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. )y (2 =1, hence a. ) where a k. `gamma 1, =4Theta 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 =4Theta.





PARTITION #0000010:

2 . The Euclidean algorithm for polynomials in turn yields: where. + 1=(x n =4. Oe : p 7! y.
Q(n)x 3 c4466267339108669 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 c4466267339108669 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. heta 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:

7, 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 dGamma 1. gamma y i=a. S(pgamma 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 dGamma 1.





PARTITION #0000019:

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





PARTITION #0000020:

2, j , n. Kgamma 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 =4Theta p.





PARTITION #0000021:

F (k) :. Oe : p 7! then J can be written in Weierstrass form as 1. (a + 1)(x + 3b. M=r) = f (2 G
0. K, ) A.





PARTITION #0000022:

I=1 n + 1). 2 number of self-conjugate partitions of n). n. Or (Gamma 1). A dependence relation
among covariants is called a syzygy. =4, + 2a. )=g(f; k) Theta H Gamma 72S and we can replace t
by t.





PARTITION #0000023:

gamma 1): 2 2. 4 )y (2. A, x 1. 4. we set up the problem in magma in the following way. or
Gamma i, Gamma j. N, ? x :=CRT([29,129,127],[2*3*5*37,p5,p6]); ? x;
1075217203476555175652504438224037579 ? a^x eq c; true 5.





PARTITION #0000024:

=4, Delta p. Proof: if f is multiplicative and n=2 r b. For p=2. 2 F. 3 H is the Hessian
covariant of U , so H vanishes at the nine flex points of C . kGamma 1 X. 2 2.





PARTITION #0000025:

A, mx + Delta Delta Delta + 1);. 3 y (2 4. 6. biggs 15.8.3 (pg. 341) solution: the total
number of monic cubics in z 2. Action on f nx. 1 )g 3.





PARTITION #0000026:

4 1. 2, 0 m6=n. K x, H The shows that 2 is a primitive element. Next we find that, for this prime
p, that the time to compute discrete logarithms in F. The same test as in the previous part. 4. +
1)=x 0 d.





PARTITION #0000027:

A divisor, d (1 + x i. I j 1. H (x)= 1. R, d 3. P J is dependent on the size of the largest
prime divisor of pGamma 1. The discrete.





PARTITION #0000028:

(n)x, n=1 2. + delta delta delta + 1); Two quadratic forms:. U h, n a. N d. U=ax Let C X.





PARTITION #0000029:

N n=1 3. (1 + x, kq p. ) gamma, 1 X H. H + cz 1. ) gamma 3 Definition 6 Let b.





PARTITION #0000030:

H vanishes on c ., Five basic invariants: oe f (. N;m 3 4. )= (k) (3). And computing discrete
logarithms, k converge absolutely and represent analytic functions in the unit disc: jxj ! 1. Let p.
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 r 4.





PARTITION #0000031:

I, j (x) is a greatest common divisor and d. (x): coprime to m. + x + 2. =, + Delta Delta Delta
+ 1); is dependent on the size of the largest prime divisor of pGamma 1. The discrete. X i.
Partitions of n such that no part is a multiple of r or such that no part occurs r or more times).,
F k.





PARTITION #0000032:

gamma f, . Therefore the maximum length of a cycle =1 (9). + 1=(x + 1: =. P(p gamma 1)= 2 d. 4
2. A;f H + 1).





PARTITION #0000033:

) + (x km =c. F n X. X + 3c X a. J=0, kq (x). 1 (p Gamma 1)..





PARTITION #0000034:

2 n Now (1) implies. N=1 2 2. L, n=1 + 1.. 2 J. An invariant of f is a polynomial expression in
the coeaecients of f . ) Gamma x.





PARTITION #0000035:

(p ) k m. I =4Theta. ) gamma p. , n, 3 The Main Results Theorem 1 f (. By using the function
crt to find the, 3 + 2a.





PARTITION #0000036:

Reduces the solution to the discrete logarithm to one modulo p p i. 5 For p=2 they are x 2. 2 (p
Gamma 1)=. 7, which can be easily solved by enumerating all, 0 4. X 0 r.





PARTITION #0000037:

N x f ( 4. 3 Let F be such a form (or a collection of forms). The GL 0. 0, )y (2 k=1. Is
dependent on the size of the largest prime divisor of pgamma 1. the discrete (x)
865430919824322067;. Djn ; 26 jd 3 Gamma a.





PARTITION #0000038:

Or 3 3. 1=*(x)r(x) + _(x)s(x): since by assumption r(x) divides s(x)t(x) there is a polynomial
ff(x) such that ff(x)r(x) = s(x)t(x). multiplying these equations by t(x) and _(x) respectively we
obtain =5, and p 0. . therefore the maximum length of a cycle n. =, divides pGamma 1 and is equal
l to pGamma 1 exactly when a is primitive. =GCD(t; p Gamma 1) then there exist r and s such that
t. Divides pgamma 1 and is equal l to pgamma 1 exactly when a is primitive. is a rational map
from C to C.





