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Math Help - System in Z3

  1. #1
    Super Member dhiab's Avatar
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    System in Z3

    Find x,y,z in Z solutions for system :
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  2. #2
    MHF Contributor red_dog's Avatar
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    \left\{\begin{array}{ll}x+y=z=12\\x^2+y^2=z^2+12\e  nd{array}\right.

    Let s=x+y, \ p=xy

    Then \left\{\begin{array}{ll}s=z+12\\s^2-2p=z^2+12\end{array}\right.

    Replace s in the second equation. Then p=12z+66. The quadratic equation having the roots such as their sum is s and their product is p is t^2-st+p=0

    That means t^2-(z+12)t+12z+66=0. The roots of this equation are x and y and are integers. Then the discriminant must be a perfect square.

    \Delta=(z-12)^2-264=k^2\Rightarrow(z-k-12)(z+k-12)=264

    \left\{\begin{array}{ll}z-k-12=2\\z+k-12=132\end{array}\right.\Rightarrow z=79

    Replace z in t-equation: t^2-91t+1014=0 with the roots 13 and 78. So, x=13, \ y=78, \ z=79 or x=78, \ y=13, \ z=79

    You can try other cases:

    \left\{\begin{array}{ll}z-k-12=4\\z+k-12=66\end{array}\right. or \left\{\begin{array}{ll}z=k-12=8\\z+k-12=33\end{array}\right. etc.
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by red_dog View Post
    \left\{\begin{array}{ll}x+y=z=12\\x^2+y^2=z^2+12\e  nd{array}\right.

    Let s=x+y, \ p=xy

    Then \left\{\begin{array}{ll}s=z+12\\s^2-2p=z^2+12\end{array}\right.

    Replace s in the second equation. Then p=12z+66. The quadratic equation having the roots such as their sum is s and their product is p is t^2-st+p=0

    That means t^2-(z+12)t+12z+66=0. The roots of this equation are x and y and are integers. Then the discriminant must be a perfect square.

    \Delta=(z-12)^2-264=k^2\Rightarrow(z-k-12)(z+k-12)=264

    \left\{\begin{array}{ll}z-k-12=2\\z+k-12=132\end{array}\right.\Rightarrow z=79

    Replace z in t-equation: t^2-91t+1014=0 with the roots 13 and 78. So, x=13, \ y=78, \ z=79 or x=78, \ y=13, \ z=79

    You can try other cases:

    \left\{\begin{array}{ll}z-k-12=4\\z+k-12=66\end{array}\right. or \left\{\begin{array}{ll}z=k-12=8\\z+k-12=33\end{array}\right. etc.
    HELLO :THANK YOU
    I'have a simpl solution :

    Find numbers in Z by divid 66.......(66=2113 )
    12 solutions LOOK HERE
    Attached Files Attached Files
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