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Question on two variables

  1. Jul 3, 2011 #1
    I found a question on a Facebook page asked by someone on that page. The question is:-

    The questioner asked to find integer solutions for it. But i think it's not possible to solve unless we have one more equation. Am i right..?
  2. jcsd
  3. Jul 3, 2011 #2
    hey, if all they are asking is to find integer solution...who cares how? Clearly, this is the equation of a circle with a radius of about 5.19...so, you know that every x,y combination that satisfies the equation is no more than that...all you have to do now is try all integer combinations of x,y from -5 to 5 and see what happens....or just the reasonable combinations, if you want, knowing that x,y need to meet at the circle with 5.19 radius...you know what I mean..
  4. Jul 3, 2011 #3
    It would have been more fun if he had asked for all the rational solutions...
  5. Jul 3, 2011 #4
    Sorry gsal!! :frown:
    I don't understand what you said.
  6. Jul 3, 2011 #5
    Just plug in all integers from -5 to 5 and see if you get a solution.
  7. Jul 3, 2011 #6
    No, i am not getting the solution.
  8. Jul 3, 2011 #7
    Well, that means that there aren't any solutions.
  9. Jul 3, 2011 #8
    Ok thanks!! :smile:
  10. Jul 3, 2011 #9


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    Reduce modulo 4. What do you know about the quadratic residues mod 4?
  11. Jul 3, 2011 #10
    Sorry!! I don't understand you.
  12. Jul 4, 2011 #11

    I like Serena

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    How do you find the rational solutions? :confused:
    Or prove that there are none?
  13. Jul 4, 2011 #12
    There are none here. Finding rational solution to [itex]x^2+y^2=27[/itex] is the same as finding rational solutions to [itex]x^2+y^2=3[/itex] (divide everything by 9. And this is the same as finding nonzero integer solutions to [itex]x^2+y^2=3z^2[/itex]. Furthermore, we can assume that x,y and z have no common factors.

    Let's say that there is a solution, then we can look at it modulo 3. Thus we would have


    but checking all elements of [itex]\mathbb{Z}/3\mathbb{Z}[/itex] gives us that x=0,y=0 have to be solutions. Thus our orignal equation


    must have a solution of the form x=3m, y=3n. But then


    This implies that z would be divisible by 3. So x,y and z are divisible by 3 and thus have common factors. This is against the assumption.

    The general result of finding rational solutions on conics is given by Legendre's theorem: www.risc.jku.at/education/courses/ss2011/caag/proj-rat-points-conic.pdf
    Last edited: Jul 4, 2011
  14. Jul 4, 2011 #13
    Everything went over my head...:uhh:
  15. Jul 4, 2011 #14
    OK, let's do this in step.

    Let's assume that x and y are rational numbers that satisfy




    are rational numbers that satisfy


    Thus it suffices to find rational solutions to [itex]u^2+v^2=3[/itex]

    Now, u and v are rational, thus they have the form [itex]u=\frac{a}{c},~v=\frac{b}{c}[/itex] with a,b,c integers with no common factors and c nonzero. Then


    is equivalent to


    So it suffices to find nonzero integers a,b,c such that


    and such that a,b,c have no common factors.

    Do you understand it until here?
    Last edited: Jul 4, 2011
  16. Jul 4, 2011 #15

    I like Serena

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    Why wouldn't a and c have a common factor? :confused:

    (Btw, you didn't define v properly.)
  17. Jul 4, 2011 #16
    a and c can have a common factor. But a,b and c can't.
    Let's say that a, b, and c have a common factor m. Then a=ma', b=mb', c=mc'. And thus




    Thus I have eliminated the common factor. That is, if I have an integer solution (a,b,c) with a common factor, then I can find an integer solution (a',b',c') without a common factor. Thus it suffices to look at


    such that x,y,z do not have common factors.

    (Btw, you didn't define v properly.)[/QUOTE]
  18. Jul 4, 2011 #17

    I'd go with x = 6, y = 3i. Do Gaussian integers count?
    Last edited: Jul 4, 2011
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