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Math Help - solve in Z

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

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

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

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

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

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

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

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

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

    Let u = \sqrt{x} :

    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 u)

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

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

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

    x = 4^2 = 16

    Conclusion : x=20-\sqrt{20-\sqrt{x}} admits one solution in \mathbb{Z} : x_1 = 16.
    Last edited by Bacterius; February 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 20 are 16, 9, 4 and 1. Denote a^2 any of these squares. If 20 - \sqrt{a^2} = 20 - a = b^2, b \in \mathbb{Z}, then b^2 is a solution for x.

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

    If a^2 = 16, then we get a perfect square, thus the only solution in \mathbb{Z} for your equation, is 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 : x=u^{2}...(x\geq 0)
    I'have :
     <br />
20-u^{2}=\sqrt{20-u}<br />
    u^{4}-40u^{2}+u+380=0
    (u-4)(u+5)(u^{2}-u-19)=0
    (u-4=0)or (u+5=0)or( u^{2}-u-19=0)=0
    conclusion :
     <br />
u=4<br />
    but u=-5 is not solution ( 20-u^{2}\geq 0 )
    the equation : u^{2}-u-19=0 have not solution in Z.
    which means the only solution your equation has in \mathbb{Z} is x=16.
    you might also consider the way "Bacterius" solved your problem,it's a very good method.
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