1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?