Proving Divisibility: How to Show b|a When b3|a2 - Helpful Tips"

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Discussion Overview

The discussion revolves around the problem of proving that if \( b^3 \) divides \( a^2 \), then \( b \) must divide \( a \). Participants explore various approaches, including prime factorization and modular arithmetic, while seeking clarification and deeper insights into the relationships between the variables involved.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using prime factorizations as a method to approach the problem.
  • Another participant proposes that since \( b^3 | a^3 \), it may be useful to explore the implications of this relationship.
  • A participant expresses uncertainty about how to proceed after establishing \( a^3 = mb^3 \) for some integer \( m \), questioning the next steps in the proof.
  • There is a query about whether it can be proven that \( m \) is a cube, indicating a potential avenue for exploration.
  • Another participant raises a question about the nature of cubes, specifically whether multiplying a cube by a non-cube results in a cube.
  • A different approach is presented where it is stated that if \( b^2 | a^2 \), then \( b | a \), with a reasoning process involving square roots and integer properties.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to prove the statement, and multiple competing views and approaches remain present throughout the discussion.

Contextual Notes

Some assumptions regarding the properties of integers and divisibility are implicit in the discussion, and the participants do not fully resolve the mathematical steps involved in their reasoning.

Who May Find This Useful

Readers interested in number theory, particularly in divisibility and properties of integers, may find the discussion relevant.

audiowize
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Hello,

If we are given that b3|a2, how do we show that b|a?

I started off looking at prime factorizations, but I could use a push in a more substantial direction.
 
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Prime factorizations are one way to go.

Or you could observe that b^3 | a^3.
 
Last edited:
I know that since [tex]b^3 | a^3[/tex]

Then [tex]a^3 = mb^3[/tex] ( [tex]\exists m \in \mathbb{Z}[/tex]). And [tex]a^3 \equiv b^3 \pmod m[/tex].

But I don't know where to go from here... :confused:
 
roam said:
I know that since [tex]b^3 | a^3[/tex]

Then [tex]a^3 = mb^3[/tex] ( [tex]\exists m \in \mathbb{Z}[/tex]). And [tex]a^3 \equiv b^3 \pmod m[/tex].

But I don't know where to go from here... :confused:

Can you prove that m is a cube?
 
hamster143 said:
Can you prove that m is a cube?

Hmm, that's a good idea but I'm not sure if it's possible to prove that...
 
roam said:
Hmm, that's a good idea but I'm not sure if it's possible to prove that...

If a cube is multiplied by a non-cube, is it still a cube?
 
[tex]b^2|b^3|a^2[/tex] so [tex]b^2|a^2[/tex]

Can you show that if [tex]b^2|a^2[/tex] then [tex]b|a[/tex]

Here's how I did it:
a and b are integers such that b^2|a^2 => mb^2 = a^2
Since the square root of a^2 = a, then we can take the sqrt(mb^2) = b*sqrt(m) = a.
Since a and b are integers, sqrt(m) must also be an integer
=> b|a

If you want to see some other ways check out https://www.physicsforums.com/showthread.php?t=337837&page=2"
 
Last edited by a moderator:

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