Date:         Mon, 13 Oct 2003 03:29:40 -0700
From: August Highland <hmfah@HOTMAIL.COM>
Subject: MONIC IRREDUCIBLE CUBICS/FLEXPOINT FACTORIZATION
To: WRYTING-L@LISTSERV.UTORONTO.CA

MONIC IRREDUCIBLE CUBICS/FLEXPOINT FACTORIZATION


PARTITION #0000001:

X, = be.  =1 (9), gives r.  X+b )( 0.  Proof: we apply proposition 1 with f (n)=n. if we let a = n
\gamma rn (the set of natural numbers not divisible by r) then (3) follows. if we let a = fn 2 n :
(m; n) = 1g, then (4) follows. gives rise to the no-.  (p ) =5, and p.





PARTITION #0000002:

N=1 CLAIM: The nine flex points form a 3-torsion packet..  N, denote the curve 4. We set up the
problem in Magma in the following way..  (1 + x + p(p \Gamma.  Definition 7 let f (a 9.  K=1 (k)q(n
\Gamma k) p.





PARTITION #0000003:

)) a 2.  + x 8.  N=0, there is a single divisor D * 0 such that (f ) 5.  (\gamma 1), 1 d.  3 m.





PARTITION #0000004:

1 n + 2a.  )=ff (2 =1, hence a.  ) where a k.  `\gamma 1, =4\Theta 2.  F ( n=0.





PARTITION #0000005:

P n.  2, 0 2.  + 1), and n.  K=1 3 =.  + t u 0.





PARTITION #0000006:

(n) denote the number of partitions of n into parts that belong to a. 3 + 1; x.  P(p \gamma 1)(p
\gamma 2) F . . ..  =2, 0 g(n; k)=2.  N=mk, \Gamma 4a + oe.  Definition 2 let oe, p n.





PARTITION #0000007:

X + oe 1.  6, (P ) ? a5 :=a^n5; ? c5 := c^n5; ? a5^129; 1106532280219457618983939634726858708298 ?
c5; 1106532280219457618983939634726858708298 ? a6 := a^n6; ? c6 := c^n6; ? a6^127;
809579285918008980133272648385832028198 ? c6; 809579285918008980133272648385832028198.  P k X.  R
Chinese remainder lifting..  Gives \Theta 4.





PARTITION #0000008:

3 3 and n.  3 1 X \Gamma.  Mx )= n2A.  2, ` Definition 3 Let p(n) denote the number of partitions of
n..  X 2 j.





PARTITION #0000009:

0, 1 0.  ) + (x oe \Gamma oe.  T u H.  K =r.  P (a =4\Theta.





PARTITION #0000010:

2 . The Euclidean algorithm for polynomials in turn yields: where.  + 1=(x n =4.  Oe : p 7! y.
Q(n)x 3 =634466267339108669 and p.  J, ? Index(-@ a^(n1*i) : i in [1..2] @"", c^n1); 1 ? Index(-@
a^(n2*i) : i in [1..3] @"", c^n2); 2 ? Index(-@ a^(n3*i) : i in [1..5] @"", c^n3); 4 ? Index(-@
a^(n4*i) : i in [1..37] @"", c^n4); 29 ? CRT([1,2,4,29],[2,3,5,37]); m.





PARTITION #0000011:

(p ) ffl for space quartics, we use a syzygy satis- X.  1 =634466267339108669 and p 3.  2, f (2 p.
N i 2.  2 Remarks: Proposition 1 is Theorem 14.8 in [1]. If we let A=N; f (n) = n, then we obtain
m6=n.





PARTITION #0000012:

+ 1; x, a 2.  \theta h \gamma 72s then the map \Delta.  F (2 1 X.  1 )= x + 3c.  Without loss of
generality we may assume that n * m. the euclidean algorithm for integers yields k rjm 0.





PARTITION #0000013:

X 0.  R =rt + s(p \Gamma 1), so.  ) + (x q(n)x.  + 4a 0 2.  6, \Delta 2.





PARTITION #0000014:

I 1 + x.  \gamma r 5 r.  This is given as theorem 1 in [2], and is a special case of theorem 1(a)
below. H \Gamma 11S + 3a.  T(x) \gamma _(x)ff(x)r(x)=*(x)r(x)t(x) or equivalently 2 (n)x.  1 with
Weil's result gives a rational map from C to a Weierstrass model. (a.





PARTITION #0000015:

I, 2 1.  Claim: the nine flex points form a 3-torsion packet. djn ; 26 jd.  2 2 (P ).  , n, 10 k=1.
1, (x) is a greatest common divisor and d Connection to curves It is via these fundamental syzygies
that we obtain Weierstrass models for the Jacobians of our curves..





PARTITION #0000016:

=37, which can be easily solved by enumerating all + oe.  (n)=, oe 2.  N=1 1 X 2.  A;f, = + \Delta
\Delta \Delta + x.  And x 2 n=1.





PARTITION #0000017:

A j d\Gamma 1.  \gamma y i=a.  S(p\gamma 1) X k.  3 2 + 1 in Z.  Z + 3c logarithm can be computed
independently for each prime divisor of p \Gamma 1 -- more correctly for prime power divisor -- and
the discrete logarithm can be recovered by the Chinese remainder theorem, as is the next example..





PARTITION #0000018:

F 6 5.  1 0 p.  Let p and q be two flex points, with tangent lines l Then (1 + x.  (p ), F (P ).  1
2. Biggs 15.6.4 (pg. 336) Solution: For (i), any divisor of a(x) and b(x) also divides the greatest
common divisor. Since d d\Gamma 1.





PARTITION #0000019:

F, 0 k\Gamma i.  N=1 (n)=q(n). Y.  J, , + 1)(x.  3 n.  F = in this list is 1,.





PARTITION #0000020:

2, j , n.  K\gamma 1 x, 3 that is,.  \gamma 2r Then `\Gamma 1.  \delta 1073741839. we repeat, 2 2.
? p :=2^32+15; ? modexp(3,(p-1) div 2,p); 4294967310 ? modexp(3,(p-1) div 3,p); 2086193154 ?
modexp(3,(p-1) div 5,p); 239247313 ? modexp(3,(p-1) div 131,p); 1859000016 ? modexp(3,(p-1) div
364289,p); 1338913740 =4\Theta p.




august highland

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