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Prize Puzzles concerning Primes and Composites of the form 20x^2 -1:

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  • aldrich617
    I offer a $25 prize to the first person who can submit a verifiable counterexample or proof for conjectures 5) or 7) below, and also a $25 prize for a proof of
    Message 1 of 2 , Dec 10, 2008
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      I offer a $25 prize to the first person who can submit a verifiable
      counterexample or proof for conjectures 5) or 7) below, and also a
      $25 prize for a proof of 6) or 8). The deadline is 1/19.

      x, A(x), k, T(k) : integers;

      Let A(x) = 20x^2 - 1;
      Let T(k) = 5*(2*k -1)^2 - 4*A(x) ;

      5) If x > 1 , then A(x) will be prime if no k value
      exists in the interval 2x < k < 3x such that T(k) is a
      square of an integer.

      Example: if x = 5, then A(x) = 499, 10< k < 15 and the
      values of T(k) are 209, 649, 1129 and 1649 .
      None of these is a square, therefore 499 is prime.

      6) If x > 1 , then A(x) will be composite if there exists at
      least one k value in the interval 2x < k < 3x such that T(k) is a
      square of an integer.

      Example: if x = 4, then A(x) = 319, 8< k < 12 and the
      values of T(k) are 169, 529 and 929 .
      Two of these are squares, therefore 319 is composite

      x, A(x),K,M,S,n,p : integers;

      Let A(x) = 20x^2 -1. For any value of x let K1 = x, K2 = 2x+1
      and let K2...Kn be consecutive integers with n denoting the
      maximal possible number of terms in a sequence for S1 = K1
      +K2...+ Kn within the interval 0 < n < x+1 and also for
      S2 = K1 + K2...+ Kn within the interval 1 < n < x+2.
      (note: 40*S1 + 9 and 40*S2 + 9 are always squares)

      Example: If x = 4, then S1 = 4 + 9 + 10 + 11 = 34 and n = 4.
      If x = 4, then S2 = 4 + 9 + 10 + 11 + 12 = 46 and n = 5.

      Using the same S1 and S2 values found for each x, find any shorter
      runs of the form S1 or S2 = M1 + M2...+ Mp, with p < n, M1 > x,
      M2 = 2M1 +1, and M2...Mp consecutive integers with p denoting
      the number of terms in the sequence.

      Example: For S1 = 34 from above it is evident that 11 + 23
      also equals 34, with M1 = 11 and p = 2.
      For S1 = 46 from above it is evident that 15 + 31
      also equals 46, with M1 = 15 and p = 2.

      7) If any value for S1 > 1, S2 or M1 found by the processes
      described above is substituted for x in A(x) = 20x^2 -1 then
      A(x) is always composite.

      Examples: if S1 = 34 then A(34) = 23119 = 61*379.
      if S1 = 34 and M1 = 11 then A(11) = 2419 = 41*59.
      if S2 = 46 then A(46) = 42319 = 101 * 419.
      if S2 = 46 and M1 = 15 then A(15) = 4499 = 11*409.

      8) If all possible composites from 7) above are removed from the
      sequence 20x^2 -1 , then all the remaining values of A(x) are prime.

      Aldrich Stevens
    • Bryan Bartlett
      Taking number 6... 6) If x 1 , then A(x) will be composite if there exists at least one k value in the interval 2x
      Message 2 of 2 , Dec 13, 2008
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        Taking number 6...
        <snip>
        6) If x > 1 , then A(x) will be composite if there exists at
        least one k value in the interval 2x < k < 3x such that T(k) is a
        square of an integer.

        Example: if x = 4, then A(x) = 319, 8< k < 12 and the
        values of T(k) are 169, 529 and 929 .
        Two of these are squares, therefore 319 is composite
        <snap>

        let x equal a composite number a * b

        therefore, A(x) = 20a^2b^2-1
        thus, T(k) = 5(2K-1)^2-80a^2b^2+4

        Simplified, T(k) = 20k^2-20k-80a^2b^2+9



        By your theory, there should be a value of k where 2ab<k<3ab where the
        above is a perfect square

        k can be redescribed as kab, where k is a number between 2 and 3

        so, 20k^2a^2b^2-20kab-80a^2b^2+9.

        we can rewrite this as follows...

        (20k^2-80)(a^2b^2)-20k(ab)+9

        now, we can begin possibly factoring this into a square.

        the 9 at the end, and the negative in the center, gives us part of the answer

        ( - 3)^2

        we will throw in another variable to represent the missing portion, called w

        (w-3)^2 will factor into w^2-6w+9

        using comparison, w^2 = (20K^2-80)(a^2b^2) and -6w = -20k(ab)

        using second comparison, -6w must equal -20k(ab), thus w = 10/3k(ab)

        so our factor must be (10/3k(ab)-3)^2

        we need to solve for k, to ensure k is between 2 and 3. if it is, then
        the premise of number 6 can be proven

        using first comparison, we know (10/3k(ab))^2=(20k^2-80)(a^2b^2)

        100/9k^2a^2b^2 = (20k^2-80)(a^2b^2)

        the a^2b^2 cancel out, giving us 100/9k^2=20k^2-80

        subtract 20k^2 from both sides to give us -80/9k^2=-80 multiply both
        sides by -9/80 gives us k^2=9, or k = 3

        With your original theory, this would prove it false. I would surmise
        that there will be some numbers that your rules as written will not
        work.

        However, adjusted so that the interval is 2x<k<=3x and the mental
        exersise above will prove it true.

        Sorry if it isnt in long proof form, its been forever since ive done
        one of those. I hope this was easy to follow
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