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  • paulmillscv@yahoo.co.uk
    Hi to all, A question was asked recently. Are there infinitely many of n, n+1 both square free? A simple bit of algebra came up with the following, BUT it
    Message 1 of 64 , Mar 30, 2001
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      Hi to all,
      A question was asked recently. Are there infinitely many of n,
      n+1 both square free? A simple bit of algebra came up with the
      following, BUT it may have important consequences. First the algebra,

      n is not a square => n = X^2 + a a not equal to 0
      n+1 is not a square => n+1 = Y^2 + b b not equal to 0

      => X^2 +a +1 = Y^2 + b. This is a 2nd order Diophantine equation
      in 2 variables and `obviously' (i.e. please look up the solution )
      has an infinite number of solutions.
      So… does this spell the beginning of the end of functional number
      theory type answers to `infinitely many of' type questions, OR was
      this just a one-off. Can anyone else think of other `infinitely
      many' conjectures which might be solved in this way? The point being
      that `infinitely many' is taken care of by the properties of the
      equation.


      Regards,
      Paul Mills
      Kenilworth,
      England.
    • Kermit Rose
      ... Thank you David. I had been confused by consideration of the following: if x y = z and y x, then if we set t = (y+x)/2 and s = (y-x)/2 then z = t^2 - s^2
      Message 64 of 64 , Oct 20, 2012
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        On 10/20/2012 10:07 AM, primenumbers@yahoogroups.com wrote:
        > 2.2. Re: Question
        > Posted by: "djbroadhurst"d.broadhurst@... djbroadhurst
        > Date: Fri Oct 19, 2012 9:02 am ((PDT))
        >
        >
        >
        > --- Inprimenumbers@yahoogroups.com,
        > Kermit Rose<kermit@...> wrote:
        >
        >> >Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
        >> >
        >> >for what values of x would
        >> >x^2 + 5 x + 6 be a perfect square.
        > Set A = 1, B = 0, C = -1, D = 5, E = 0, F = 6.
        >
        > and you will get the obvious answer from Dario:
        >
        > x = -2
        > y = 0
        > and also:
        > x = -3
        > y = 0
        > Calculation time: 0h 0m 0s
        >
        > David

        Thank you David.

        I had been confused by consideration of the following:

        if x y = z and y > x,
        then if we set t = (y+x)/2 and s = (y-x)/2

        then z = t^2 - s^2
        and x = y - 2 t.

        So z = x y = (y - 2t) y = y^2 - 2 t y

        y^2 - 2 t y - z = 0

        Transforming w = y - k yields

        (w + k)^2 - 2 t (w + k) - z = 0

        w^2 + (2 k - 2 t) w + (k^2 - 2 t k - z) = 0

        which has integral solution if and only if

        (2k - 2 t)^2 + (z + 2 t k - k^2) is an integral square

        In all that derivation, I forgot that I still did not know what value t was.

        I confused myself too easily.

        Kermit








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