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Math Help - Equation

  1. #1
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    Equation

    I need to find the integer solutions to this equation.
    (x+y)^2 = x^2y^2
    Any Idea?
    Thanks
    everk
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  2. #2
    MHF Contributor Amer's Avatar
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    xy = x+y or x+y = -xy
    the only solution is when x and y = 2 or x and y = -2

    suppose y>x
    then we can say y=kx+r k is an integer r<x

    x(kx +r ) = x + kx+r

    x (kx +r) = x(k+1) +r the left side is a multiple of x so the right should be so should x divides r but x>r then r=0

    so y = kx

    x(kx) = x + kx divde both sides by x

    kx = 1+k

     x = \frac{1+k}{k} but x is an integer so k should equal 1

    x = 2 and  y = 1x + 0 = x =2

    other case x+y = -xy same

    so xy both 2 or both -2
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  3. #3
    MHF Contributor harish21's Avatar
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    Hi Amer,

    Apart from x=y=2 and x=y=-2, wouldn't x=y=0 count as well?
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