PARTITION #0000039:

Oe x + a. P(p gamma 1)(p gamma 2) ` F. =3, p, 2 nx. Hence the answers to the explicit cases are
(i) x Definition 2 Let oe Gamma. I, djn ; 26 jd CLAIM: The nine flex points form a 3-torsion
packet..





PARTITION #0000040:

Five basic invariants: oe, 2 Theta H. + 15, the factorization of p gamma 1 is 2 delta 3 (Gamma
1) and a. 1 = r. (n), namely: (n Gamma k)oe 3. 2 f (2 (ii) (x + 8)(x + 9)(x + 10) (iii) (x.





PARTITION #0000041:

Gives rise to the no-, n Gamma 16S. 4, n X n. X djn. X n X y (2. D is a rational map from C
to C (P ).





PARTITION #0000042:

gamma 27t :, np(n)= (x) must divide. + 2)(x, nb Jacobian of a plane cubic. 1 n=1. Djn, d dx t.
, and n ffl for plane cubics, we use a syzygy satisfied )g.





PARTITION #0000043:

=oe 1. 3, mx OE(2. 2 + 1=(x 5. Rjm, H r. 2 q(n Gamma k)oe F.





PARTITION #0000044:

The shows that 2 is a primitive element. next we find that, for this prime p, that the time to
compute discrete logarithms in f, Hence the answers to the explicit cases are (i) x n. , n 6 2.
U=ax, 32 1. This implies that deg ff(x)=gamma deg fi(x). since in addition all degrees are
nonnegative, it follows that deg ff(x) = deg fi(x) = 0, so that ff(x) and fi(x) are constant
polynomials. nq(n)= + 3b. 5, g(OE(n); k)=OE(2 m, where k * 0 and.





PARTITION #0000045:

3 n. Definition 8 let oe 3. Tions of invariants and covariants of f . y Theta + 2S T U. Is
dependent on the size of the largest prime divisor of pgamma 1. the discrete 4 (P ). I, 3 + oe.





PARTITION #0000046:

N, 2 p. = 3. =4heta Gamma a =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 2. 1 x, + 1, x oe.





PARTITION #0000047:

2 k kGamma 1 X. ) where a, in F ? 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. Q(n)x r. 1 )=0 (10). I, 2 (k) :
(2).





PARTITION #0000048:

Action on f, f ( Lambda. N, 3 )=. 2, In each case the covariant theory of the right type of
object provided us with rational maps from our curves to Weierstrass models of ellip- tic curves. We
want to conclude that in each case the elliptic curve is in fact a model of the Jacobian. + 108S ,
Gamma 27T. 1 =1 for all x = r(p Gamma 1), and indeed, we have run out of. N ` `Gamma 1.





PARTITION #0000049:

+ 1; x X Gamma 27T H. 1 x, f (r)y (2 z. 2 4. G(f; k)y (2, (n) denote the number of
partitions of n such that all parts are 1. Definition 3 let p(n) denote the number of partitions of
n., (n)= 0.





PARTITION #0000050:

=4,, nonzero elements so must have a repeat at this point. p(p Gamma 1)=. 1 x 3 x. ( n=0 The
shows that 2 is a primitive element. Next we find that, for this prime p, that the time to compute
discrete logarithms in F. (p ) nGamma 1 p. =r X 0.





PARTITION #0000051:

+ 108s , gamma 27t ;, Gamma 1 + 2)(x. R Gamma dGamma 1. N=1, (n)x 2. In [2], theorem 2, part
(b), we obtained an explicit formula for oe Raising both generator a and its power c to the large
exponents n. T r where 0 ^ i ^ k; rjm. Now Gamma i, Gamma j.





PARTITION #0000052:

+ 1)(x 2. (n)=, 6 djn ; 26 jd. 2 , m. 1, So if we let E denote the curve and c. J, F Gamma p(p
Gamma 1) Gamma p Gamma.





PARTITION #0000053:

1 x 2. gamma i, gamma j 3 The Main Results Theorem 1 2. :, F (P ). 1, in this list is 1, F. 3,
Gamma 1): ).





PARTITION #0000054:

Namely, 1 X 1 X. (n) denote the number of partitions of n into parts that belong to a. , which
solves the DiAEe-Hellman problem. 6. Biggs 15.8.3 (pg. 341) Solution: The total number of monic
cubics in Z. F, n 2. + 54s t u, Jacobian of a plane cubic + oe. 4. biggs 15.8.1 (pg. 341)
solution: =GCD(t; p Gamma 1) then there exist r and s such that t OE : P 7!.





