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Math Help - Find positive integers

  1. #1
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    Find positive integers

    Hello, need your help thanks.

    Find all positive integers x,y,z satisfying simultaneously two conditions.

    i) \frac{x-y\sqrt{2009}}{y-z\sqrt{2009}} is a rational number .

    ii) x^2 + y^2+z^2 is a prime .
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  2. #2
    Super Member Showcase_22's Avatar
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    <br /> <br />
\frac{x-y\sqrt{2009}}{y-z\sqrt{2009}}=\frac{p}{q}<br />

    q(x-y\sqrt{2009})=p(y-z \sqrt{2009})

    qx-qy \sqrt{2009}=py-pz \sqrt{2009}

    qx+pz \sqrt{2009}=py+qy \sqrt{2009}

    qx+pz \sqrt{2009}=y(p+q \sqrt{2009})

    y=\frac{qx+\sqrt{2009}}{p+q \sqrt{2009}}

    but since y has to be an integer, then p=kqx and q=k where k \in \mathbb{N}.

    I'm not sure how this helps =S
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  3. #3
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    Hello, maria18!

    I found something, but I too haven't solved it yet.


    Find all positive integers x,y,z satisfying simultaneously two conditions:

    (1)\;\;\frac{x-y\sqrt{2009}}{y-z\sqrt{2009}} is a rational number.

    (2)\;\;x^2 + y^2+z^2 is a prime.

    Rationalize: . \frac{p}{q} \;=\;\frac{x-y\sqrt{2009}}{y-z\sqrt{2009}} \cdot \frac{y+z\sqrt{2009}}{y+z\sqrt{2009}} \;=\; . \frac{xy + xz\sqrt{2009} - y^2\sqrt{2009} - 2009yz}{y^2-2009z^2}

    . . . . . . . . . \frac{p}{q} \;=\;\frac{xy - 2009 + (xz - y^2)\sqrt{2009}}{y^2-2009z^2}


    Since p is an integer: . xz-y^2 \:=\:0 \quad\Rightarrow\quad xz \:=\:y^2


    Can we exploit that fact? . . . I don't know.

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  4. #4
    Moo
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    Quote Originally Posted by Showcase_22 View Post
    <br /> <br />
\frac{x-y\sqrt{2009}}{y-z\sqrt{2009}}=\frac{p}{q}<br />

    q(x-y\sqrt{2009})=p(y-z \sqrt{2009})

    qx-qy \sqrt{2009}=py-pz \sqrt{2009}

    qx+pz \sqrt{2009}=py+qy \sqrt{2009}

    qx+pz \sqrt{2009}=y(p+q \sqrt{2009})

    y=\frac{qx+{\color{red}pz}\sqrt{2009}}{p+q \sqrt{2009}}
    but since y has to be an integer, then p=kqx and q=k where k \in \mathbb{N}.

    I'm not sure how this helps =S
    I think it rather means that there exists k such that qx=pk and pz=qk
    I'm not sure either how this helps lol, but it has a good look !
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, maria18!

    I found something, but I too haven't solved it yet.



    Rationalize: . \frac{p}{q} \;=\;\frac{x-y\sqrt{2009}}{y-z\sqrt{2009}} \cdot \frac{y+z\sqrt{2009}}{y+z\sqrt{2009}} \;=\; . \frac{xy + xz\sqrt{2009} - y^2\sqrt{2009} - 2009yz}{y^2-2009z^2}

    . . . . . . . . . \frac{p}{q} \;=\;\frac{xy - 2009 + (xz - y^2)\sqrt{2009}}{y^2-2009z^2}


    Since p is an integer: . xz-y^2 \:=\:0 \quad\Rightarrow\quad xz \:=\:y^2


    hello and thanks Soroban

    I find==> with xz \:=\:y^2 we have x^2+y^2+z^2 = (x+z)^2 - xz = (x+z)^2 - y^2 = (x+z-y)(x+z+y)

    so (x+z-y)(x+z+y) is prime if ...
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  6. #6
    Super Member Showcase_22's Avatar
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    Thanks Moo for pointing that out.

    This question is actually harder than it looks!

    <br /> <br />
(x+z-y)(x+z+y)<br />
can only be prime if only one of the expressions in the brackets is 1.

    ummm.......
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  7. #7
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    Quote Originally Posted by Showcase_22 View Post

    (x+z-y)(x+z+y) can only be prime if only one of the expressions in the brackets is 1.
    we have obligatorily x+z-y=1 so x+z=1+y and we have xz= y^2
    so x and y are the roots of equation X^2-(y+1)X+y^2=0 so we get y \in [-\frac{1}{3} ; 1] or y=0 or y=1
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  8. #8
    Super Member Showcase_22's Avatar
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    You can use that to find all the other integers though can't you?

    (I haven't tried it myself, but by substituting them into the equationswe can obtain values for x and z. They're going to be different depending on the y value).
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