Divisibility Question

  • Thread starter audiowize
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  • #1
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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.
 

Answers and Replies

  • #2
907
2
Prime factorizations are one way to go.

Or you could observe that b^3 | a^3.
 
Last edited:
  • #3
1,266
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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:
 
  • #4
907
2
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?
 
  • #5
1,266
11
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...
 
  • #6
841
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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?
 
  • #7
22
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[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"
 
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