Factorization of Large Numbers via Factorization of Small Numbers

Transcription

1 Global Journal of Pure and Applied Mathematics. ISSN Volume, Number 6 (6), pp Research India Publications Factorization of Large Numbers via Factorization of Small Numbers Xingbo Wang Department of Mechatronics Foshan University, Foshan City, Guangdong Province, PRC, 58 Abstract The article puts forward an elementary, easily-applicable and parallelable approach to factorize big odd composite numbers. The approach can turn factorization of a big odd composite number into factorization of certain small numbers. Mathematical deductions are exhibited in detail and numerical experiments are presented via factorization of the Mersenne numbers and Fermat numbers. Experiments show that, the more factors the big number contains, the faster the approach works. Keywords: Integer factorization; Odd composite number; Multiplication table; Algorithm MSC : A5,A5. INTRODUCTION Factorization of big integers has been a key topic in the field of number theory and cryptography. To meet the demands of information security, factorizing a big number still attracts a lot of researchers to contribute their efforts in developing kinds of approaches, as summarized in literatures [] to [4]. Among the various approaches, the multiplication-table based one is most easilyunderstood and easily-applicable. This was demonstrated by Jongsoo Park and Mathology Sys in their paper[5], as well as in WANG's paper[6]. This article, as a following work of the paper[6], makes a further investigation on multiplication-table based and tail-dependent factorization of odd composite numbers in a way that taking both the tail (the ones-digit) and the form (in term of 6n ) of a odd composite number into account. It first proves several equivalent relations between an odd composite number and its factors through their tails that are subordinate to the

2 558 Xingbo Wang multiplication-table, then digs out a transitional law among odd composite numbers of same form with the same tail, and then designs a parallelable algorithm that factorizes a big odd number by factorizing a certain small one.. PRELIMINARIES. Definitions and Notations In this whole article, symbol Z denotes all the positive integers and symbol Z denotes the positive integers together with. Symbol A Bmeans that result B is derived from condition A; symbol x Z means there exists an x in set Z +. Let k k integer n s decimal representation be n nk nk... n n ( nk nk... nn ) then the unit s digit n is called tail of n. This whole article only concerns integer s decimal representation. Symbol a b means b can be divided by a; symbol ( ab, ) and [ abare, ] to express respectively the greatest common divisor (GCD) and the least common multiple (LCM) of integers a and b. Suppose a set S is composed of elements s, s,..., namely, S { s, s,...} ; then symbol Ssmeans. i the i th term s i of S. In this whole article, odd number means that is bigger than 3 and are of the form 6k (k>)unless special remark is made.. Lemmas Lemma.. An arbitrary non-negative integers N can be expressed by N 6 n r, r 5, and an odd number bigger than 3 must be of the form either N 6n or N 6n, where n Z. If N is of the form 6n+ then so it is with N+6 for arbitrary integer ; if N is of the form 6n+ then so it is with N 6 for arbitrary integer n. If N is of the form 6n then so it is with N+6 for arbitrary integer ; if N is of the form 6n then so it is with N 6 for arbitrary integer n. Lemma. If N=(6k+)(6l+) is composite, then N6(6l+), N6(6k+) are all composite numbers for <k, <l. If N=(6k-)(6l-) is composite, then N6(6l-), N+6(6k-) are all composite numbers, for <k, <l. If N=(6k+)(6l-) is composite, then N+6(6l-), N+6(6k+) are all composite numbers, for <k, <l. Lemma.3. Integer n s decimal tail is n if and only if n n (mod). Lemma.4. The following statements on odd number N >3 are true for n k Z. (i) N is of the form 6n with tail if and only if it is of the form 3k 3. (ii) N is of the form 6n with tail 3 if and only if it is of the form 3k 3. (iii) N is of the form 6n with tail 7 if and only if it is of the form 3k 7. Z and

