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?

Let $\operatorname{lcm}(a,b)=\frac{ab}{\gcd(a,b)}$. Since the $\gcd$ divides both $a$ and $b$, it's clear from the definition that the $\operatorname{lcm}$ is an integer divisible by both $a$ and $b$. And if $a$ does not divide $b$, then the $\operatorname{lcm}$ is strictly greater than $b$, since $a\neq \gcd(a,b)$. By this question, the squareroot of an integer is either an integer or irrational, so since $a^2b^2|n^2$, $ab|n$.
Pick $x$ and $y$ so that $ax+by=\gcd(a,b)$. Then $\frac{n}{(\operatorname{lcm}(a,b))^2}=\frac{n}{a^2b^2}\cdot (ax+by)^2=\frac{n}{a^2b^2}\cdot (a^2x^2+2abxy+b^2c^2)=\frac{nx^2}{b^2}+\frac{2nxy}{ab}+\frac{ny^2}{a^2}$ is an integer.
