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Homework Help: Does a solution exist?

  1. Apr 7, 2005 #1
    I've been asked to find all possible pairs (a,b) of integers such that [tex]a^3 = 5b^3[/tex].

    I started by considering the cases where a and b share a common factor, so that [tex]a = kl[/tex] and [tex]b = km[/tex]. But this can be reduced to the form [tex]l^3 = 5m^3[/tex], so we can assume WLOG that a and b are coprime.

    If this is the case, then [tex]5 \mid a^3[/tex], so [tex]5 \mid a[/tex]. Let [tex]a = 5x[/tex].

    [tex](5x)^3 = 5b^3[/tex]
    [tex]125x^3 = 5b^3[/tex]
    [tex]25x^3 = b^3[/tex]

    So [tex]25 \mid b^3[/tex] and [tex]5 \mid b[/tex]. Let [tex] b = 5y[/tex].

    This contradicts our earlier assumption that both a and b are coprime, and hence no solution exists.

    Have I made any mistakes?
  2. jcsd
  3. Apr 7, 2005 #2
    Well, there's one solution!

    [tex]0^3 = 5(0)^3[/tex]

    Other than that, your answer is fine. You just need to modify it for the zero case (if a=b=0 then one of your steps doesn't work).
    Last edited: Apr 7, 2005
  4. Apr 7, 2005 #3
    Ah, of course...there's always that pesky number called zero!!!

    Thanks for pointing that out Data!
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