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Thread: Puzzle with numbers

  1. #1
    Contributing Member
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    May 2006
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    75

    Puzzle with numbers

    There are two real numbers and the sum of their reciprocals is equal to -1 and the sum of their cubes is equal to 4. Can you find the numbers?


  2. #2
    Expert Member
    Join Date
    Jun 2006
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    410

    Re: Puzzle with numbers

    Could you publish the solution to this problem?

    -- James


  3. #3
    Expert Member
    Join Date
    Jun 2006
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    410

    Re: Puzzle with numbers

    Found the solution to this problem here,

    http://www.qbyte.org/puzzles/p009s.html

    Solution:

    1/x + 1/y = -1 (1)
    x3 + y3 = 4 (2)

    (1) implies x + y = -xy
    (2) implies (x + y)3 - 3xy(x + y) = 4
    Hence -(xy)3 + 3(xy)2 - 4 = 0

    By inspection, xy = -1 is a solution of this cubic equation.
    Factorizing, we have (xy + 1)(xy - 2)2 = 0.
    Hence xy = -1, x + y = 1, or xy = 2, x + y = -2.

    If xy = -1 and x + y = 1, then x, y are roots of the quadratic equation u2 - u - 1 = 0.
    (Consider the sum and product of the roots of (u - A)(u - B) = u2 - (A + B)u + AB = 0.)
    Hence u = (1 ± root 5)/2.

    If xy = 2 and x + y = -2, then x, y are roots of u2 + 2u + 2 = 0.
    This has complex roots: u = -1 ± i.

    Therefore the real solutions are x = (1 ± root 5)/2, y = (1 minus or plus root 5)/2.

    -- James.


  4. #4
    Contributing Member
    Join Date
    Jul 2006
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    74

    Re: Puzzle with numbers

    A very interesting puzzle which gave me idea to solve an other puzzle. Thanks for posting this.


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