm\:a=1,\:$ so $\rm\:A = (A,B),\:$ i.e. $\rm\:A\:|\:B.$

Remark $\ $ Above is $(1\Rightarrow 2)$ in the $5$ characterizations of squarefree that I gave here.

The Bezout-based proof in Brett's answer can be expressed more concisely as follows:
$$\rm a^2\:|\:b^2c\:\Rightarrow\:a^2\:|\:(bc)^2\:\Rightarrow\: a\:|\:bc\Rightarrow\: a^2\:|\:a^2c,abc,b^2c\:\Rightarrow\: a^2\:|\:(a,b)^2c\:\Rightarrow\: (a/(a,b))^2\:|\:c$$