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Thread: integers

  1. #1
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    integers

    If x and y are integers such that $\displaystyle (x-y)^2 +2y^2 = 27$ 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 $\displaystyle (x-y)^2 +2y^2 = 27$ 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:

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

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

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

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

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

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