if $3^3 2^2 \ | a^2$ then $3^2 2 \ | a $ where a is integer

Question

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.

Answer 1

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 $2$

Answer 2

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.

Answer 3

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 $2\cdot3^2\mid a$.

Answer 4

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 } $.