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5x^2 - 5x +1 revisited

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  • aldrich617
    Perhaps someone would enjoy applying some Gaussian concepts to this basic stuff!: Previously, in September, 08 it was conjectured that: x, A(x), B(x), k,
    Message 1 of 1 , Nov 28, 2009
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      Perhaps someone would enjoy applying some
      Gaussian concepts to this basic stuff!:

      Previously, in September, '08 it was
      conjectured that:

      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,
      otherwise A(x) will be composite.

      The second part of this conjecture
      was soon proved to be true, but
      the first part was only shown to be
      a necessary condition for primality in
      this situation and it remains inconclusive as to
      whether it is also a sufficient condition.

      Since then a bit more information has
      come to light about the composites of
      5x^2 -5x + 1, though not about its
      primes. It can be empirically observed
      that if the notion of the 'k' values, hitherto
      consistantly used to denote integers generating
      odd squares under the right conditions
      (when factors are present) , is expanded
      to include all of the numbers k -1/2
      across the same interval, that these 'half'
      values of k turn out to also indicate
      compositeness when they generate (even)
      square values of T(k). These squares
      invariably come paired with odd squares
      on the other side of the divide that is at
      a point approximately 1/4 of the way
      from the lower limit of the interval where
      T(k) applies. All squares come in pairs
      with the sole exception of when the A(x)
      value is itself a square that is also equal
      to a value of T(k).Such pairs can be in
      odd-even, even-odd and odd-odd
      combinations but never even-even.

      If there are several factors in A(x) many
      squares are generated, and their exact
      total is always the same for composites with
      the same numbers of distinct and redundant
      prime factors.Each case has its own formula
      and the simplest of these occurs when all of
      the prime factors of A(x) are distinct . If we
      let 'f ' denote this number of prime factors
      then the formula for the number of squares
      that will be present in the simple case is 2^f - 2;
      eg : two factors makes two squares, three make
      six, four make fourteen etc. Half of these will
      occur in in the first quadrant of the interval, and
      half later. The first square is always paired with
      the last, the second with the second to last and
      so forth.

      To illustrate:
      At x = 65, 5x^2 - 5x + 1 = 20801 = 11*31*61,
      and the relevant interval has a total of 63 terms
      beginning at k = 65.5 and ending at 91.5. The
      significant values are as follows:
      k : 65.5 66 68 / 79.5 86 89
      2k -1 : 130 131 135 / 158 171 177
      sqrt(T(k)) : 36 51 89 / 204 251 271

      Let us call the 2k-1 values above R1,R2,R3,R4,
      R5,R6 and the sqrt(T(k)) values r1,r2,r3,r4,r5,r6.
      Actually, only R1,r1,R2,r2,R3,r3 have to be found
      before the other values can be easily calculated from
      them, and A(x) factored as well. Note that
      R1 + R6 = r1 + r6, R2 + R5 = r2 + r5, and
      R3 + R4 = r3 + r4. This relationship means that
      if say R1 is known, then its paired number R6
      can be found by means of a fast binary search
      pattern. Then it is an easy calculation to find
      two factors of A(x) because
      gcd((x - 1)(R1 + R6) + R1, A(x)) gives
      one and gcd((x - 1)(R1 + R6) + R6, A(x)) another :
      Specifically for 20801, 64 * (130 + 177) + 130 =
      11*31*58 and 64 * (130 + 177) + 177 = 61*325.
      This process can also be done for the other paired
      values of 2k - 1 so that 64 * (131 + 171) + 131 =
      11*61*129 and 64 * (131 + 171) + 171 = 31*629,
      and 64 * (135 + 158) + 135 = 11*1717 and 64 *
      (135 + 158) + 158 = 31*61*101.

      Locating even one square by trying out all of the
      possibilities of 5*(2*k -1)^2 - 4*A(x) for an
      interval approximately x/4 in length can be difficult
      for large values of x. There are fortunately ways to
      eliminate most of these by exploiting the fact that
      valid sqrt(T(k)) solutions occur on other quadratic
      equations whose middle three terms can be expressed
      as ...A(x), A(x) + 10*k, A(x) + 2*10*k + 10...
      For k = 79.5 in our example this would work out to
      20801, 21596, 22401....41616 = 204^2. The terms
      of these equations with solutions are composed
      exclusively of primes +/- 1 mod 10 and also squares
      of any sort, whereas all of the other possible
      equations of the same form will have some terms
      with odd powers of +/-3 mod 10 primes that
      can thus be eliminated with a sieving scheme.

      I have developed a crude Pascal demo program
      that will show that the idea works well for
      A(x) < 1e19 and anyone interested can have
      a copy upon request. It is hard to gauge at this point
      just how large a number with non-trivial factors
      could be really cracked in this manner. Perhaps
      with top-notch equipment, a sophisticated sparse-
      matrix mathematic,and advanced software array
      forms (plus a better programmer than myself)
      the technique could be competitive as a brute-
      force algorithm.

      Aldrich Stevens
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