Conjecture about a new LLT-like Primality Criterion based on Cycles of the DiGraph under x2 − 2 modulo a prime used for Mersenne numbers Tony Reix [email protected] First version: 2007, 10th of May Updated: 2009, 21th of February and 2010, 3rd of January. I Version 0.14 J This paper presents ideas dealing with a possible new primality criterion. The method is studied for Mersenne numbers, but it is expected that it could be used also for other numbers, like Fermat numbers, Wagstaff numbers, and others. The main idea of the method is: Consider a number N . Study the DiGraph under x2 − 2 modulo N , which is made of Trees and Cycles. Find a Universal Seed S0 for the family of the number, and then find a cycle such that Se ≡ Ss (mod N ) , with : Si+1 = Si2 − 2 (mod N ). Then find a Lucas sequence (Un , Vn ) that fits the Sn sequence and prove that if N is prime , then Se ≡ Ss (mod N ) for some starting s and ending e values. Now, study the period π(N ) of the Lucas Sequence (Un , Vn ) and prove that the period divides NQ − 1 if and only if N is prime. This can be done by proving that, if N = fi , the period of (Un , Vn ) (mod N ) is equal to lcm(πi (fi )), where π(fi ) is the period of the Lucas Sequence modulo fi , and that lcm(πi (fi )) does not divide π(N ). This DRAFT paper shows ideas for a proof for Mersenne numbers. It is crystal clear that this test cannot speed up the proof of primality of a Mersenne number. However the method used for proving it (once proved !!) could be used to prove that a Mersenne is not prime faster than the classical LLT. And the same method could be used for Fermat and Wagstaff numbers. This method makes use of the properties of the cycles of length q − 1 that appear in the Digraph under x2 − 2 modulo a Mersenne prime (the classical LLT makes use of the properties of the tree of the same Digraph). I thank Dr H.C. Williams, who supported me and provided me some help when I forgot the Little Theorem of Fermat. The fundamental ideas of this paper are based on the work of Lucas, who imagined to use the (now-called) Lucas sequences (Un (P, Q), Vn (P, Q)), which are defined as (with (P, Q) integers): Un+1 (P, Q) = P Un (P, Q) − QUn−1 (P, Q) with: U0 (P, Q) = 0, U1 (P, Q) = 1 Vn+1 (P, Q) = P Vn (P, Q) − QVn−1 (P, Q) with: V0 (P, Q) = 2, V1 (P, Q) = P 1

for proving that a number is prime, using the fastest sequence: Si+1 = Si2 −2 (mod N ), with S0 being the seed of the sequence. However, it is clear that E. Lucas never imagined to use a Cycle (ending by: S0 (mod N )) rather than a Tree (ending by: 0 (mod N )) . This is the first main new idea of the test I imagined. The second main new idea deals with the method used for trying to prove the theorem. Since the law of apparition that E. Lucas imagined does not work here (ω | N 2−1 instead of ω | N − 1 as usual), I try to use the properties of the period of Q the Lucas Sequence, showing that the period of a non-prime number N = fi is the least common multiple of the periods of the factors fi of N , and showing that this period divides N −1 if and only if N is prime. I used the books of P. Ribenboim and H.C. Williams as reference for the properties of the Lucas (Un , Vn ) sequence. I also found some good ideas in the thesis of Dr Hinkel, who describes and proves the link between the first apparition of Un ≡ 0 (mod N ) and the period of the (Un , Vn ) sequence. The first part of the attempt for providing a proof makes use of the Lucas Sequence method, as described in many papers and books, like ”The Little ´ Book of Bigger Primes” by Paulo Ribenboim or like ”Edouard Lucas and Primality Testing” by Hugh C. Williams. ¡ ¢ Here after, (a | b) or a/b is the Legendre symbol. All references to theorems apply to properties of Lucas Sequences as given by P. Ribenboim in his book ”The Litlle Book of Bigger Primes” in 2.IV pages 44-etc . Conjecture 1 (Lucas-Reix) Mq = 2q − 1 . S0 = 32 + 1/32 , Si+1 = Si2 − 2 (mod Mq ) Mq is a prime iff: Sq−1 ≡ S0 (mod Mq ) Qq−2 and iff: Si ≡ 1 (mod Mq ) 0 and iff: Si  S0 (mod Mq ), 0 < i < q − 1