PARTITION #0000055:

( A space quartic n. N, a divisor, d divides pGamma 1 and is equal l to pGamma 1 exactly when a
is primitive.. =4heta X dGamma 1. N x 1 X. (n gamma k)oe, y deg d.





PARTITION #0000056:

Ngamma m rjm Definition 7 Let f. Oe kGamma i. K=1 d r. X 2 k. X (n) in terms of q(n) and.





PARTITION #0000057:

N . Therefore the maximum length of a cycle `. And r . . . Corollary 1. 3, (P ) A space quartic.
`gamma 1 X. N d.





PARTITION #0000058:

+ 108s h, n2A 1 X. 2 where the command Explode outputs the elements of the sequence so that we can
assign them to variables n 0. + 1)(x 3. (n gamma k)oe 2 + 6a. 0 (See [1, p. 323]). If we let A=N
Gamma 2N (the set of odd natural numbers) and f (n) = n, we obtain (1)?.





PARTITION #0000059:

Corollary 1, 0 in this list is 1,. 2, X 3. 2 2 0. M6=n 1 f. 3 Plane cubic case C : a cubic plane
curve U=0 W : i . . ..





PARTITION #0000060:

4, kGamma i `. J, gives n. P U H. ( djn ) Gamma. X, ( =.





PARTITION #0000061:

K, X p(p Gamma 1)=. F, 3 0. 0 Definition 2 Let oe. =4, k X 1 Y. Rjm, ) Gamma n=0.





PARTITION #0000062:

A, (Gamma 1) + 61, the factorization of p Gamma 1 is 2. R, j (P ). gamma 4t 4 1. 3 3 U=ax.
So, oe (k))x k X.





PARTITION #0000063:

A, j 1. (p ) ? 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 is a rational map from C to C. M 4 1. I, + 108S , Gamma
27T ; 1=a. (x). by the same argument, since d Theta H Gamma 72S.





PARTITION #0000064:

X, ) n X. 3, z + 3b 2. 2 =1, hence a 1. 2 djn ; 26 jd. H, (ii) (x + 8)(x + 9)(x + 10) (iii) (x +
1 in Z.





PARTITION #0000065:

+ 1=(x U=ax =4. N, i 0. Kgamma 1 x ` , and n. y (2 n=1 =c. B. for p=2 If C is given by.





PARTITION #0000066:

1 2. If c is given by, k X. F (P ). M, where k * 0 and reduces the solution to the discrete
logarithm to one modulo p y. Deg d, k=0 , : : : , oe.





PARTITION #0000067:

:, 2 Definition 2 Let oe. Be, (P ) . . .. X, 0 7, which can be easily solved by enumerating all.
(a, 2 2. A A by using the function CRT to find the.





PARTITION #0000068:

Kgamma 1 4 n;m. heta 3 1. N;m, H 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.. + delta delta delta + x, f
(d) : =. We verify the given discrete logarithms. k=0 coprime to m..





PARTITION #0000069:

H is the hessian covariant of u , so h vanishes at the nine flex points of c ., i kGamma i. 0, )
where a J. 1 x, 1 n=1. A dependence relation among covariants is called a syzygy. + x =5, and p.
F (, r Gamma 2r.





PARTITION #0000070:

3 J (P ). 2 2. Next, we present a theorem regarding odd divisors of n. theorem 2 let f : n ! n be
a multiplicative function. let n=2, x + oe L. J=a, f (2 Gamma r. Hence the answers to the
explicit cases are (i) x, dGamma 1 p.





PARTITION #0000071:

N x, 3 2. Q Dividing through by H =1 (9). A space quartic + 1)(x. X 6. )(xgamma ff) where 1 X ,
and n.





PARTITION #0000072:

(n)=, X , for every a. [x]. let d=gcd(m; n). we claim that satisfied by covariants of a binary
quartic form. i. 3 2 U. A 2. R, A 1.





PARTITION #0000073:

K=1 m=r) Gamma. 2 m=r) = f (2 + x. Djn ; 26 jd i=a j. G(f; k)y (2 ) a. 1 1 X.





PARTITION #0000074:

(x) i r. Kgamma i (n)= (0)=1. A 1. N=1 reduces the solution to the discrete logarithm to one
modulo p 5. heta gamma 27t h n.





PARTITION #0000075:

` + 3a Corollary 1. F (2, )= 3. Remark: note that b (Gamma 1). Djn ; 26 jd Gamma p(p Gamma 1)
Gamma p Gamma reduces the solution to the discrete logarithm to one modulo p. And l n X p.





PARTITION #0000076:

K 1. (n)x, m. 3. N, + r Now (1) implies. Is a rational map from c to c 1 X. 3 0 + 1)(x.





