Loading ...
Sorry, an error occurred while loading the content.
 

Diophantine equation

Expand Messages
  • Sebastian Martin Ruiz
    Hello all: Let c an even positive integer number fixed. Consider the quadratic equation in two variables: -3x^2+y^2-2xy-4cx+4cy+4=0 Can anyone prove that for
    Message 1 of 5 , May 24, 2013
      Hello all:

      Let c an even positive integer number fixed.


      Consider the quadratic equation in two variables:


      -3x^2+y^2-2xy-4cx+4cy+4=0

      Can anyone prove that for every c large enough there is at least one other than the trivial solution (1,1)  of the equation with x and y both positive integers? i.e. (y>x>1 integers)


      Sincerely

      Sebastián Martín Ruiz

      P.S. (It is related to the prime numbers. I'll tell you soon.)


      [Non-text portions of this message have been removed]
    • WarrenS
      ... --Let z=y-x. Then your equation is equivalent to 4x^2 - z^2 - 4cz = 4 with x 1 and z 0 integer unknowns. Now make the further change of variable t=z+2c.
      Message 2 of 5 , May 25, 2013
        > Hello all:
        >
        > Let c an even positive integer number fixed.
        >
        >
        > Consider the quadratic equation in two variables:
        >
        >
        > -3x^2+y^2-2xy-4cx+4cy+4=0
        >
        > Can anyone prove that for every c large enough there is at least one other than the trivial solution (1,1)  of the equation with x and y both positive integers? i.e. (y>x>1 integers)
        >
        >
        > Sincerely
        >
        > Sebastián Martín Ruiz

        --Let z=y-x. Then your equation is equivalent to
        4x^2 - z^2 - 4cz = 4
        with x>1 and z>0 integer unknowns.
        Now make the further change of variable t=z+2c.
        Then the equation is
        (2x)^2 - t^2 = 4+4c^2
        with x>0 and t>2c integer unknowns,
        which forces t to be even. Letting t=2r
        we finally transform the equation to the form
        x^2 - r^2 = 1+c^2
        in integer unknowns x>1 and r>c.

        This in turn may be written
        (x-r)*(x+r) = 1+c^2.

        Does a solution of this always exist if c is large enough?
        If c is even, then x=r+1 where 2r=c^2 will work.
        If c is odd, then there is no solution because
        right hand side is 2 mod 4 and since x and r must have same parity,
        the left hand side is divisible by 4.

        So, in conclusion, your conjecture is DISPROVEN.
        Assuming I did not make any stupid errors.

        --Warren D. Smith.
      • WarrenS
        ... --Let z=y-x. Then your equation is equivalent to 4x^2 - z^2 - 4cz = 4 with x 1 and z 0 integer unknowns. Now make the further change of variable t=z+2c.
        Message 3 of 5 , May 25, 2013
          > Hello all:
          >
          > Let c an even positive integer number fixed.
          >
          >
          > Consider the quadratic equation in two variables:
          >
          >
          > -3x^2+y^2-2xy-4cx+4cy+4=0
          >
          > Can anyone prove that for every c large enough there is at least one other than the trivial solution (1,1)  of the equation with x and y both positive integers? i.e. (y>x>1 integers)
          >
          >
          > Sincerely
          >
          > Sebastián Martín Ruiz

          --Let z=y-x. Then your equation is equivalent to
          4x^2 - z^2 - 4cz = 4
          with x>1 and z>0 integer unknowns.
          Now make the further change of variable t=z+2c.
          Then the equation is
          (2x)^2 - t^2 = 4+4c^2
          with x>0 and t>2c integer unknowns,
          which forces t to be even. Letting t=2r
          we finally transform the equation to the form
          x^2 - r^2 = 1+c^2
          in integer unknowns x>1 and r>c.

          This in turn may be written
          (x-r)*(x+r) = 1+c^2.

          Does a solution of this always exist if c is large enough?
          If c is even, then x=r+1 where 2r=c^2 will work.
          If c is odd, then there is no solution because
          right hand side is 2 mod 4 and since x and r must have same parity,
          the left hand side is divisible by 4.

          So, in conclusion, your conjecture is DISPROVEN.
          Assuming I did not make any stupid errors.

