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Math Help - Diophantine Equation

  1. #1
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    Diophantine Equation

    I need help solving this Diophantine Equation I have. Its X^2+Y^2+1=3XY
    The answers must be integers and the only one I found was x=1, y=1.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Liu997 View Post
    I need help solving this Diophantine Equation I have. Its X^2+Y^2+1=3XY
    The answers must be integers and the only one I found was x=1, y=1.
     x^2+y^2+1=3xy \implies y=\frac{3x\pm\sqrt{5x^2-4}}{2}

    This means  5x^2-4 must be square and  3x\pm\sqrt{5x^2-4} must be even to have a solution.

    So,  x=\pm 1, \pm 2, \pm 5 are some solutions...
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  3. #3
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    Quote Originally Posted by chiph588@ View Post
     x^2+y^2+1=3xy \implies y=\frac{3x\pm\sqrt{5x^2-4}}{2}

    This means  5x^2-4 must be square and  3x\pm\sqrt{5x^2-4} must be even to have a solution.

    So,  x=\pm 1, \pm 2, \pm 5 are some solutions...
    But I need the x and y variables that work even though they are interchangeable in this equation. I found another one using excel x=5, y=2

    I bow down to who can give me like a reason for this
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Liu997 View Post
    But I need the x and y variables that work even though they are interchangeable in this equation. I found another one using excel x=5, y=2

    I bow down to who can give me like a reason for this
    I don't quite know what you just said here...


    We can narrow down some possible solutions by looking modulo  n .

    Looking modulo  3 :
     3xy\equiv0\mod{3}\implies x^2+y^2+1\equiv0\mod{3}\implies x^2,y^2\equiv1\mod{3}\implies x and  y can't be multiples of  3 .

    I'll let you know of any more progress I make.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chiph588@ View Post
     x^2+y^2+1=3xy \implies y=\frac{3x\pm\sqrt{5x^2-4}}{2}

    This means  5x^2-4 must be square and  3x\pm\sqrt{5x^2-4} must be even to have a solution.

    So,  x=\pm 1, \pm 2, \pm 5 are some solutions...
    I claim  3x\pm\sqrt{5x^2-4} will always be even (can you see why?), so we don't need to worry about that.

    So solving our equation is equivalent to solving  5x^2-n^2=4 (This almost looks like Pell's equation!).
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Here's all the solutions for  |x| \leq 1000 :

     x= \pm610,\pm233,\pm89,\pm34,\pm13,\pm5,\pm2,\pm1
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Break through!

    Let's just look at the solutions when  x is positive.

    Let  \{x_1,x_2,\cdot\cdot\cdot,x_n,\cdot\cdot\cdot\} be all ordered  x solutions i.e.  x_1=1, x_2=2, x_3=5,...

    I claim the solutions for  y paired with  x_n is  y=x_{n-1},x_{n+1} . (Can you see why?)

    This is big because now we can find all solutions quite easily!

     x_{n+1} = \frac{3x_n+\sqrt{5x_n^2-4}}{2} . Let's see if you can find the formula for  x_n now.
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  8. #8
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    Here's all the solutions for  |x| \leq 1000 :

     x= \pm610,\pm233,\pm89,\pm34,\pm13,\pm5,\pm2,\pm1
    Those are the even-labeled Fibonacci numbers!
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Bruno J. View Post
    Those are the even-labeled Fibonacci numbers!
    Precisely! (Well almost: You mean odd-labeled Fibonacci numbers.)

    Now I guess all that's left is to show the odd-labeled Fibonacci numbers satisfy  x_{n+1} = \frac{3x_n+\sqrt{5x_n^2-4}}{2} .

    I'm having a lot of trouble showing this though... what's your take on this?
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  10. #10
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    Precisely! (Well almost: You mean odd-labeled Fibonacci numbers.)
    I count from zero!


    Now I guess all that's left is to show the odd-labeled Fibonacci numbers satisfy  x_{n+1} = \frac{3x_n+\sqrt{5x_n^2-4}}{2} .

    I'm having a lot of trouble showing this though... what's your take on this?
    I haven't tried yet; I've got too many things to do tonight! No good idea comes to mind though. I'll give it a shot tomorrow if I can.
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  11. #11
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Bruno J. View Post
    I count from zero!



    I haven't tried yet; I've got too many things to do tonight! No good idea comes to mind though. I'll give it a shot tomorrow if I can.
    Yeah, I've been too busy to hit this head on too...
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  12. #12
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    Precisely! (Well almost: You mean odd-labeled Fibonacci numbers.)

    Now I guess all that's left is to show the odd-labeled Fibonacci numbers satisfy  x_{n+1} = \frac{3x_n+\sqrt{5x_n^2-4}}{2} .

    I'm having a lot of trouble showing this though... what's your take on this?
    One more thing: you need to show this is the only sequence that generates positive solutions. I've figured out a way, but let's see you try first.



    We want to show  F_{2n+3}=\frac{3F_{2n+1}+\sqrt{5F_{2n+1}^2-4}}{2} .

    Now  F_{2n+3} = F_{2n+2}+F_{2n+1} = F_{2n+1}+F_{2n}+F_{2n+1} = 2F_{2n+1}+F_{2n} .

    Therefore it suffices to show  \sqrt{5F_{2n+1}^2-4} = F_{2n+1}+2F_{2n} . (Can you see why?)


    So...  \sqrt{5F_{2n+1}^2-4} = F_{2n+1}+2F_{2n} \iff F_{2n+1}F_{2n}+F_{2n}^2 = F_{2n+1}^2-1 .

    This last result can be shown inductively (I omit the details, but have verified it).

    Since we know this is the only sequence of positive integers that satisfies this equation, we have all solutions to  x^2+y^2+1=3xy with  x>0 :
     (x,y) = \{(F_{2n+1},F_{2n-1}),\; (F_{2n+1},F_{2n+3}) \;|\; n\geq0\} . (When  n=0 ,  F_{2n-1} = F_{-1} = 1 .)
    The solutions with  x<0 follow analogously.
    Last edited by chiph588@; March 25th 2010 at 02:07 PM.
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  13. #13
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    Thanks guys. I also see that this equation is very related to
    X^2+Y^2+2=XYZ which was what I actually need to solve and find relations to the Fibonacci sequence.
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  14. #14
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Liu997 View Post
    Thanks guys. I also see that this equation is very related to
    X^2+Y^2+2=XYZ which was what I actually need to solve and find relations to the Fibonacci sequence.
    Well excel's telling me there's only solutions when  z=\pm4 . If only this was a proof!

    So I'd say this is going to be a two step process:

    1.) Show  z=\pm4 generates the only solutions.

    2.) Find these solutions.
    Last edited by chiph588@; March 29th 2010 at 06:53 PM.
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  15. #15
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    I already did use the excel to show the answers for that equation. Since I received this Equation as an assignment, I could really use some help. My teacher pointed me to the direction of assigning Fibonnacci numbers as algebraic expressions. Sorry for the late responses I usually don't have access to a computer regularly.
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