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Re: Prize Puzzles : Two more Primality Conjectures

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  • Mike Oakes
    ... You don t say so, but I assume you are requiring K1 = K ? -Mike Oakes
    Message 1 of 5 , Oct 3, 2008
      --- In primenumbers@yahoogroups.com, "aldrich617" <aldrich617@...>
      wrote:
      >
      > 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)

      You don't say so, but I assume you are requiring K1 = K ?

      -Mike Oakes
    • Mike Oakes
      ... Thanks for presenting this extraordinarily interesting challenge. [And you confirmed offline that indeed K1=K.] After spending most of the last 2 days on
      Message 2 of 5 , Oct 5, 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.
        >
        > 2) If all possible composites from 1) above are removed from the
        > sequence
        > 5x^2 - 5x +1 , then all the remaining values of A(x) are prime.
        >
        > Aldrich Stevens
        >

        Thanks for presenting this extraordinarily interesting challenge.
        [And you confirmed offline that indeed K1=K.]

        After spending most of the last 2 days on it, I have just proved
        conjecture 1).

        I don't know if other people are working on it, in which case it
        would spoil their fun to present the proof; but if no-one objects,
        then I will publish it here in 24 hours' time.

        -Mike Oakes
      • 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 3 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 4 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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