          --Warren D. Smith.
        • Sebastian Martin Ruiz
          you made a sign error (+-) in(*****) ________________________________ De: WarrenS Para: primenumbers@yahoogroups.com Enviado: Domingo 26
          Message 4 of 5 , May 26, 2013
            you made a sign error (+-) in(*****)



            ________________________________
            De: WarrenS <warren.wds@...>
            Para: primenumbers@yahoogroups.com
            Enviado: Domingo 26 de Mayo de 2013 6:15
            Asunto: [PrimeNumbers] Re: Diophantine equation



             

            > Hello all:
            >
            > Let c an even positive integer number fixed.
            >
            >
            > Consider the quadratic equation in two variables:
            >
            >
            > -3x^2+y^2-2xy-4cx+4cy+4=0
            >
            > Can anyone prove that for every c large enough there is at least one other than the trivial solution (1,1)  of the equation with x and y both positive integers? i.e. (y>x>1 integers)
            >
            >
            > Sincerely
            >
            > Sebastián Martín Ruiz

            --Let z=y-x. Then your equation is equivalent to
            4x^2 - z^2 - 4cz = 4
            with x>1 and z>0 integer unknowns.
            Now make the further change of variable t=z+2c.
            Then the equation is
            (2x)^2 - t^2 = 4+4c^2 (*******)
            with x>0 and t>2c integer unknowns,
            which forces t to be even. Letting t=2r
            we finally transform the equation to the form
            x^2 - r^2 = 1+c^2
            in integer unknowns x>1 and r>c.

            This in turn may be written
            (x-r)*(x+r) = 1+c^2.

            Does a solution of this always exist if c is large enough?
            If c is even, then x=r+1 where 2r=c^2 will work.
            If c is odd, then there is no solution because
            right hand side is 2 mod 4 and since x and r must have same parity,
            the left hand side is divisible by 4.

            So, in conclusion, your conjecture is DISPROVEN.
            Assuming I did not make any stupid errors.

            --Warren D. Smith.




            [Non-text portions of this message have been removed]
          • WarrenS
            ... --you are correct... trying again below. Hopefully no stupid errors this time, but I again am not checking carefully. ... --Let z=y-x. Then your
            Message 5 of 5 , May 26, 2013
              On 5/26/13, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
              > you made a sign error (+-) in(*****)

              --you are correct... trying again below. Hopefully no stupid errors this time, but
              I again am not checking carefully.


              >> Hello all:
              >>
              >> Let c an even positive integer number fixed.
              >>
              >>
              >> Consider the quadratic equation in two variables:
              >>
              >>
              >> -3x^2+y^2-2xy-4cx+4cy+4=0
              >>
              >> Can anyone prove that for every c large enough there is at least one other
              >> than the trivial solution (1,1) of the equation with x and y both
              >> positive integers? i.e. (y>x>1 integers)
              >>
              >>
              >> Sincerely
              >>
              >> Sebastián Martín Ruiz
              >
              --Let z=y-x. Then your equation is equivalent to
              4x^2 - z^2 - 4cz = 4
              with x>1 and z>0 integer unknowns.
              Now make the further change of variable t=z+2c.
              Then the equation is
              (2x)^2 - t^2 = 4-4c^2 (****have corrected sign error here***)
              with x>0 and t>2c integer unknowns,
              which forces t to be even. Letting t=2r
              we finally transform the equation to the form
              x^2 - r^2 = 1-c^2
              in integer unknowns x>1 and r>c.

              This in turn may be written
              (x-r)*(x+r) = 1-c^2.
              or equivalently
              (r-x)*(r+x) = c^2 - 1 = (c-1)*(c+1)
              Does a solution of this always exist if c is large enough?

              If c is even, then r=x+1 where 2r+2=c^2 will work.

              If c is odd (then my old erroneous disproof with the sign error
              fails to work) then since r>c and x>1 are demanded, the
              obvious solution r=c, x=1 is not eligible.

              But since the right hand side is divisible by 4 since c-1 and c+1 both are even,
              r=x+2, 4x+4=(c-1)*(c+1) which implies x=(c^2-5)/4 works.

              So, now your conjecture is PROVEN -- solutions always exist for c large enough.

              --
              Warren D. Smith
              http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
            Your message has been successfully submitted and would be delivered to recipients shortly.