Conjecture 2 (Reix) Mq = 2q − 1 is a prime iff: 3

Mq −1 2

≡ 1 (mod Mq )

The search for a prime should be done in two steps: First, find PRPs by using: Mq is prime ⇒ Sq−1 ≡ S0 (mod Mq ). Q Then, check that the number is a prime by verifying: q−2 Si ≡ 1 (mod Mq ). 0 2

1

Notations

2

General

For any N , let’s call: With gcd(n, N ) = 1, order(a, N ) is the least n > 0 such that an ≡ 1 (mod N ).

2.1

Lucas sequence

Since we use the same Lucas sequences (Un (P, Q), Vn (P, Q)) in the whole paper, it will be named as: Lucas sequences (Un , Vn ). Instead of writing: Un ≡ X (mod N ) and Vn ≡ Y (mod N ) for the same n and N , we will write: (Un , Vn ) ≡ (X, Y ) (mod N ). For any N prime, let’s call: ω = ω(N ) is the least n > 0 such that Un ≡ 0 (mod N ). Ω = Ω(N ) is the least n > 0 such that (Un , Vn ) ≡ (0, 2) (mod N ). π = π(N ) is the least n > 0 such that (Un , Vn ) ≡ (0, 2) (mod N ) and (Un+1 , Vn+1 ) ≡ (1, P ) (mod N ) . However, since, with N odd and (Un , Vn ) ≡ (0, 2) (mod N ), we have by Un n IV.5b: Un+1 = Vn +P ≡ 1 (mod N ) and by IV.5a: Vn+1 = P Vn +DU ≡P 2 2 (mod N ) , and thus: (Un+1 , Vn+1 ) ≡ (1, P ) (mod N ) . It means that, for any (P, Q) : π = Ω . And we have: (Uπ+i , Vπ+i ) ≡ (Ui , Vi ) (mod N ) for any i ≥ 0.

2.2

Properties of Mersenne composites

With Mq =

Q

fi , it is well known that:  Mq ≡ 1 (mod 6q)    Mq ≡ 7 (mod 24) fi ≡ ±1 (mod 8)    fi ≡ 1 (mod 2q)

In another paper, I’ve proved: Theorem 1 (Reix) Let Mq = 2q −1 (q prime > 3) be a Mersenne number. For each pair (a, b) of positive integers such that: Mq = F1 F2 , there exist unic pairs (x, y) and (S, D) of positive integers such that: I: Mq = (8x)2 − (3qy)2

II: Mq = (1 + Sq)2 − (Dq)2

3

Where: Fi = 1 + 2qAi , i = 1, 2 .  And:  S = A1 + A2 

D = A2 − A1 (A2 > A1 ) P = A1 A2

This entails for S: 1 + qS ≡ 0 (mod 8) and thus: S ≡ −q (mod 8) since 1/q ≡ q (mod 8) ; and for D: Dq = 3qy and thus: D ≡ 0 (mod 3). We have: Mq = (1 + 2qA1 )(1 + 2qA2 ) = 1 + 2q(S + 2qP) = 1 + 6qA. This implies: S + 2qP ≡ 0 (mod 3) and, since 1/q ≡ q (mod 3), we have: P ≡ qS (mod 3).  (mod 8)  S ≡ −q D≡ 0 (mod 3)  P ≡ qS (mod 3)

2.3

ωi , Ωi , πi for Mersenne composites

Mq not prime (q prime), let’s call fi the n > 1 prime factors of Mq : Mq = Q n 1 fi . Sometimes, we group the n factors fi in 2 sets: N = F1 F2 . fi prime, let’s call: ωi = ω(fi ) is the least n > 0 such that Un ≡ 0 (mod fi ). Ωi = Ω(fi ) is the least n > 0 such that (Un , Vn ) ≡ (0, 2) (mod fi ). πi = π(fi ) is the least n > 0 such that (Un , Vn ) ≡ (0, 2) (mod fi ) and (Un+1 , Vn+1 ) ≡ (1, P ) (mod fi ) .

