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Re: Prize Puzzles : Two more Primality Conjectures

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  • Mike Oakes
    ... [snip] Proof of Conjecture 1): If K is even, n = K/2. If K is odd, n = (K-1)/2. Let s work in terms of n. Given n, denote the even-valued K by Ke = 2*n,
    Message 1 of 5 , Oct 6, 2008
      --- In primenumbers@yahoogroups.com, "aldrich617" <aldrich617@...>
      wrote:
      >
      > I offer a $50 prize to the first person who can submit a verifiable
      > counterexample or proof by 11/1/8 for either conjecture 1) or 2)
      > below:
      >
      > x,A(x),K,M,S,n,p : integers;
      >
      > Let A(x) = 5x^2 - 5x +1
      > with K > 3,
      > and n or p denoting the number of terms in a sequence.
      >
      > For any value of K first find the sum of the consecutive integers
      > K1 + K2.....+ Kn = S such that Kn < (3K -1)/2.
      > Example: if K = 9 then S = 9 + 10 + 11 + 12 = 42, and n = 4.
      > (note: 40*S + 1 is always a square)
      > Using the same value of S find any shorter runs of consecutive
      > integers
      > M1 + M2.....+ Mp = S where p < n, and M1 > K1.
      > Example: For S = 42 from above it is evident that 13 + 14 + 15
      > also equals 42, with M1 = 13 and p = 3.
      >
      > Conjectures:
      >
      > 1) If any value for S or M1 found by the process described above is
      > substituted for x in A(x) = 5x^2 - 5x +1 then
      > A(x) is always composite .
      > Examples: if S = 42 then A(42) = 8611 = 79*109.
      > if M1 = 13 then A(13) = 781 = 11 * 71.
      >
      [snip]

      Proof of Conjecture 1):

      If K is even, n = K/2.
      If K is odd, n = (K-1)/2.
      Let's work in terms of n.

      Given n, denote the even-valued K by
      Ke = 2*n,
      and the odd-valued K by
      Ko = 2*n+1.

      Denote the corresponding values of S by:
      Se = (2*n) + ... + (3*n-1)
      = (3*n-1)+3*n/2 - (2*n-1)*2*n/2
      = (5*n^2-n)/2
      and
      So = (2*n+1) + ... + 3*n
      = (3*n+1)+3*n/2 - (2*n+1)*2*n/2
      = (5*n^2+n)/2

      Let 1 <= p < n, and suppose that
      Se = M + ... + (M+p-1)
      = (M+p-1)*(M+p)/2 - (M-1)*M/2
      = (2*M-1+p)*p/2

      Inserting the formula for Se gives
      2*M-1+p = (5*n^2-n)/p
      i.e.
      (2*M-1) = (5*n^2-n-p^2)/p

      Iff the RHS is an odd integer, this gives a solution.

      For any such solution we find
      A(M) = 5*M^2-5*M+1
      = (5/4)*(4*M^2-4*M) + 1
      = (5/4)*(2*M-1)^2 - 1/4
      = 5*(5*n^2-n-p^2)^2/(4*p^2) - 1/4
      = [125*n^4 - 50*n^3 + (5-50*p^2)*n^2 + 10*p^2*n +(5*p^4-p^2)]/(4*p^2)
      = f1 * f2 / (4*p^2)
      where
      f1 = 5*n^2-p^2
      and
      f2 = 25*n^2-10*n+1-5*p^2

      [Remark: this surprising algebraic factorisation of a quartic
      polynomial into a product of 2 quadratics is the core of the proof,
      and was found by examining a lot of special cases with, in
      particular, p=2, 3 and n.]

      Now n > p, so
      f1 > (4*p^2)
      and n >= (p+1), so
      f2 = (5*n-1)^2 - 5*p^2
      >= (5*p+4)^2 - 5*p^2
      > (4*p^2)

      So, however the factors of (4*p^2) divide f1*f2, the result must be a
      product of 2 integer factors each > 1.
      Therefore, A(M) is composite, which was to be proved.

      The limiting case p = 1 gives M = Se,
      so A(Se) is composite, which was also to be proved.

      If, instead, we choose the "odd-K" case (Ko and So), then all the
      above goes through with just the simple change n => -n.

      This concludes the proof of conjecture 1).

      -Mike Oakes
    • Phil Carmody
      ... [snip] ... Nowhere is it specified that n is maximal. Of course, it s underspecified if you don t assume that, but all that means is that the problem s
      Message 2 of 5 , Oct 6, 2008
        --- On Mon, 10/6/08, Mike Oakes <mikeoakes2@...> wrote:
        > --- In primenumbers@yahoogroups.com, "aldrich617" wrote:
        > > Let A(x) = 5x^2 - 5x +1
        > > with K > 3,
        > > and n or p denoting the number of terms in a sequence.
        > >
        > > For any value of K first find the sum of the
        > consecutive integers
        > > K1 + K2.....+ Kn = S such that Kn < (3K -1)/2.
        [snip]

        > Proof of Conjecture 1):
        >
        > If K is even, n = K/2.
        > If K is odd, n = (K-1)/2.

        Nowhere is it specified that n is maximal. Of course, it's underspecified if you don't assume that, but all that means is that the problem's underspecified.

        Phil
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