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Math Help - integers

  1. #1
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    integers

    If x and y are integers such that (x-y)^2 +2y^2 = 27 find the possible values of x.
    p.s that's the complete question.

    Any help would be so much appreciated!
    Thank you in advance!
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  2. #2
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    Quote Originally Posted by z1llch View Post
    If x and y are integers such that (x-y)^2 +2y^2 = 27 find the possible values of x.
    p.s that's the complete question.

    Any help would be so much appreciated!
    Thank you in advance!
    Solve this equation for x:

    x = y-\sqrt{27-2y^2}~\vee~ x = y+\sqrt{27-2y^2}

    Since 27-2y^2 \geq 0 the values of y are in |y| \in \{1,2,3\}

    Plug in these values and check if x is an integer. I've got 4 solutions:

    (4, -1),\ (-4, 1),\ (6, 3),\ (-6, -3)
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  3. #3
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    hi earboth. thanks for helping!
    i understand till the x=y-\sqrt{27-2y^2}
    can please explain why is 27-y^2 \geq 0
    and how to get the four solutions please?
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  4. #4
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    Quote Originally Posted by z1llch View Post
    hi earboth. thanks for helping!
    i understand till the x=y-\sqrt{27-2y^2}
    can please explain why is 27-y^2 \geq 0
    and how to get the four solutions please?
    Hi

    (x-y)^2 = 27 - 2y^2 and (x-y)^2 \geq 0
    Therefore you must have 27 - 2y^2 \geq 0
    Otherwise there are no integer solutions

    27 - 2y^2 \geq 0 for |y| = 0 or 1 or 2 or 3
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