3 Factorization of Large Numbers via Factorization of Small Numbers 559 (iv) N is of the form 6n with tail 9 if and only if it is of the form 3k 9. (v) N is of the form 6n with tail if and only if it is of the form 3k. (vi) N is of the form 6n with tail 3 if and only if it is of the form 3k 3. (vii) N is of the form 6n with tail 7 if and only if it is of the form 3k 7. (viii) N is of the form 6n with tail 9 if and only if it is of the form 3k 9. Proof. Take the cases (i) and (vi) as examples. (i) Sufficiency. Let x3k 3; then it is of course of the form 6n under the condition of n Z and k Z. Note that, x 3(mod3) x 3(mod3) x (mod5), x (mod 6); x (mod3), x (mod) By Lemma.3, the sufficiency holds. Necessity. By Lemma.4, odd number N 6n with tail leads to N (mod) Hence 6n m 3n 5m. Since n Z and (3,5), it yields n 5 s, m 3 t, s Z, t Z. Consequently, N 3s 3k 3with k Z. (vi) Let N 6nwith tail 3; then N 3(mod) 6n m 3 3n 5m n 4 5 t, m 3t 6n 6(4 5 t) 3t 3 Lemma.5. Suppose m, n, s and t be non-negative integers; then N (3 m s)(3 n t) is of the form 3k if and only if st (mod3). Proof. N (3 m s)(3 n t) 3 mn 3tm 3sn st 3(3 mn tm sn) st Sufficiency. st (mod3) st 3u N 3(3 mn tm sn u) 3k. Necessity. N (3 m s)(3 n t) being of the form 3k results in N (mod3) 3(3 mn tm sn) st (mod3) st (mod3) ( st, ) Lemma.6. Let M m s n t where m, n, s and t are non-negative integers; if ( mn, ) (3 )(3 ) st (mod3) then for arbitrary q Z, it hold Proof.(Omitted) ( st, ) ( st, ) M( mn, ) 3 q (mod3), M( m q, n) (3( m q) s)(3 n t) (mod3) and ( st, ) M( m, nq) (3 m s)(3( n q) t) (mod3).

4 56 Xingbo Wang 3. MAIN RESULTS AND PROOFS Main results presented in this section include four parts. The first part is to show characteristics of factors in odd composite numbers 6n, the second is to construct sieves to select the odd composite numbers 6n according to the numbers tails, the third is to introduce a general approach to factorize an odd composite number according to the number s tail, and the fourth is to show the approach to factorize big odd composite numbers through factorization of certain small ones. 3. Factors in Odd Composite Numbers 6n Theorem 3.. Let N=6n+ be a composite number; then for certain s, t Z, N can be factorized into one of the following forms that are given in (3.), (3.), (3.3) and (3.4) in accordance with its tail being,3,7 or 9, respectively. m (3s 7)(3t 3) m (3s 3)(3t 7) m3 (3s 9)(3t 9) m4 (3s 3)(3t 3) ; s, t Z m5 (3s )(3t ) m6 (3s 7)(3t 3) m7 (3s 3)(3t 7) m8 (3s 9)(3t 9) m3 (3s 9)(3t 7) m3 (3s 7)(3t 9) m33 (3s 3)(3t 3) m34 (3s 3)(3t 3) ; s, t Z m35 (3s 3)(3t ) m36 (3s )(3t 3) m37 (3s 9)(3t 7) m38 (3s 7)(3t 9) m7 (3s 3)(3t 7) m7 (3s 7)(3t 3) m73 (3s 9)(3t 3) m74 (3s 3)(3t 9) ; s, t Z m75 (3s 7)(3t ) m76 (3s )(3t 7) m77 (3s 9)(3t 3) m78 (3s 3)(3t 9) m9 (3s 7)(3t 7) m9 (3s 3)(3t 3) m93 (3s 3)(3t 9) m94 (3s 9)(3t 3) ; s, t Z m95 (3s 7)(3t 7) m96 (3s 3)(3t 3) m97 (3s 9)(3t ) m98 (3s )(3t 9) (3.) (3.) (3.3) (3.4)

