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.
 A: Let the highest power of $3$ in $a$ is $b$
$\implies$ the highest power of $3$ in $a^2$ is $2b$
Here $\displaystyle2b\ge3\implies b\ge\frac32$
As $b$ is an integer, $b\ge2$
Similarly, for the power of $2$
A: Hint $\rm\ 2^{\large\color{#c00}2} 3^{\large\color{#0a0}3} \mid 2^{\large\color{#c00}{2I}} 3^{\large\color{#0a0}{2J}} 5^{\large2K}\!\cdots \overset{\,}\Rightarrow \color{#c00}{I\ge 1},\ \color{#0a0}{J\ge 2}\ \ $ by existence and uniqueness of prime factorizations.
A: Here is a Bezout's Identity approach.

Suppose $9\nmid a$. Then $\gcd(a,9)\mid3$. Thus, there exist $x,y$ so that
$$
ax+9y=3\tag{1}
$$
Then
$$
a^2x^2+27\left(2y-3y^2\right)=9\tag{2}
$$
Thus, $\gcd\!\left(a^2,27\right)\mid9$. Then $27\nmid a^2$. By contraposition, we have
$$
27\mid a^2\implies9\mid a\tag{3}
$$

Suppose $2\nmid a$. Then $\gcd(a,2)=1$. Thus, there exists $x,y$ so that
$$
ax+2y=1\tag{4}
$$
Then
$$
a^2x^2+4\left(y-y^2\right)=1\tag{5}
$$
Thus, $\gcd\!\left(a^2,4\right)=1$. Then $4\nmid a^2$. By contraposition, we have
$$
4\mid a^2\implies2\mid a\tag{6}
$$

$$
\frac a{18}=\frac a2-4\frac a9\in\mathbb{Z}\tag{7}
$$
Therefore, $(3)$, $(6)$, and $(7)$ imply that if $2^2\cdot3^3\mid a^2$, then $2\cdot3^2\mid a$.
A: 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 }
$.