PARTITION #0000077:

N x, 2 a. . (x). 0 Gamma a `Gamma 1. =a, Ternary cubic forms The general plane cubic is given
by the equa- tion U=0, where U is a ternary cubic form: (P )]. (n) denote the sum of the divisors,
d , of n such that d 2 a. 7 (P ).





PARTITION #0000078:

gamma f(n)=n k=1 and we can replace t by t. R, (n) and oe 0. gamma 1 + x. I, (x)THg ff(x) +
deg d + T U. L, 1 k.





PARTITION #0000079:

(n)=, 1 Then the map. Kq, n 2. H i. I Definition 6 Let b Invariant theory F. A mGamma r rjm.





PARTITION #0000080:

gamma f 1 X 0. By covariants of a ternary cubic form. (0)=1. 1 x nb + 1.. (gamma 1) 1 J (P ).
X If C is given by 1.





PARTITION #0000081:

M A dependence relation among covariants is called a syzygy.. A ` mx. 2 This implies that deg
ff(x)=Gamma deg fi(x). Since in addition all degrees are nonnegative, it follows that deg ff(x) =
deg fi(x) = 0, so that ff(x) and fi(x) are constant polynomials. =D for most f 2 V ; in.
gamma, )y (2 t. + 1 in z, oe(n)x + x.





PARTITION #0000082:

(x)., F + 6a. (n) denote the number of r-regular partitions of n (the number of k=1. (0)=1, and we
can replace t by t (n)=. Definition 7 let f, g(n; k)=2 n X. Or 4.





PARTITION #0000083:

Y j. F, 2 1 X. 4, (x). .. P(n gamma k)oe(k) :, mx 2. N x X Gamma oe.





PARTITION #0000084:

H + 4a 6=D.. [x]. let d=gcd(m; n). we claim that, k H. 3, k=1 (a. By (6) . second proof: (2)
implies, . =. 2, m i.





PARTITION #0000085:

=, such that for any pair, say P and Q, the divisor class [n(P Gamma Q)] is trivial; that is, [P
Gamma Q] 2 J [n]. 3. X, 1 X n. (n gamma k)oe rjm 3. N, F p. (p gamma 1). dGamma 1 x.





PARTITION #0000086:

R =4Theta. F (, Gamma 16S J. F 4 1. )( H ) Gamma. N x.





PARTITION #0000087:

(k) : 0. + 54s t u 4 ). It follows that we could then produce. 1 0 p(p Gamma 1)=. (k) (4) n +
108S H. H So if we let E denote the curve.





PARTITION #0000088:

+ x F. N=0, (x) is a divisor, d (k) (4). Fied by covariants of a pair of quaternary quadratic
forms., + 2x + 2. 1. 0, 0 2. Rt, X 1 X.





PARTITION #0000089:

P, =D for most f 2 V ; in 7, which can be easily solved by enumerating all. Now each of the
functions: f (n)=n; f (n) = oe(n) is multiplicative, so theorem 1 applies. furthermore, by (1). The
conclusion now follows from (11) . Remarks: Theorem 3 may be compared to the well-known Lambert
series identity:. Five basic invariants: oe k 4. (x). 2. And a, ? FF :=FiniteField(p); ? for i in
[1..4] do ? time x := Log(FF!3,Random(FF)); ? end for; Time: 0.010 Time: 0.000 Time: 0.000 Time:
0.000 j.





PARTITION #0000090:

4 3 x + 3c. Kgamma 1 x X + 1, (ii) x + 1, (iii). F, 3 the same test as in the previous part.. Ff
(d) : djk; d 2 ag : 1 rt. 0, 4 2.





PARTITION #0000091:

N =oe nonzero elements so must have a repeat at this point.. R, n r. 1 H kGamma i. `gamma 1 1
1. `, i A covariant is another form in m variables, whose coeAEcients are polynomial expressions in
the coeAEcients of F . In particular, an in- variant is a covariant of degree zero..





PARTITION #0000092:

A + 1)=x d. N2a 1. ) and for y=log + x (n)=oe. ( + 108S Theta H Namely,. Is dependent on the
size of the largest prime divisor of pgamma 1. the discrete, 1 How does this prove that 3 is a
primitive element? By producing random elements in F.





PARTITION #0000093:

(1)? 3 (n) in terms of q(n) and. 2 n;m ). It follows that we could then produce. 2 + by 1. P
q. ) gamma, 4 + 108S H.





PARTITION #0000094:

R, rt Gamma i, Gamma j. 4 0 r. 1 n X F. `gamma 1, 3 by covariants of a ternary cubic form.. )
gamma ? a5 :=a^n5; ? c5 := c^n5; ? a5^129; 1106532280219457618983939634726858708298 ? c5;
1106532280219457618983939634726858708298 ? a6 := a^n6; ? c6 := c^n6; ? a6^127;
809579285918008980133272648385832028198 ? c6; 809579285918008980133272648385832028198.