3

Definition of the (Un , Vn ) sequence

With q prime > 3, we have: Mq ≡ 1 (mod 6q) . Let: β = 32 and α e ≡ 1/β (mod Mq ) . Since Mq is prime, there is only one integer α e such that 0 < α e < Mq . There are an infinity of α > Mq such that α ≡ α e (mod Mq ). Here below, we explain how to compute α e and some α (Mq prime or not). When q ≡ 1 (mod 3) and since q is odd, we have: q ≡ 1 (mod 6) . Thus Mq = 2q − 1 = 26k+1 = 2(23 )2k − 1 ≡ 1 (mod 9) . Thus 8Mq + 1 ≡ 0 8M +1 (mod 9) = 9e α and α e = 9q ≡ 1/9 (mod Mq ) . When q ≡ 2 (mod 3) and since q is odd, we have: q ≡ 5 (mod 6) . Thus Mq = 2q − 1 = 26k+5 = 32(23 )2k − 1 ≡ 4 (mod 9) . Thus 2Mq + 1 ≡ 0 2M +1 (mod 9) = 9e α and α e = 9q ≡ 1/9 (mod Mq ) . With q > 5, we always have: β < α e 1, then all such divisors are grouped in the same set F1 or F2 . Hummmm Am I correct ???????????? Let say: m = lcm(π1 , π2 ) . We have: m = k1 π1 = k2 π2 . ½ Um = Uk1 π1 ≡ Uπ1 ≡ U0 ≡ 0 (mod F1 ) Thus: Um = Uk2 π2 ≡ Uπ2 ≡ U0 ≡ 0 (mod F2 ) Since gcd(F1 , F2 ) = 1 then: Um ≡ 0 (mod F1 F2 ) . 10

Also:

½

Um+1 = Uk1 π1 +1 ≡ Uπ1 +1 ≡ U1 ≡ 1 (mod F1 ) Um+1 = Uk2 π2 +1 ≡ Uπ2 +1 ≡ U1 ≡ 1 (mod F2 )

Since gcd(F1 , F2 ) = 1 then: Um+1 − 1 ≡ 0 and thus: Um+1 ≡ 1 (mod F1 F2 ). Now, since Um+1 = And, since Vm+1 =

Vm U1 +Um V1 ≡ V2m (mod F1 F2 ) , thus Vm 2 Vm V1 +DUm U1 ≡ P V2m (mod F1 F2 ) , thus 2

≡ 2 (mod F1 F2 ). Vm+1 ≡ P .

Which shows that m is a period of (Un , Vn ) (mod F1 F2 ) . Now, since m = lcm(π1 , π2 ) is the least number that divides both π1 and π2 , there is no number t lower than m such that t is a period of (Un , Vn ) (mod F1 F2 ). As a conclusion, the combination of each period πi of (Un , Vn ) modulo a factor fi of Mq implies that the period π of (Un , Vn ) modulo Mq is equal to the least common multiple of the periods πi : π = lcm(πi ) .