5 Factorization of Large Numbers via Factorization of Small Numbers 56 where the upper index + means the factorization is corresponding to N 6n and the first lower index i,3,7,9 means N is tailed by i, and the second lower index j,,...,8 means the j th one of the 8 possible factorizations. Proof. Here only take N of tail as an example. Let N ab with a, b Z. Since N is an odd composite number of the form 6n+, by Lemma, there are two possible cases matching to a and b. (i) a 6k and b6l, k, l Z ; (ii) a6kand b6l, k, l Z. By Lemma.4, pair (a, b) in case (i) satisfy ( a, b) {(3x 7,3 y 3),(3 x 3,3 y 7),(3x 9,3 y 9),(3 x 3,3 y 3) x, y Z } Hence the product N (6k )(6 l ) with tail must be in the set P given by P x y x y x y x y x y Z {(3 7,3 3),(3 3,3 7),(3 9,3 9),(3 3,3 3), } Similarly, the product N (6k )(6 l ) with tail must be in the set P given by P x y x y x y x y x y Z {(3,3 ),(3 7,3 3),(3 3,3 7),(3 9,3 9), } Theorem 3.. Let N be an odd composite number of the form 6n-with tail, 3,7 and 9; then for certain s, t Z, N can be factorized into one of the following forms that are given in (3.5), (3.6), (3.7) and (3.8) in accordance with its tail being,3,7 or 9, respectively. m (3s 3)(3t 7) m (3s 7)(3t 3) m3 (3s 7)(3t 3) m4 (3s 3)(3t 7) ; s, t Z m5 (3s 9)(3t 9) m6 (3s 9)(3t 9) m7 (3s )(3t 3) m8 (3s 3)(3t ) m3 (3s 9)(3t 7) m3 (3s 7)(3t 9) m33 (3s )(3t 3) m34 (3s 3)(3t ) ; s, t Z s35 (3s 7)(3t 9) m36 (3s 9)(3t 7) m37 (3s 3)(3t 3) m38 (3s 3)(3t 3) (3.5) (3.6)

6 56 Xingbo Wang m7 (3s )(3t 7) m7 (3s 7)(3t ) m73 (3s 9)(3t 3) m74 (3s 3)(3t 9) ; s, t Z m75 (3s 3)(3t 9) m76 (3s 9)(3t 3) m77 (3s 7)(3t 3) m78 (3s 3)(3t 7) (3.7) m9 (3s 7)(3t 7) m9 (3s 7)(3t 7) m93 (3s 3)(3t 3) m94 (3s 3)(3t 3) ; s, t Z m95 (3s )(3t 9) m96 (3s 9)(3t ) m97 (3s 9)(3t 3) m98 (3s 3)(3t 9) where the upper index - means the factorization is corresponding to N 6n and the first lower index i,3,7,9 means N is tailed by i, and the second lower index j,,...,8 means the j th one of the 8 possible factorizations. Proof. (Omitted) (3.8) 3. Sieve of Odd Composite Numbers of 6k According to Theorem 3. and 3., each odd composite number has its own form of factorization in correspondence to its tail and in terms of its form of 6n. Therefore a sieve can be constructed by means of that both an odd number's tail and its form of 6n are taken into account. This section constructs such a sieve. First, construct a sieve for odd composite numbers of the form 6k+ with tail by defining set S such that with S { s, s, s, s, s, s, s, s } (3.9) s { x x s ( k ) d, s 5l l 65, d 5(3( l ) 7); l, k Z } s { x x s ( k ) d, s 5l l 65, d 5(3( l ) 3); l, k Z } s3 { x x s ( k ) d, s 5l l, d 5(3( l ) 9); l, k Z } s4 { x x s ( k ) d, s 5l l, d 5(3( l ) 3); l, k Z } s5 { x x s ( k ) d, s 5l l 5, d 5(3( l ) 7); l, k Z } s6 { x x s ( k ) d, s 5l l 5, d 5(3( l ) 3); l, k Z } s7 { x x s ( k ) d, s 5l 9l 6, d 5(3( l ) ); l, k Z } s8 { x x s ( k ) d, s 5l l, d 5(3( l ) 9); l, k Z } Then the following theorem 5 ensures that S form a sieve for odd numbers of the form 6n+ and of the tail. (3.)

