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

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  • 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 1 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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