6.13

− + − lcm(A+ 1 , A2 ) - F1 F2 − 1

− − + Let say: F1+ − 1 = 2qA+ 1 and F2 − 1 = 2qA2 with F1 ≡ +1 (mod 8) and F2− ≡ −1 (mod 8) . − + − + − − Now, suppose that lcm(A+ 1 , A2 ) | F1 F2 − 1 = (F1 − 1)F2 + F2 − 1 = − + + (F2 − 1)F1 + F1 − 1 . + − − + − Since we have: A+ 1 | lcm(A1 , A2 ) and A2 | lcm(A1 , A2 ) , that implies that + − − + A1 | F2 − 1 and A2 | F1 − 1 . + + Now, since F1+ = 1+8k1 = 1+2qA+ 1 , we have: qA1 = 4k1 and thus: A1 ≡ 0 (mod 4) since q is odd. − + − Now, A+ 1 = 4K1 and F2 − 1 = −2 + 8K2 shows that A1 - F2 − 1. − So there is a contradiction, showing that the hypothesis (lcm(A+ 1 , A2 ) | F1+ F2− − 1) is wrong.

6.14

Contradiction between: lcm(πi ) = π and π |

We have: πi | fi − 1 for all i and π |

6.15

Mq −1 2

Mq −1 2

= 3qa | Mq − 1 .

This is the conclusion I’d like to reach !!

π(fi ) | fi − 1 . π = lcm(πi ) . Q lcm(fi − 1) = lcm(2qai ) . lcm(ai ) - fi − 1 = Mq − 1 and 2q | Mq − 1 thus lcm(2qai ) - Mq − 1 and lcm(fi − 1) - Mq − 1 . What can we conclude with πi = π(fi ) | fi − 1 ???????????? 3 | 15 - 6 but 3 | 6 ........ 11

........ So, if Mq is not a prime, then its period π does not divide Mq − 1.

6.16

Some properties..

(D. Shanks, ”Solved and Unsolved... 1978”: Theorem 20 page 32.) With fi = 2qai + 1 then: ½ qa 3 i ≡ +1 (mod fi ) iff fi ≡ ±1 (mod 12) 3qai ≡ −1 (mod fi ) iff fi ≡ ±5 (mod 12) (W. Stein: Elementary Number Theory, 2004/11, page 73 , 4.5). p+1 If (a | p) = 1 and if p ≡ 3 (mod 4) then a 4 is a square root of a . (W. Stein: Elementary Number Theory, 2004/11, page 36 , Lemma 2.5.4) Suppose that a and b have orders r and s respectively and that gcd(r, s) = 1, then ab has order rs . W. Stein: Elementary Number Theory, 2004/11, page 74, Exercise 4.2 p ≥ 5 prime . ½ +1 iff p ≡ ±1 (mod 12) (3 | p) = −1 iff p ≡ ±5 (mod 12) D. Burton: Elementary Number Theory, 6th Edition, Chapter 8, Problem 4 page 151. Let say that the order of a (mod Q n) uisi r and that the order of b (mod n) is s, with gcd(r, s) = d and d = m i=0 pi . Then we have: order(ab, n) = Qm

rs

vi i=0,m pi

with 0 ≤ vi ≤ ui .

Thus order(ab, n) | rs , and rs d ≤ order(ab, n) ≤ rs . Obviously, if gcd(r, s) = 1 then order(ab, n) = rs. D. Burton: Elementary Number Theory, 6th Edition, Chapter 8, Corollary of 8.1 page 149. If a has order r modulo n, then the integers a, a2 , ..., ar are incongruent modulo n. W. Stein: Elementary Number Theory, 2004/11, page 73 p+1 If (a | p) = 1 and if p ≡ 3 (mod 4) then a 4 is a square root of a modulo p.

12

6.17 πi |

An idea...

fi −1 2

= qai .