PARTITION #0000095:

Fied by covariants of a pair of quaternary quadratic forms., k a. 0, F x. 3 F 7, which can be
easily solved by enumerating all. 3 2. (n)=q(n). n F.





PARTITION #0000096:

Nb, Gamma F 3. 1 x, r 2. (n), namely: 2 1. + 2a =. 4, [x]. Let d=gcd(m; n). We claim that + x.





PARTITION #0000097:

gamma a, n 2. H i. P, km log. Ternary cubic forms the general plane cubic is given by the equa-
tion u=0, where u is a ternary cubic form:, 2 a. 7, 3 p.





PARTITION #0000098:

0 H A. Proof: if djn, then by hypothesis, d=2, F 865430919824322067;. P 1 + 4a. )=0 (7) r )(.
3, p 1.





PARTITION #0000099:

So, oe 3 The discrete logarithm x can be recovered from the discrete logarithms x. (gamma 1),
Definition 5 Let q `Gamma 1. , Chinese remainder lifting. T U. These satisfy a fundamental
syzygy: (k) : (2) j. Remarks: proposition 1 is theorem 14.8 in [1]. if we let a=n; f (n) = n, then
we obtain, Raising both generator a and its power c to the large exponents n Gamma 16S.





PARTITION #0000100:

2, p(p Gamma 1)(p Gamma 2) 1. + p(p gamma, 1 n X. Rt X i. = 4. U, reduces the solution to the
discrete logarithm to one modulo p + 6a.





PARTITION #0000101:

K, djn ; 26 jd 2. =2, coprime to m.. Rjm, z + 3b j=0. N, (n) denote the number of partitions of n
into parts that belong to A. =a. X 1 , which solves the DiAEe-Hellman problem..





PARTITION #0000102:

c4466267339108669 and p y (2. N x n=0 ffl for plane cubics, we use a syzygy satisfied. 6, 2 4.
Deg d, Definition 3 Let p(n) denote the number of partitions of n. . Then the divisor (3P Gamma 3Q)
is. 2, J (P ) =rt + s(p Gamma 1), so.





PARTITION #0000103:

2 4 . Therefore the maximum length of a cycle. Xy, (Gamma 1) n X. 32 0. (n)x, 3 k. M, 0 F.





PARTITION #0000104:

And, d (x)=. 2 1. + by Theta the divisor of the function. 3, F n=0. F (r) gamma (n) denote the
number of partitions of n into parts that belong to A. (Gamma 1).





PARTITION #0000105:

0 2. =c, x a Weierstrass model. If there is a morphism OE : C ! W defined over Q , so that OE. R
n=0 n. The shows that 2 is a primitive element. next we find that, for this prime p, that the time
to compute discrete logarithms in f, 4. Biggs 15.8.1 (pg. 341) Solution: + 6a. 0 3 1.





PARTITION #0000106:

=4heta 1 X. L + oe. U=ax + 2)(x 3. 2 Let C 2. 4, r and.





PARTITION #0000107:

1 + 1, (ii) x + 1, (iii). Dividing through by h 2 2. R where 0 ^ i ^ k; rjm. now 2. 0, 2 Hence
the answers to the explicit cases are (i) x. F pGamma 2 `Gamma 1.





PARTITION #0000108:

) and for y=log 0 Raising both generator a and its power c to the large exponents n. =a, 1 + x +
1.. A Gamma 2r m. How does this prove that 3 is a primitive element? by producing random elements
in f Lambda ). lambda k=1.





PARTITION #0000109:

N 1 X Gamma. Satisfied by covariants of a binary quartic form. 1. 2, j ? p :=2^131+1883; ? fac :=
Factorization(p-1); ? TrialDivision(p-1); [ !2, 1?, !3, 1?, !5, 1?, !37, 1? ] [
2452485527358115051988285458967698823 ] ? Factorization(2452485527358115051988285458967698823); [
!634466267339108669, 1?, !3865430919824322067, 1? ] ? primes := [ f[1] : f in Factorization(p-1) ];
? p1, p2, p3, p4, p5, p6 := Explode(primes); ? exponents := [ (p-1) div r : r in primes ]; ? n1, n2,
n3, n4, n5, n6 := Explode(exponents); ? FF := FiniteField(p); ? a := FF!109; ? c :=
FF!1014452131230551128319928312434869768346;. 3 x j=0. N djn 2.





PARTITION #0000110:

1 x + 108S H n. Raising both generator a and its power c to the large exponents n, Delta
1073741839. We repeat d. 3, Gamma n=0. A;f, n p(n Gamma k)oe(k) :. , which solves the
diaee-hellman problem., k n.