7 Factorization of Large Numbers via Factorization of Small Numbers 563 Theorem 3.3. Let S be defined and given by (3.9) and (3.); then (i) odd number 6x ( xs, i,,...,8) is a composite number and is tailed by l ; i (ii) arbitrary odd composite number of the form 6x with tail can find a seed in Proof. (i) Direct calculations yield S. where s3k3 and t 3l 3 6s (3l 3 3k 3 3)(3( l ) 7) (6( s t) 3)(6t 7) 6s (3l 3 3k 3 7)(3( l ) 3) (6( s t) 7)(6t 3) 6s3 (3l 3 3k 3 9)(3( l ) 9) (6( s t) 9)(6t 9) 6s4 (3l 3 3k 3 3)(3( l ) 3) (6( s t) 3)(6t 3) ; k, l Z 6s5 (3l 3 3k 3 3)(3( l ) 7) (6( s t) 3)(6t 7) 6s6 (3l 3 3k 3 7)(3( l ) 3) (6( s t) 7)(6t 3) 6s7 (3l 3 3k 3 )(3( l ) ) (6( s t) )(6t ) 6s8 (3l 3 3k 3 9)(3( l ) 9) (6( s t) 9)(6t 9). (3.) It can immediately see that all the numbers 6x ( xs i, i,,...,8) are composite and are of the form 3n and are of tail. (ii) By Theorem 3, arbitrary odd composite number of the form 6x with tail can be factorized into one of the 8 forms in (3.). Since (3.) is equivalent to (3.), which is derived from S, it is sure that arbitrary odd composite number of the form 6x with tail must find a seed in S. Similarly, with S, S, S, S, S, S and S are constructed as follows. 7 9 S { s, s, s, s, s, s, s, s } (3.) s3 { x x s ( k ) d, s 5l 7l 4, d 5(3( l ) 7); l, k Z } s3 { x x s ( k ) d, s 5l 7l 4, d 5(3( l ) 9); l, k Z } s33 { x x s ( k ) d, s 5l 8l 3, d 5(3( l ) 3); l, k Z } s34 { x x s ( k ) d, s 5l 8l 3, d 5(3( l ) 3); l, k Z } s35 { x x s ( k ) d, s 5l 3l, d 5(3( l ) ); l, k Z } s36 { x x s ( k ) d, s 5l 3l, d 5(3( l ) 3); l, k Z } s37 { x x s ( k ) d, s 5l 7l, d 5(3( l ) 7); l, k Z } s38 { x x s ( k ) d, s 5l 7l, d 5(3( l ) 9); l, k Z } (3.3) S { s, s, s, s, s, s, s, s } (3.4)