Let suppose that lcm(π1 , π2 ) | F1+ F2− − 1 . Then it entails: π1 | F2− − 1 and π2 | F1+ − 1. We know : F1+ ≡ +1 (mod 8) and F1+ = 1 + 2qa1 . Thus: a1 ≡ 0 (mod 4). And: F2− ≡ −1 (mod 8) and F2− = 1 + 2qa2 . Thus: qa2 ≡ −1 (mod 4). Now, π1 | F2− − 1 entails π1 | −1 + 4k2 or π1 = q . If π1 is even, then this is impossible. How to prove that π1 is even ???? as all experimental data show ! WRONG ! That’s wrong for q = 43 with all combinations of f1 , f2 , f3 ! Check again !! By Hinkel 5.35, we have: Vω2i ≡ 4 (mod fi ) . Then: if Vωi ≡ 2 (mod fi ) then πi = ωi . And: if Vωi ≡ −2 (mod fi ) then πi = 2ωi . √ By Hinkel page 18: Vn − DUn = 2β n . With: n = ωi : Vωi ≡ 2β ωi (mod fi ) . So we have to solve: β ωi ≡ ±1 (mod fi ). Since β = 32 then: β ωi = (32 )ωi = (3ωi )2 ≡ ±1 (mod fi ) . Since (−1 | p)2 = 1 when p ≡ 1 (mod 4) and -1 otherwise, then fi ≡ 1 (mod 8) is a necessary condition in order to have πi = 2ωi . So, still with fi ≡ 1 (mod 8) then x2 ≡ −1 (mod fi ) has a solution. How can we prove that x = 3ω ?? We have the following experimental data in table 2, from q in table 3 (− means −1. + means +1): fi (mod 8) − − + +

fi (mod 3) − + − +

πi /ωi 1 1 2 2

ρi /πi 1 2 2 2

fi (mod 24) 23 7 17 1

q (mod 3) ± + − ±

Table 2: Relationships between ωi , πi , ρi for the factors fi of Mq

6.18

Observed properties of period of Mersenne composites

When Mq has exactly 2 divisors, with π1 and π2 being the period of (Un , Vn ) modulo f1 and f2 respectiveley, let define: s = π1 + π2 , d = π2 − π1 (with π2 > π1 ), and p = π1 π2 . Note that we have: q | s , q | d and q 2 | p . 13

And we have: 1/q ≡ q (mod 8) and 1/q ≡ q (mod 3). We observe the following properties (for   s ≡ −1 d ≡ 0  p ≡ qs Showing that we (should...) have:   s/q ≡ S d/q ≡ D  p ≡ P

q = 11, 23, 37, 41): (mod 8) (mod 3) (mod 3)

(mod 8) (mod 3) (mod 3)

For q = 11, 23, 41 (−1 (mod 3)), we have: s = qS , d = D , p = q 2 P. Let call: M = Mq . For q = 37 (+1 (mod 3)), we have: 2s = q(S +3) , 2d = q(D −3) , 2p = q 2 P. Let call: M = 2Mq − fi . And it seems that: M = (1 + s)2 − d2 = (8x)2 − (3qy)2 ≡ 1 (mod 6q) , like Mq ...

7

Examples           q=5

P   Q        D

M5 β α = α+β = αβ gcd(P, Q) = P 2 − 4Q 1

2

= 31 =9 = 102 = 109 = 900 =1 = 912

≡ 7 (mod M5 ) ≡ 16 (mod M5 ) ≡ 1 (mod M5 ) ≡ 4 (mod M5 )

4=q−1

3

(mod M5 ) S0 = P ≡ 16 7→ 6 7→ 3 7→ 7 7→ 16 Q3 0 Si ≡ 16 ∗ 6 ∗ 3 ∗ 7 ≡ 1 (mod M5 )           q=7

        

P Q D

M7 β α = α+β = αβ gcd(P, Q) = P 2 − 4Q

= 127 =9 = 422 = 1773 = 15876 =9 = 17552

14

≡ 113 (mod M7 ) ≡ 122 (mod M7 ) ≡1 (mod M7 ) ≡ 21

(mod M7 )

1

2

3

4

6=q−1

5

(mod M7 ) S0 = P ≡ 122 7→ 23 7→ 19 7→ 105 7→ 101 7→ 39 7→ 122 Q5 0 Si ≡ 1 (mod M7 )           q = 11

P   Q        D

M11 β α = α+β = αβ gcd(P, Q) = P 2 − 4Q 1

= 2047 =9 = 6822 = 465133 = 4186116 =1 = 4651152

2

≡ 455 (mod M11 ) ≡ 464 (mod M11 ) ≡1 (mod M11 )

