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Please help with finding all integers to an equation!

  1. Jan 10, 2009 #1
    How can i find all integers to this equation?


    I allready found just by tring that if x=y then its 6 and x=12 ,y=4 (x=4 and y=12)

    I think that there are more, but how can i find all of them?

    PS. the first task was: 1/x + 1/y = 1/3 and then i had to find all integers.

    Thank you for your help!

    Sorry for my bad english.
  2. jcsd
  3. Jan 10, 2009 #2


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    I would write it in the form x=3*y/(y-3). For large values of y thats a little larger than 3 and not an integer. So once you tried a y value large enough that 3*y/(y-3)<4, there aren't any larger solutions. Then I'd just try all y values less than that to see if x is also an integer. Do you need negative solutions as well?
  4. Jan 10, 2009 #3
    [tex]\frac{1}{x} + \frac{1}{y} = \frac{1}{3} \Rightarrow 3x + 3y = xy \Rightarrow xy - 3x - 3y + 9 = 9 \Rightarrow (x-3)(y-3) = 9[/tex].

    The number 9 has three positive divisors: 1, 3, and 9. You just have to solve each of the systems of equations to get all the solutions.
  5. Jan 10, 2009 #4


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    Very clever.
  6. Jan 11, 2009 #5
    Ok, thanks Dick and Snipez90 for your help!

    I have one question. xy-3x-3y+9=9 > (x-3)(y-3) How did you factor that one?

    No, but thanks for asking.

    EDIT:OH, No need for help on this question any more. Just figured it out how to do that :D
    Last edited: Jan 11, 2009
  7. Jan 11, 2009 #6
    By inspection I would imagine, you have a cross term xy so you know that if you will have a product of linear factors it will be of the form (x-alpha)*(y-beta) [to give you that xy term] and then the choices for alpha and beta are obvious once you write out the product.
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