<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] 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 $2\cdot3^2\mid a$.
</s>