8 564 Xingbo Wang with with with with s7 { x x s ( k ) d, s 5l l 4, d 5(3( l ) 7); l, k Z } s7 { x x s ( k ) d, s 5l l 4, d 5(3( l ) 3); l, k Z } s73 { x x s ( k ) d, s 5l 4l 3, d 5(3( l ) 3); l, k Z } s74 { x x s ( k ) d, s 5l 4l 3, d 5(3( l ) 9); l, k Z } s75 { x x s ( k ) d, s 5l 6l 4, d 5(3( l ) ); l, k Z } s76 { x x s ( k ) d, s 5l 6l 4, d 5(3( l ) 7); l, k Z } s77 { x x s ( k ) d, s 5l 4l, d 5(3( l ) 3); l, k Z } s78 { x x s ( k ) d, s 5l 4l, d 5(3( l ) 9); l, k Z } (3.5) S { s, s, s, s, s, s, s, s } (3.6) s9 { x x s ( k ) d, s 5l 3l 88, d 5(3( l ) 7); l, k Z } s9 { x x s ( k ) d, s 5l 7l 48, d 5(3( l ) 3); l, k Z } s93 { x x s ( k ) d, s 5l 5l, d 5(3( l ) 9); l, k Z } s94 { x x s ( k ) d, s 5l 5l, d 5(3( l ) 3); l, k Z } s95 { x x s ( k ) d, s 5l 3l 8, d 5(3( l ) 7); l, k Z } s96 { x x s ( k ) d, s 5l 7l 8, d 5(3( l ) 3); l, k Z } s97 { x x s ( k ) d, s 5l l 3, d 5( 3( l ) ); l, k Z } s98 { x x s ( k ) d, s 5l l 3, d 5(3( l ) 9); l, k Z } (3.7) S { s, s, s, s, s, s, s, s } (3.8) s { x x s ( k ) d, s 5l 5l 7, d 5(3( l ) 7); l, k Z } s { x x s ( k ) d, s 5l 5l 7, d 5(3( l ) 3); l, k Z } s3 { x x s ( k ) d, s 5l 5l 37, d 5(3( l ) 3); l, k Z } s4 { x x s ( k ) d, s 5l 5l 37, d 5(3( l ) 7); l, k Z } s5 { x x s ( k ) d, s 5l 6l, d 5(3( l ) 9); l, k Z } s6 { x x s ( k ) d, s 5l 6l, d 5(3( l ) 9); l, k Z } s7 { x x s ( k ) d, s 5l 9l 3, d 5(3( l ) 3); l, k Z } s8 { x x s ( k ) d, s 5l 9l 3, d 5(3( l ) ); l, k Z } (3.9) S { s, s, s, s, s, s, s, s } (3.) s3 { x x s ( k ) d, s 5l l 4, d 5(3( l ) 7); l, k Z } s3 { x x s ( k ) d, s 5l l 4, d 5(3( l ) 9); l, k Z } s33 { x x s ( k ) d, s 5l 8l 54, d 5(3( l ) 3); l, k Z } s34 { x x s ( k ) d, s 5l 8l 54, d 5(3( l ) ); l, k Z } s35 { x x s ( k ) d, s 5l l 4, d 5(3( l ) 9); l, k Z } s36 { x x s ( k ) d, s 5l l 4, d 5(3( l ) 7); l, k Z } s37 { x x s ( k ) d, s 5l 3l, d 5(3( l ) 3); l, k Z } s38 { x x s ( k ) d, s 5l 3l, d 5(3( l ) 3); l, k Z } (3.) S { s, s, s, s, s, s, s, s } (3.)

