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?