TITLE: If $b^2$ is the largest square divisor of $n$ and $a^2 \mid n$, then $a \mid b$.
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: if you want to do it by contradiction starting with supposing $b^2$ is the largest square divisor of $n$, $a^2|n$ and $a \nmid b$. 
Then let $c=\frac{a}{gcd(a,b)}$. Since $a \nmid b$ we know that $a \neq b$ so $c>1$. Now, we can say that $c|a \implies c^2|a^2 \implies c^2|n$. Also, since $c$ and $b$ are relatively prime (by how we defined $c$), we can say $n=xb^2c^2=x(bc)^2$  since we are dealing in positive integers, it follows that $b < bc \implies b^2 < (bc)^2$ thus we have a contradiction.
${\bf Note:}$ You must divide out the gcd of a and b before multiplying to ensure they are relatively prime, otherwise you run the risk of getting a number larger than the original $b$
For example, say we have the number $2^23^25^2=900$ so I can say $15^2|900$ and $6^2|900$ but  $900\neq 15^26^2x$ where $x$ is an integer