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Divisibility Question

  1. Oct 28, 2009 #1

    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.
  2. jcsd
  3. Oct 28, 2009 #2
    Prime factorizations are one way to go.

    Or you could observe that b^3 | a^3.
    Last edited: Oct 28, 2009
  4. Oct 29, 2009 #3
    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:
  5. Oct 29, 2009 #4
    Can you prove that m is a cube?
  6. Oct 29, 2009 #5
    Hmm, that's a good idea but I'm not sure if it's possible to prove that...
  7. Oct 29, 2009 #6
    If a cube is multiplied by a non-cube, is it still a cube?
  8. Oct 29, 2009 #7
    [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: Apr 24, 2017
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