9 Factorization of Large Numbers via Factorization of Small Numbers 565 with with s7 { x x s ( k ) d, s 5l l 73, d 5(3( l ) 7); l, k Z } s7 { x x s ( k ) d, s 5l l 73, d 5(3( l ) ); l, k Z } s73 { x x s ( k ) d, s 5l 9l 3, d 5(3( l ) 3); l, k Z } s74 { x x s ( k ) d, s 5l 9l 3, d 5(3( l ) 9); l, k Z } s75 { x x s ( k ) d, s 5l 9l 3, d 5(3( l ) 9); l, k Z } s76 { x x s ( k ) d, s 5l 9l 3, d 5(3( l ) 3); l, k Z } s77 { x x s ( k ) d, s 5l 6l, d 5(3( l ) 3); l, k Z } s38 { x x s ( k ) d, s 5l 6l, d 5(3( l ) 7); l, k Z } (3.3) S { s, s, s, s, s, s, s, s } (3.4) s9 { x x s ( k ) d, s 5l 8l 5, d 5(3( l ) 7); l, k Z } s9 { x x s ( k ) d, s 5l 8l 5, d 5(3( l ) 7); l, k Z } s93 { x x s ( k ) d, s 5l l, d 5(3( l ) 3); l, k Z } s94 { x x s ( k ) d, s 5l l, d 5(3( l ) 3); l, k Z } s95 { x x s ( k ) d, s 5l 5l 35, d 5(3( l ) 9); l, k Z } s96 { x x s ( k ) d, s 5l 5l 35, d 5(3( l ) ); l, k Z } s97 { x x s ( k ) d, s 5 l, d 5(3( l ) 3); l, k Z } s98 { x x s ( k ) d, s 5 l, d 5(3( l ) 9); l, k Z } (3.5) 3.3 General Approach To Factorize Odd Composite Numbers Based on sieves (3.9) to (3.5), new approaches to factorize odd composite numbers can be found as introduced in the following subsections Theorem with Proof Theorem 3.4. Suppose N is an odd composite number; s. s s, s. d are defined in sieves S and S respectively, where,3,7,9 following statements are true. i i s. s,, s, s. d i, j,,...,8 n (i) For N 6nwith tail i if there exists an l such that 5 s. d ( ) N 5. s. d ( n s. s ) n (ii) For N 6nwith tail i if there exists an l such that 5 s. d ( ) N 5. Proof. Taking s 3 as an example, a direct calculation yields s. d n s. s and ; then the, then, then

10 566 Xingbo Wang and s. s 5l 7l 4 3 s3. d 5(3( l ) 7) n k( s3. d) s3. s s3. d n s3. s n k(5(3( l ) 7)) 5l 7l 4; k Z, l Z 6n 6(5l 7l 4 k(5(3( l ) 7))) 3 l 3 34l 3 k(3( l ) 7) 3 (3l 3k )(3( l ) 7) s. d n s. s ( s. d) n s. s (3 l 3 3l 3 ) ( n 5l 7l 4) (5 5) l ( ) l ( n 5 3 4) l ( ) (( )) 455 ( n 5 3 4) ( 5 5) n n 55 5 The other cases are similarly calculated and all the results are list as follows. 6s (3l 3 3k 3)(3( l ) 7) 6s (3l 3 3k 7)(3( l ) 3) 6s3 (3l 3 3k 9)(3( l ) 9) 6s4 (3l 3 3k 3)(3( l ) 3) ; k, l Z 6s5 (3l 3 3k 3)(3( l ) 7) 6s6 (3l 3 3k 7)(3( l ) 3) 6s7 (3l 3 3k )(3( l ) ) 6s8 (3l 3 3k 9)(3( l ) 9) 6s3 (3l 3k 3 9)(3( l ) 7) 6s3 (3l 3k 3 7)(3( l ) 9) 6s33 (3l 3k 3 3)(3( l ) 3) 6s34 (3l 3k 3 3)(3( l ) 3) ; k, l Z 6s35 (3l 3k 3 3)(3( l ) ) 6s36 (3l 3k 3 )(3( l ) 3) 6s37 (3l 3k 3 9)(3( l ) 7) 6s38 (3l 3k 3 7)(3( l ) 9)