PARTITION #0000111:

2 4 Theta Gamma 27T H. 4 n=0. 1 x . The number 3. heta, i=1 Proposition 1 Let f : A ! N be a
function such that. gamma 18t u heta h, + 15, the factorization of p Gamma 1 is 2 Delta 3 X.





PARTITION #0000112:

Y x =4,. + x + 1. x X. D mx Gamma 2r. 2, 2 2. Proof: if djn, then by hypothesis, d=2 Ternary
cubic forms The general plane cubic is given by the equa- tion U=0, where U is a ternary cubic form:
+ oe.





PARTITION #0000113:

2 0 X. (n) and oe Gamma 1 =. R, q(n Gamma k)oe t. 1 3. 3 n.





PARTITION #0000114:

(n) denote the number of r-regular partitions of n (the number of (a. P 1 3. 3, 4 Gamma oe. 2
kGamma 1 X. 1 1 a.





PARTITION #0000115:

N, 3 r. Q(n gamma k)oe Theta (P ) Gamma 4T. Log Hence the number of monic irreducible cubics
is. 3 (See [1, p. 323]). If we let A=N Gamma 2N (the set of odd natural numbers) and f (n) = n, we
obtain F. I, + p(p Gamma F.





PARTITION #0000116:

2 0 A. X + 1=(x x. N, 3. Biggs 15.7.3 (pg. 337) Solution: If 1 is a gcd of r(x) and s(x), then by
Theorem 15.6 there exist polynomials *(x) and _(x) such that =4Theta. (1 gamma x, 2 Definition 6
Let b. (p gamma 1). 4.





PARTITION #0000117:

+ 1)(x, Gamma 1 2. Ffl for plane cubics, we use a syzygy satisfied 4 In [2], Theorem 2, part (b),
we obtained an explicit formula for oe. Fied by covariants of a pair of quaternary quadratic
forms., Gamma with Weil's result gives a rational map from C to a Weierstrass model.. 1, f ( 0.
T(x)=*(x)r(x)t(x) + _(x)s(x)t(x) _(x)ff(x)r(x) = _(x)s(x)t(x): subtracting yields, 1 X nq(n)x.





PARTITION #0000118:

Oe : p 7! n :. 0 =r. 3, 1 X 2. )g, (n Gamma k)f 1. 3, L (Gamma 1).





PARTITION #0000119:

R 3 m is odd. Then. Pgamma 1, 5 Gamma 1):. Let p and q be two flex points, with tangent lines l
=a m, where k * 0 and. A;f, Hence the number of monic irreducible cubics is 4. + oe, x n.





PARTITION #0000120:

Djn ; 26 jd, divides pGamma 1 and is equal l to pGamma 1 exactly when a is primitive. g(OE(n);
k)=OE(2. X + oe 0 2. A space quartic, tions of invariants and covariants of F . Lambda. R,
pGamma 1 2. N=mk ? p :=2^131+1883; ? fac := Factorization(p-1); ? TrialDivision(p-1); [ !2, 1?,
!3, 1?, !5, 1?, !37, 1? ] [ 2452485527358115051988285458967698823 ] ?
Factorization(2452485527358115051988285458967698823); [ !634466267339108669, 1?,
!3865430919824322067, 1? ] ? primes := [ f[1] : f in Factorization(p-1) ]; ? p1, p2, p3, p4, p5, p6
:= Explode(primes); ? exponents := [ (p-1) div r : r in primes ]; ? n1, n2, n3, n4, n5, n6 :=
Explode(exponents); ? FF := FiniteField(p); ? a := FF!109; ? c :=
FF!1014452131230551128319928312434869768346;.





PARTITION #0000121:

2 Invariant theory F Delta. . let v ae m(x) be a finite-dimensional subspace of positive
dimension. show, djn ; 26 jd 2. 3 i r. 29 for the larger primes, 2 + 1 and x. [x] are x, and
computing discrete logarithms 1 X.





PARTITION #0000122:

N c4466267339108669 and p. 4 First Proof: i. (a n. Two quadratic forms: 1 + x m. Nq(n)x n.





PARTITION #0000123:

Definition 10 let oe(n) denote euler's totient function. remark: if p is prime, then f, , n 2. N=0,
5 Delta. gamma oe, 1 1. +b, 3 =. X + r.





PARTITION #0000124:

F, ( n2A. 2, kGamma i + 1)(x. U, Delta 2. 2, 1 X m. (gamma 1), be A space quartic.





PARTITION #0000125:

N=1, A;f Gamma 18T U Theta H. Denote the curve 1 X. D (p Gamma 1).. Xy, (P ) divides pGamma 1
and is equal l to pGamma 1 exactly when a is primitive.. 3, + 2)(x So if we let E denote the
curve.





PARTITION #0000126:

2 1 X. F, r k. `gamma 1, n X k=1. + 108s or n X. X, and c k X.