3

≡ 446

(mod M11 )

4

5

6

7

(mod M11 ) S0 = P ≡ 464 7→ 359 7→ 1965 7→ 581 7→ 1851 7→ 1568 7→ 175 7→ 8

10=q−1

9

1965 7→ 581 7→ 1851 7→ 1568 Q9 0 Si ≡ 1312 (mod M11 )           q = 13

M13 β α = α+β = αβ gcd(P, Q) = P 2 − 4Q

P   Q        D

= 8191 =9 = 27302 = 7452909 = 67076100 =9 = 74528912

1

2

10

11

≡ 7281 (mod M13 ) ≡ 7290 (mod M13 ) ≡1 (mod M13 ) ≡ 888

3

4

(mod M13 ) 5

6

(mod M13 ) S0 = P ≡ 7290 7→ 890 7→ 5762 7→ 2519 7→ 5525 7→ 5957 7→ 7

8

9

2435 7→ 7130 7→ 3552 7→ 2562 7→ 2851 7→ 2727 Q11 0 Si ≡ 1 (mod M13 )

8

12=q−1

7→

7290

Periods (mod Mq )

Mq = 1 + 2qα =

Q

(1 + 2qαi ) .

In the following table (3), α[i] is α whenQMq is prime, and Q αi when Mq is not prime. When Mq is not prime, π/q α always is π/ αi . This shows that the period π depends clearly on Mq being prime or not. Note that: 1) q | πq , and: 2) πq is odd when Mq is prime (in bold), though πq is odd or even when Mq is not prime. Now consider Qthe following table (4) (for Mq prime), and compare with the column π/q α[i] for q of the previous table with Mq prime. 15

q 5 7 11 13 17 19 23 29 31 37 41 43 47 53

π 15 63 44 455 65535 262143 89240 19836 357913941 154079544 13407675188 118642203 165806600 4007499600

Factors of π 3q 32 q 22 q 5.7q 3.5.q.257 33 .7.q.73 23 .5.q.97 22 .32 .19.q 3.7.11.q.151.331 23 .3.q.167.1039 22 .q.59.163.8501 32 .q.113.2713 23 .52 .31.q.569 24 .3.52 .13.37.q.131

(Mq − 1)/π 2 2 93/2 18 2 2 182361/1940 3085465/114 6 619094385/694052 2199023255551/13407675188 22728922538/306569 1497207322929/1763900 5664905191661/2520440

q

Q

Table 3: Periods modulo Mq The first part of the table has been built by computing the order order(3, Mq ). The second part of the table has been built by the following process: 1) Find I the greatest i such that Mq −1 ≡ 0 (mod 3i ) ; 2) With n = (Mq −1)/3i , i = 0..I compute 3n (mod Mq ) and find O the greatest i such that 3n ≡ 1 (mod Mq ). We see: O ≤ I . Based on the data in table 4, we have the conjecture: Mq − 1 = 3n with n = 0, 1, 2 order(3, Mq ) I have been warned that the conjecture is wrong for: q = 3217 and many others.... Nuts !!!! q = 3217; Mq = 2q − 1; Mq − 1 = 3 × 13 × order(3, Mq )

9

Period (mod fi )

Table 5. The + or − sign before each q indicates if it is ±1 (mod 3). The + or − sign before each fi indicates if it is ±1 (mod 8). The + or − sign after each fi indicates if it is ±1 (mod 3). Notice that πi = ωi iff fi = −1 (mod 8) and πi = 2ω iff fi = +1 (mod 8) . The last column in table 5 shows the relationship between the order(3, fi ) i) is 1 or 2 accordingly to and the period πi (mod fi ) of (Un , Vn ) : order(3,f πi 16

