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Prize Puzzles : Two more Primality Conjectures

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  • aldrich617
    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 :
    Message 1 of 5 , Oct 3, 2008
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      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

      ps.

      I have it on good authority that the puzzle reprinted below
      from 9/8/8 is probably true but very difficult to prove, so
      I am extending the contest deadline until 11/1/8.

      Prize Puzzle : Primality Conjecture
      I offer a $50 prize to the first person who can submit a verifiable
      counterexample or proof for the following primality conjecture:

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

      Let A(x) = 5x^2 - 5x +1;
      Let B(x) = 25x^2 - 40x + 16;
      Let T(k) = 5*(2*k -1)^2 - 4*A(x) ;

      If x > 3 and k > x , then A(x) will be prime if no value
      T(k) exists such that 1 < T(k) < B(x) is a square.
    • Mike Oakes
      ... You don t say so, but I assume you are requiring K1 = K ? -Mike Oakes
      Message 2 of 5 , Oct 3, 2008
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        --- 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 3 of 5 , Oct 5, 2008
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          --- 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 4 of 5 , Oct 6, 2008
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            --- 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 5 of 5 , Oct 6, 2008
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              --- 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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