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Equation with two variables (integers)

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve the equation (x & y are integers):
    [itex](x^3+4)(xy^2-x^2y+3y^2-12)=x^6[/itex]


    2. Relevant equations
    3. The attempt at a solution

    [tex]xy^2-x^2y+3y^2-12=\frac{x^6}{x^3+4} \\

    xy^2-x^2y+3y^2-12=x^3-4 + \frac{16}{x^3+4} \\

    16 \geq x^3+4 \\
    x^3 \leq 12

    [/tex]

    That's all I can think of to do. I've tried expanding it and it doesn't seem to help. Any hints, please? Thanks.
     
  2. jcsd
  3. Apr 24, 2012 #2

    NascentOxygen

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    Staff: Mentor

    Hi staples! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]
    Continuing on with your approach, we see that with all other terms being integers, then 16/(x3 +4) must also be a whole number, +ve or -ve, so that narrows it down to just a few possibilities for you to try for x candidates. :smile: Then try these one at a time to see whether you can find any corresponding y solution/s.
     
    Last edited by a moderator: May 5, 2017
  4. Apr 25, 2012 #3
    Thanks :)
     
  5. May 2, 2012 #4

    tiny-tim

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    ooh, nice welcome-smilie, NascentOxygen! :smile:

    (is that a self-portrait? o:))
     
    Last edited by a moderator: May 6, 2017
  6. May 4, 2012 #5

    NascentOxygen

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    Staff: Mentor

    Indeed. It's the spitting image. Smiley25-1.gif
     
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