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  1. #1
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    Number Theory (1)

    This interesting question is taken from an Iranian () competetion (high school level): (Have fun!)

    Suppose $\displaystyle x,y,z$ are positive integers with $\displaystyle xy=z^2+1.$ Prove that there exist integers $\displaystyle a,b,c,d$ such that $\displaystyle x=a^2+b^2, \ y=c^2+d^2,$ and $\displaystyle z=ac+bd.$
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  2. #2
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    you mean

    $\displaystyle (a^2 + b^2 )(c^2 + d^2 ) = (ac + bd )^2 + 1 \implies$

    $\displaystyle (ad - bc )^2 = 1 $

    $\displaystyle ad - bc = 1 $ or $\displaystyle bc - ad = 1 $

    I remember that If $\displaystyle (A,B) = 1 $ , then there exists integers

    $\displaystyle A,B $ such that $\displaystyle Ax + By = 1$ , $\displaystyle (x,y) \in \mathbb{N} $

    Is it correct if $\displaystyle (a,b) = (a,c) = (b,d) = (c,d) = 1 $

    there also exists integers $\displaystyle a,b,c,d $ satisfying

    $\displaystyle ad - bc = 1 $ or $\displaystyle bc - ad = 1 $ ?


    For exmaple , set $\displaystyle a = 5 , c = 3 $

    $\displaystyle 5d - 3b = 1 $ , i obtain $\displaystyle b = 3 + 5t , d = 2 + 3t $

    let $\displaystyle t = 0$

    therefore

    $\displaystyle x = a^2 + b^2 = 25 + 9 = 34 $

    $\displaystyle y = c^2 + d^2 = 9 + 4 = 13 $

    $\displaystyle z = ac + bd = 15 + 6 = 21 $

    $\displaystyle 34 \times 13 = 442 = z^2 + 1 $

    It is also true when $\displaystyle t = 1 $

    $\displaystyle a = 5 , b= 8 , c=3 , d=5 $

    $\displaystyle (5^2 + 8^2 )( 3^2 + 5^2 ) = \left ( (5)(3) + (8)(5) \right)^2 + 1$
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Oops
    Last edited by Bruno J.; Nov 22nd 2009 at 10:04 PM.
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  4. #4
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    Quote Originally Posted by Bruno J. View Post

    ...and $\displaystyle z=ac+bd$ follows by simple computation.
    are you sure about that?
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