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

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The discussion centers on proving that if \( a^n \) divides \( b^n \), then \( a \) must divide \( b \). The initial approach involves expressing \( a \) and \( b \) in terms of their greatest common divisor \( d \) and coprime factors \( a_1 \) and \( b_1 \). The conclusion drawn is that since \( (a_1, b_1) = 1 \), it follows that \( a_1 \) divides \( b_1 \), leading to the assertion that \( a \) divides \( b \). However, there is a clarification needed regarding the implications of \( d \cdot b_1 = l \cdot d \cdot a_1 \), which does not imply \( b \) divides \( a \). Ultimately, the solution confirms that \( a \mid b \) is indeed correct.
evinda
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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) :confused:
 
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evinda said:
But...is it right?? (Thinking) We show that $(a_1,b_1)=1$ and then we conclude that $a_1 \mid b_1$... (Wasntme) :confused:

Hi! (Happy)

Then that must mean that $a_1=1$ doesn't it? (Thinking)
 
I like Serena said:
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)
 
evinda said:
$$d \cdot b_1=l \cdot d \cdot a_1 \Rightarrow b \mid a \Rightarrow a \mid b$$

evinda said:
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)
 
I like Serena said:
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)
 
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