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Thread: solve in Z

  1. #1
    Super Member dhiab's Avatar
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    solve in Z

    Solve in Z :
    $\displaystyle

    x=20-\sqrt{20-\sqrt{x}}
    $
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  2. #2
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    Quote Originally Posted by dhiab View Post
    Solve in Z :
    $\displaystyle

    x=20-\sqrt{20-\sqrt{x}}
    $
    hi
    you can put $\displaystyle x=u^2$.
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  3. #3
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    Or just say $\displaystyle x=16$.
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  4. #4
    Super Member dhiab's Avatar
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    Hllo : $\displaystyle x=u^{2}...(x\geq 0)$
    I'have :
    $\displaystyle
    20-u^{2}=\sqrt{20-u}
    $
    $\displaystyle u^{4}-40u^{2}+u+380=0$
    $\displaystyle (u-4)(u+5)(u^{2}-u-19)=0$
    (u-4=0)or (u+5=0)or($\displaystyle u^{2}-u-19=0$)=0
    conclusion :
    $\displaystyle
    u=4
    $
    but u=-5 is not solution ( $\displaystyle 20-u^{2}\geq 0$ )
    the equation : $\displaystyle u^{2}-u-19=0 $ have not solution in Z.
    Last edited by dhiab; Feb 16th 2010 at 01:38 AM.
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  5. #5
    Super Member Bacterius's Avatar
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    $\displaystyle x=20-\sqrt{20-\sqrt{x}}$

    $\displaystyle x - 20=-\sqrt{20-\sqrt{x}}$

    $\displaystyle 20 - x=\sqrt{20-\sqrt{x}}$

    $\displaystyle (20 - x)^2=20-\sqrt{x}$

    $\displaystyle 400 - 40x + x^2=20-\sqrt{x}$

    $\displaystyle 400 - 40x + x^2 - 20 + \sqrt{x}=0$

    $\displaystyle 380 - 40x + x^2 + \sqrt{x}=0$

    $\displaystyle x^2 - 40x + \sqrt{x} + 380=0$

    Let $\displaystyle u = \sqrt{x}$ :

    $\displaystyle u^4 - 40u^2 + u + 380=0$

    Solve this fourth degree polynomial (I can't do this yet so just assume you solve this for $\displaystyle u$)

    And you find that $\displaystyle u = -5$, $\displaystyle u = 4$, $\displaystyle u = -\frac{\sqrt{77}-1}{2}$, $\displaystyle u = \frac{\sqrt{77}+1}{2}$.

    If $\displaystyle u < 0$ is negative, then $\displaystyle \sqrt{x} < 0$, which is not possible, so if $\displaystyle u < 0$, it is not a solution in $\displaystyle \mathbb{Z}$. The last solution is positive, but when squared does not give a relative number, so it is not a solution in $\displaystyle \mathbb{Z}$.

    Substitute back the only potential solution into $\displaystyle x$ ($\displaystyle x = u^2$) to find out that the only solution in $\displaystyle \mathbb{Z}$ is :

    $\displaystyle x = 4^2 = 16$

    Conclusion : $\displaystyle x=20-\sqrt{20-\sqrt{x}}$ admits one solution in $\displaystyle \mathbb{Z}$ : $\displaystyle x_1 = 16$.
    Last edited by Bacterius; Feb 16th 2010 at 02:20 AM.
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  6. #6
    Super Member Bacterius's Avatar
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    Another approach :

    Note that the only perfect squares under $\displaystyle 20$ are $\displaystyle 16$, $\displaystyle 9$, $\displaystyle 4$ and $\displaystyle 1$. Denote $\displaystyle a^2$ any of these squares. If $\displaystyle 20 - \sqrt{a^2} = 20 - a = b^2$, $\displaystyle b \in \mathbb{Z}$, then $\displaystyle b^2$ is a solution for $\displaystyle x$.

    $\displaystyle 20 - \sqrt{1} = 20 - 1 = 19$
    $\displaystyle 20 - \sqrt{4} = 20 - 2 = 18$
    $\displaystyle 20 - \sqrt{9} = 20 - 3 = 17$
    $\displaystyle 20 - \sqrt{16} = 20 - 4 = 16 = 4^2$

    If $\displaystyle a^2 = 16$, then we get a perfect square, thus the only solution in $\displaystyle \mathbb{Z}$ for your equation, is $\displaystyle x_1 = 16$.



    This is far simpler than my other fully algebra´c method, but it is conceptually harder to grasp, so it needs to be written a little better. Any tips ?
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  7. #7
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    Quote Originally Posted by dhiab View Post
    Hllo : $\displaystyle x=u^{2}...(x\geq 0)$
    I'have :
    $\displaystyle
    20-u^{2}=\sqrt{20-u}
    $
    $\displaystyle u^{4}-40u^{2}+u+380=0$
    $\displaystyle (u-4)(u+5)(u^{2}-u-19)=0$
    (u-4=0)or (u+5=0)or($\displaystyle u^{2}-u-19=0$)=0
    conclusion :
    $\displaystyle
    u=4
    $
    but u=-5 is not solution ( $\displaystyle 20-u^{2}\geq 0$ )
    the equation : $\displaystyle u^{2}-u-19=0 $ have not solution in Z.
    which means the only solution your equation has in $\displaystyle \mathbb{Z}$ is $\displaystyle x=16$.
    you might also consider the way "Bacterius" solved your problem,it's a very good method.
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