<s>[INST] if $3^3 2^2 \ | a^2$ then $3^2 2 \ | a $ where a is integer if $3^3 2^2 \ | a^2$ then  $3^2 2 \ |a $ where a is integer.
I just cannot see it. please explain this trivial remark.
 [/INST] I answered a generalization
of this here:
If $n \mid a^2 $, what is the largest $m$ for which $m \mid a$?
Here is the question
and my answer:
Given $n$, what is the largest $m$
such that $m \mid a$ for all $a$ with $n \mid a^2$?
Let
$n = \prod p_i^{n_i}
$,
$m = \prod p_i^{m_i}
$,
and
$a
=\prod p_i^{a_i}
$.
$n | a^2$
means that
$n_i \le 2a_i
$
or
$a_i
\ge \lceil \frac{n_i}{2} \rceil 
$.
$m | a$
means that
$m_i \le a_i$.
If $m$ is as large as possible,
$m_i = a_i$,
so
$m_i 
=  \lceil \frac{n_i}{2} \rceil 
$.
In this case,
$n = 3^32^2$,
so
$m 
= 3^{\lceil \frac{3}{2} \rceil }2^{\lceil \frac{2}{2} \rceil }
= 3^{2}2^{1 }
$.
</s>