α[i] π

1 1 1 9 1 1 1 4 3 6 1 5 48 100

q 3 5 7 13 17 19 31 61 89 107 127 521 607 1279 2203 2281 3217 4253 4423 9689 9941 11213 19937 21701 23209 44497 86243 110503

3O =

Mq −1 order(3,Mq )

O

1 1 1 9 1 1 3 9 1 1 3 1 3 3 1 1 3 1 3 1 3 3 3 1 9 1 1 3

0 0 0 2 0 0 1 2 0 0 1 0 1 1 0 0 1 0 1 0 1 1 1 0 2 0 0 1

I = max i/ Mq − 1 ≡ 0 (mod 3i ) 1 1 2 2 1 3 2 2 1 1 3 1 2 3 2 2 2 1 2 1 1 1 1 1 2 4 1 3

Table 4: (Mq − 1)/order(3, Mq ) .

17

order(3,Mq ) π

2 2 2 2 2 2

the following conjectured relationship is: order(3, fi ) = πi iff: fi ≡ −1 (mod 8) and fi ≡ −1

(mod 3)

order(3, fi ) = 2πi otherwise

10

Details for q = 5

See table 6.

11

Useful ???

− And, since F2− = −1 + 8k2 = 1 + 2qA− 2 , we have: qA2 = 4k2 − 1 .

When q ≡ +1 (mod 4), then A− 2 ≡ −1 (mod 4) . When q ≡ −1 (mod 4), then A− 2 ≡ +1 (mod 4) . And thus: A− ≡ ±1 (mod 4) . 2 − − So, with A− 2 = ±1 + 4K2 we have : F2 − 1 = 2qA2 = 2q(±1 + 4K2 ) ≡ ± − 2 (mod 8) .

18

± q − 11

order(3,fi ) πi

±fi

ωi

πi

πi

fi −1 πi

− 23 − + 89 −

11 22

ω 2ω

q 22 q

2 2

1 2

− 47 − + 178481 −

23 44620

ω 2ω

q 23 .5.q.57

2 2

1 2

− − + + − −

58 551 261 1102 522 4859

2ω ω 2ω 2ω 2ω 2ω

4q 19q 2.32 q 22 .19q 22 .32 q 2.32 .19q

2 2 4

2 1 2 2 2 2

− 223 + + 616318177 +

111 77039772

ω 2ω

3q 23 .3.q.167.1039

2 4

2 2

− 13367 − + 164511353 −

6683 41127838

ω 2ω

q163 22 q59.8501

2 2

1 2

− 431 − 9719 − 2099863 + f1 × f2 + f1 × f3 + f2 × f3

− − + + − −

43 4859 1049931 4859 1049931 118642203

ω ω ω ω ω ω

q q113 32 q2713 q113 2 3 q2713 32 q113.2713

10 2 2

1 1 2 1 2 2

− 2351 + 4513 + 13264529 − f1 × f2 − f1 × f3 + f2 × f2

− + − − + −

1175 94 3316132 2350 82903300

ω 2ω 2ω 2ω 2ω

52 q 22 q 23 .31.q.569 22 .52 q 3 2 2 .5 .31q569

2 24 2

1 2 2 2 2

+ 6361 + − 69431 − + 13264529 −

795 34715 5098600

2ω ω 2ω

2.3.5q 5q131 24 .52 .13.37q

4 2 2

2 1 2

− 23

− 29 + 233 − 1103 + 2089 − f1 × f2 + f1 × f3 − f2 × f3 + 37

− 41

+ 43

− 47

− 53

Table 5: Periods modulo fi

19

i 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

Ui (mod M5 ) 0 1 16 7 3 10 2 22 9 29 21 28 24 15 30 0 1 16 7 3 10 2 22 9 29 21 28 24 15 30 0 1 16 7

Vi (mod M5 ) 2 16 6 18 3 30 12 7 7 12 30 3 18 6 16 2 16 6 18 3 30 12 7 7 12 30 3 18 6 16 2 16 6 18

Table 6: (Ui , Vi ) (mod M5 )

20