11 Factorization of Large Numbers via Factorization of Small Numbers 567 6s7 (3l 3 3k )(3( l ) 7) 6s7 (3l 3k 3 3)(3( l ) 3) 6s73 (3l 3k 3 )(3( l ) 3) 6s74 (3l 3k 3 7)(3( l ) 9) ; k, l Z 6s75 (3l 3k 3 3)(3( l ) ) 6s76 (3l 3k 3 9)(3( l ) 7) 6s77 (3l 3k 3 )(3( l ) 3) 6s78 (3l 3k 3 7)(3( l ) 9) 6s9 (3l 3k 3 3)(3( l ) 7) 6s9 (3l 3k 3 7)(3( l ) 3) 6s93 (3l 3k 3 )(3( l ) 9) 6s94 (3l 3k 3 )(3( l ) 3) ; k, l Z 6s95 (3l 3k 3 3)(3( l ) 7) 6s96 (3l 3k 3 7)(3( l ) 3) 6s97 (3l 3k 3 )(3( l ) ) 6s98 (3l 3k 3 9)(3( l ) 9) 6s (3l 3k 3 7)(3( l ) 7) 6s (3l 3k 3 3)(3( l ) 3) 6s3 (3l 3k 3 3)(3( l ) 3) 6s4 (3l 3k 3 7)(3( l ) 7) ; k, l Z 6s5 (3l 3k 3 )(3( l ) 9) 6s6 (3l 3k 3 )(3( l ) 9) 6s7 (3l 3k 3 9)(3( l ) 3) 6s (3l 3k 3 )(3( l ) ) 8

12 568 Xingbo Wang 6s3 (3l 3k 3 )(3( l ) 7) 6s3 (3l 3k 3 3)(3( l ) 9) 6s33 (3l 3k 3 9)(3( l ) 3) 6s34 (3l 3k 3 7)(3( l ) ) ; k, l Z 6s35 (3l 3k 3 3)(3( l ) 9) 6s36 (3l 3k 3 )(3( l ) 7) 6s37 (3l 3k 3 7)(3( l ) 3) 6s (3l 3k 3 )(3( l ) 3) 38 6s7 (3l 3k 3 9)(3( l ) 7) 6s7 (3l 3k 3 3)(3( l ) ) 6s73 (3l 3k 3 )(3( l ) 3) 6s74 (3l 3k 3 7)(3( l ) 9) ; k, l Z 6s75 (3l 3k 3 7)(3( l ) 9) 6s76 (3l 3k 3 )(3( l ) 3) 6s77 (3l 3k 3 3)(3( l ) 3) 6s (3l 3k 3 )(3( l ) 7) 38 6s9 (3l 3k 3 3)(3( l ) 7) 6s9 (3l 3k 3 3)(3( l ) 7) 6s93 (3l 3k 3 7)(3( l ) 3) 6s94 (3l 3k 3 7)(3( l ) 3) ; k, l Z 6s95 (3l 3k 3 9)(3( l ) 9) 6s96 (3l 3k 3 )(3( l ) ) 6s97 (3l 3k 3 )(3( l ) 3) 6s98 (3l 3k 3 )(3( l ) 9) 3.3. Algorithm and Examples Based on Theorem 3.4, algorithm, which is called Algorithm I, to factorize an odd composite number N can be designed by the following critical steps.

13 Factorization of Large Numbers via Factorization of Small Numbers 569 ==========Algorithm I ============ Step. Judge if N is a multiple of 3; if it is, do N=(N div 3) until N is not a multiple of 3; then go to Step 3. Step. Judge if N is a multiple of 5; if it is, do N=(N div 5) until N is not a multiple of 5; then go to Step 3. Step 3. Judge N s form, 6n+ or 6n-? Step 4. Extract n and the tail i from N; n Step 5. If N=6n+with tail i, find l such that 5 s. d ( n s. s ). n If N=6n-with tail i, find l such that 5 Step 6. If l + s is found, then. d ( ) ( N 6n ) 5 s. d n s. s. ; if l - s is found, then. d ( ) ( N 6n ) Example. Let N ; then N with tail 7. By Theorem 6 it yields shows l 664 and computation 5 l l l l 5999 (3( ) 7995) ( ) hence 7995 (( ) ) 64 Example. Let N ; then N with tail 7. By Theorem 6 it yields l l 6894 and computation shows 5 l l l l 94 (3( ) ) ( ) Hence 7477 (( ) ) 3.4 Approach to Turn Factorization of Big Numbers into Small Ones Theorem 3.4 and Algorithm I provide a general approach to factorize an odd composite number according to the number s tail. The approach can certainly reduce a lot of searching time. However, it can see that the approach is just a little better than that introduced in article [6]. For a very big number, it still costs a lot of time when the big number s factor is also big. This section presents an approach that turns the factorization of big number to small ones.

