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X^2 + y^2

  1. Feb 9, 2007 #1
    i have to find x2 + y2 if x and y are positive integers and
    xy + x + y = 71
    xy(x + y) = 880

    this is what i have done:
    since x and y divides 880, x or y must be 2, 5 or 11, because 880 = 24.5.11

    from the first equation i can see that
    x = 11 and y = 5 or x = 5 and y = 11

    so x2 + y2 = 146

    but is there a way to do it algebraically?

    thanks in advance.
  2. jcsd
  3. Feb 9, 2007 #2
    one thing you could do is from equation 1 set x+y=71-xy, plug that into equation 2 to get xy(71-xy)=880. Set xy=z and you have a quadratic equation in which one solution is 55, therefore xy=55 and you get the same solution (you also get a solution of xy=16...but that doesn't fit your first equation)
  4. Feb 9, 2007 #3

    D H

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    You can completely solve for [itex]x^2+y^2[/itex] algebraically, and eliminate the non-integer solutions as a final step. Make the substitutions [itex]u=xy[/itex], [itex]v=x+y[/itex]. Then [itex]x^2+y^2=v^2-2u[/itex]. With those substitutions, the two given equations become

    [tex]u+v = 71[/tex]

    [tex]uv = 880[/tex]

    The two solutions to the given equations are [itex]u=55[/itex] and [itex]u=16[/itex]. The goal is to solve for [itex]x^2+y^2[/itex]. Note that
    [tex]x^2+y^2=v^2-2u = (71-u)^2-2u = 71^2-144u+u^2[/tex]

    Applying the two solutions for [itex]u[/itex] yields [itex]x^2+y^2 = 146 (u=55)[/itex] and [itex]x^2+y^2 = 2993 (u=16)[/itex].

    As a final step, note that the former solution, [itex]x^2+y^2 = 146[/itex], corresponds to [itex]u=55,v=16[/itex], which has integer solutions in [itex]x[/itex] and [itex]y[/itex]. The solution [itex]x^2+y^2 = 2993[/itex] corresponds to [itex]u=16,v=55[/itex], which has irrational solutions in [itex]x[/itex] and [itex]y[/itex]. Thus [itex]x^2+y^2 = 146[/itex].
    Last edited: Feb 9, 2007
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