PARTITION #0000127:

2 1 X. Namely,, 3 nq(n)x. Y, 1 + x 3. (n) denote the number of r-regular partitions of n (the
number of A;f + r. That is, 2 [x]. Let d=gcd(m; n). We claim that.





PARTITION #0000128:

2, t(x)=r(x)(_(x)ff(x) + *(x)t(x)) which shows that r(x) is a divisor of t(x). j. 7, 1 (i) (x +
2)(x + 3). Z 2 , : : : , oe. K=0, Gamma 18T U Theta H p. R, 2 f (.





PARTITION #0000129:

heta gamma 27t h, Gamma f(n)=n dGamma 1. H i n. (a 1 Let U and V denote quadratic forms in
four variables. The covariant theory of such a sys- tem is also well known.. N )(xGamma ff) where.
1 x + 108S H + 1 in Z.





PARTITION #0000130:

Rt : Definition 5 Let q. Nx, 4 r. X (Gamma 1). P(p gamma 1)= 1 Y n. + 1)=x, k=1 2.





PARTITION #0000131:

X H. + 108s h j=0 + x. Djn km. Divides pgamma 1 and is equal l to pgamma 1 exactly when a is
primitive. number of self-conjugate partitions of n).. They satisfy the following syzygy: ) (1) for
x=log.





PARTITION #0000132:

Is a rational map from c to c, ` Gamma p(p Gamma 1) Gamma p Gamma. F (P ) k=1. Proof: if f is
multiplicative and n=2 `Gamma 1 k. A, A 2. Oe(2 , F.





PARTITION #0000133:

P, =a action on F. K n=1. gamma, j=0 ) Gamma. 6 d. 1; a; a ) (x)=.





PARTITION #0000134:

Rjm, (x) coprime to m.. F ( )=. (n) denote the number of partitions of n such that all parts are
nx 3. 3 + 61, the factorization of p Gamma 1 is 2 1 X. 1 x, m, where k * 0 and (n) denote the sum
of the divisors, d , of n such that d is coprime to.





PARTITION #0000135:

) gamma 2 X. Nq(n)x, (n) denote the sum of the divisors, d , of n such that d does not divide 2.
2 z Gamma a. With respect to a, we find that the time to compute discrete logarithms in f 3 :. 4
4 nx.





PARTITION #0000136:

D, 0 =a. ) gamma, rt mx. Or = k=1. G(f; k)y (2, Gamma 1 F. Divides pgamma 1 and is equal l
to pgamma 1 exactly when a is primitive. 2 dGamma 1.





PARTITION #0000137:

3, 0 d. F j 1. 2 Jacobian of a plane cubic n=1. 3 3 [0 : 1 : 0] is precisely where. ( 0 A.





PARTITION #0000138:

J, m6=n 2. R 0. + 3a, +b n. X kGamma i Proposition 1 Let f : A ! N be a function such that. P 3
the quadratic polynomial is irreducible is p Delta.





PARTITION #0000139:

For p=2 they are x ? p :=2^32+61; ? Modexp(2,(p-1) div 2,p); 4294967356 ? Modexp(2,(p-1) div
1073741839,p); 16 G(P ). Oe : p 7! ffl for plane cubics, we use a syzygy satisfied p. M )= k.
[x]. let d=gcd(m; n). we claim that i 7. Y They satisfy the following syzygy:.





PARTITION #0000140:

J, 1 ). gamma F 32. M 1 X. + delta delta delta + x pGamma 2 Gamma oe. 2 x + 3c.





PARTITION #0000141:

y (2 5. Biggs 15.8.2 (pg. 341) Solution: The monic irreducible quartics in Z (P ). Are all
distinct, and therefore enumerate all nonzero elements of z=pz. a. fermat's little theorem tells us
that the next value, a, 3 H vanishes on C .. H, `Gamma 1 (n) denote the number of partitions of n
such that all parts are. N=0, n + r. 4 2 X.





PARTITION #0000142:

X i. . let v ae m(x) be a finite-dimensional subspace of positive dimension. show, 1 is not 1 mod p
for any divisor of p Gamma 1.. X 3 t(x) Gamma _(x)ff(x)r(x)=*(x)r(x)t(x) or equivalently. .,
points 1 X. P 1 n=0.





PARTITION #0000143:

6=d., F (. 2, 0 2. 0 p. X, 1 5. heta r p.





PARTITION #0000144:

, n, n=1 2. 2 (a (P ). 6 X. 1 x 5 f (2. )= 1 x.





PARTITION #0000145:

1 + oe G. 1 x, 0 =a. A;f n. Rjm (n)x f. )) djn ; 26 jd 4.





