MHB Proving Divisibility: Does a^n Divide b^n Imply a Divides b?

[evinda]

Hey! (Wave)

I am looking at the following exercise:

Show that $a^n \mid b^n \Rightarrow a \mid b$.

According to my notes,it is like that:

Let $a^n \mid b^n$.

Let $d=(a,b)$

Then, $a=d \cdot a_1 \\ b=d \cdot b_1$

$$(a_1,b_1)=1$$

$$b^n=k \cdot a^n, \text{ for a } k \in \mathbb{Z}$$

$$d^n \cdot b_1^n=k \cdot d^n \cdot a_1^n \Rightarrow b_1^n=k \cdot a_1^n$$

Therefore, $$ a_1 \mid b_1^n=\underset{n}{\underbrace{b_1 \cdot b_1 \cdots b_1}} \overset{(a_1,b_1)=1}{\Rightarrow} a_1 \mid \underset{n-1}{\underbrace{b_1 \cdot b_1 \cdots b_1}} \Rightarrow a_1 \mid \underset{n-2}{\underbrace{b_1 \cdot b_1 \cdots b_1}} \Rightarrow a_1 \mid b_1$$

So,we conclude that $b_1=l \cdot a_1, l \in \mathbb{Z}$

$$d \cdot b_1=l \cdot d \cdot a_1 \Rightarrow b \mid a \Rightarrow a \mid b$$

But...is it right?? (Thinking) We show that $(a_1,b_1)=1$ and then we conclude that $a_1 \mid b_1$... (Wasntme)

 

[I like Serena]

But...is it right?? (Thinking) We show that $(a_1,b_1)=1$ and then we conclude that $a_1 \mid b_1$... (Wasntme)

Hi! (Happy)

Then that must mean that $a_1=1$ doesn't it? (Thinking)

 

[evinda]

Hi! (Happy)

Then that must mean that $a_1=1$ doesn't it? (Thinking)

Oh,yes! (Nod) So,the solution is right,or not?? (Thinking)

 

[I like Serena]

$$d \cdot b_1=l \cdot d \cdot a_1 \Rightarrow b \mid a \Rightarrow a \mid b$$

Oh,yes! (Nod) So,the solution is right,or not?? (Thinking)

Almost.
But from $d \cdot b_1=l \cdot d \cdot a_1$ we cannot conclude that $b \mid a$. (Doh)
It's a good thing we can conclude that $a \mid b$! (Mmm)

 

[evinda]

Almost.
But from $d \cdot b_1=l \cdot d \cdot a_1$ we cannot conclude that $b \mid a$. (Doh)
It's a good thing we can conclude that $a \mid b$! (Mmm)

Oh,sorry! (Blush)(Blush) I accidentally wrote it like that.I wanted to write:

$$d \cdot b_1=l \cdot d \cdot a_1 \Rightarrow b=l \cdot a \Rightarrow a \mid b$$

Thank you very much! (Smile)