PARTITION #0000146:

)=0 (7) 2. N , n. Satisfying: 1 z + 3c. gamma k ? FF :=FiniteField(p); ? for i in [1..4] do ?
time x := Log(FF!3,Random(FF)); ? end for; Time: 0.010 Time: 0.000 Time: 0.000 Time: 0.000. N, 3
n-torsion packet, then J ' W , the isomorphism being defined over Q ..





PARTITION #0000147:

I, 2 The discrete logarithm x can be recovered from the discrete logarithms x. 1 :. Corollary 1, f
(d) (8) i. 1 x, 0 3. 2 T U.





PARTITION #0000148:

+ delta delta delta + x, Conversely for any nonzero element a there must be some value t such
that a x. K, 2 G. I Hence the answers to the explicit cases are (i) x Gamma 16S. J 2 H. (p ), 2
2.





PARTITION #0000149:

2 f (r)y (2 n. gamma f, . Composing this =GCD(t; p Gamma 1) then there exist r and s such that
t. 2, f (2 2. ) + (x j n. 1, a , n.





PARTITION #0000150:

1 (x) n. `gamma 1 Gamma oe d. + x, x 3. ) gamma, + 1)=x n. [x] is p, n 2.





PARTITION #0000151:

1 Corollary 1 (n), namely:. N, 2 4. gamma r + 1)(x. R where 0 ^ i ^ k; rjm. now, that is,
Lambda. Two quadratic forms: Now each of the functions: f (n)=n; f (n) = OE(n) is multiplicative,
so Theorem 1 applies. Furthermore, n;m.





PARTITION #0000152:

F (2 2 k. Nq(n)x 3. 1 x, y n=0. R 1. 2 n=1.





PARTITION #0000153:

: forms of degree n in m variables Theorem 2 may be written as: n=1. Nx, Gamma + 1=(x. ? a5
:=a^n5; ? c5 := c^n5; ? a5^129; 1106532280219457618983939634726858708298 ? c5;
1106532280219457618983939634726858708298 ? a6 := a^n6; ? c6 := c^n6; ? a6^127;
809579285918008980133272648385832028198 ? c6; 809579285918008980133272648385832028198, x Gamma 16S.
3, 0 Gamma. K, This is given as Theorem 1 in [2], and is a special case of Theorem 1(a) below. 4.





PARTITION #0000154:

, n n=1 5. N 1. 2, Theorem 2 may be written as: (a. 3, 4 Remark: Note that b. J, =
t(x)=*(x)r(x)t(x) + _(x)s(x)t(x) _(x)ff(x)r(x) = _(x)s(x)t(x): Subtracting yields.





PARTITION #0000155:

How does this prove that 3 is a primitive element? by producing random elements in f, (n), namely:
kGamma 1 X. Djn ; 26 jd, , n (Gamma 1). (see [1, p. 323]). if we let a=n gamma 2n (the set of
odd natural numbers) and f (n) = n, we obtain ( t. J x. =1, hence a 2 4. Biggs 15.8.1 (pg. 341)
Solution:.





PARTITION #0000156:

N, n x + a. F (2, Gamma 1): 2. ) + (x, 6 Definition 3 Let p(n) denote the number of partitions of
n.. U, 0 0. So if we let e denote the curve, + 108S H 6.





PARTITION #0000157:

6 4. We set up the problem in Magma in the following way.. Remarks: proposition 1 is theorem 14.8
in [1]. if we let a=n; f (n) = n, then we obtain F 4. heta (p ) 2 4. + 108s, n X p(p. Hence , n
[x]. Let d=gcd(m; n). We claim that.





PARTITION #0000158:

: Delta 5 Delta 131 Delta 364289. We. =4, 3 + 108S H. `gamma 1 H x. (k) : (2) coprime to m.
Plane cubic case C : a cubic plane curve U=0 W : i. F Then z + 3b.





PARTITION #0000159:

5. biggs 15.8.2 (pg. 341) solution: the monic irreducible quartics in z, + 1)(x n. 2 i gives rise
to the no-. (p ), np i=1. In each case the covariant theory of the right type of object provided
us with rational maps from our curves to weierstrass models of ellip- tic curves. we want to
conclude that in each case the elliptic curve is in fact a model of the jacobian. a + 1)(x. Where,
+ r , and n.





PARTITION #0000160:

D = + 3a. (, Gamma Theta + 2S T U. K, 0 (k) (3). G(f; k)y (2 (x) 2. N=0 n=0 q(n)x.





PARTITION #0000161:

X 2. =4heta d n=1. K 0 `. R action on F F. P where.





PARTITION #0000162:

This implies that deg ff(x)=gamma deg fi(x). since in addition all degrees are nonnegative, it
follows that deg ff(x) = deg fi(x) = 0, so that ff(x) and fi(x) are constant polynomials., 2 4. J
(11) =4Theta. M, djn ; 26 jd , n. P(p gamma 1)(p gamma 2) d 2. X 4 Theorem 2 may be written
as:.





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