14 57 Xingbo Wang 3.5. Theorems and Algorithm By Lemma. and.4, it is easy to derive out the following Theorem 3.5. Theorem 3.5 Let m pa and n m 3pb be two odd numbers; then m and n are of the same tail and are of the same form in term of 6n. Seen from (3.) to (3.8), the following Theorem 3.6 holds. Theorem 3.6. Arbitrary composite number of the form 6n can always be a product of the form a( a 6 b) such that N a N 7 (3.6) N b 7 Arbitrary composite number 6n can always be a product of the form a( a 6b ) or a( a 6b 4) N a N 7 N b 7 where symbol means the floor function. Proof. Only for the inequalities. Now consider N 6n and N a( a 6 b). (3.7) 6 ab ( N a ) N a N a 6 a 7 N a N b b 6 7 N a N a a 6 7 N a N b b 6 7 Consider b is mandatory to be in the case of N a, it is reasonable to make it a rule that N a N 7 N b 7

15 Factorization of Large Numbers via Factorization of Small Numbers 57 Similarly, when N 6n, it will have N a N 7 N b 7 Theorem 3.5 and 3.6 directly deduce the following Algorithm II. =============Algorithm II================ Step. Judge if N is a multiple of 3; if it is, do N=(N div 3) until N is not a multiple of 3; then go to Step 3. Step. Judge if N is a multiple of 5; if it is, do N=(N div 5) until N is not a multiple of 5; then go to Step 3. Step 3. Judge N s form, 6n+ or 6n-? Step 4. Extract the tail i of N; Step 5. Let S N 7 Step 6. If N=6n+with tail i, find from X a composite number S such that S S ; If N=6n-with tail i, find from X i a composite number S such that S S ; where i=,3,7,9. Step 7. Judge if N-S is a multiple of 3; if it is not, go to Step 6. Step 8. Calculate g=gcd(s, N-S); if g= go to Step 6; else g N. i Obviously, the selected composite number S is a key element in Algorithm II. Generally speaking, S ought to make it easier to compute GCD(S, N-S). For convenience of programming, S can be produced by X i or X i list in Corollary, which is derived from Theorem 3.

16 57 Xingbo Wang Corollary 3.. The following i Xi i composite number of 6n+ and 6n- with tail i. X and (,3,7,9) X {9t 6t 9,9 t 66t,9 t t 39,9 t 4t 3 36,9 t 74t 84,9 t 86t 96 t Z } respectively produce odd X t t t t t t t 3 { , ,9 53,9 7 38t 493 t Z } X {9t 4t 7,9t 96t 47,9t 84t 87,9t 56t tz } X {9t 4t 49,9t 78t 69,9t 5t 589,9t 56t 899,9t t 89,9t 38t 59,9t t 39 t Z } X {9t 9t 6,9 t 9t,9 t 44t 55,9 t 6t 3 34 tz } X {9t 8t 3,9t 7t 43,9t 8t 33,9t 6t 73 tz } X t t t t t t t t 7 { , , , ,9t 44t 57 t Z } X t t t t t t t t {9 7 9,9 8 99,9 9 9, tz } In addition, it can see that both Algorithm I and Algorithm II are parallelable Numerical Experiments Table shows the results of numerical experiments on some of the Mersenne and Fermat Numbers. The experiments are made via C++ gmp big number library on a PC with a Intel Xeon E545 CPU and 4GB memory. Table is a comparison of Algorithm I, II, III and IV. It indicates that the Algorithm IV is faster than the other three. Table : Experiments on Mersenne and Fermat Numbers N S GCD(S,N-S) Factor of N

17 Factorization of Large Numbers via Factorization of Small Numbers CONCLUSION It is both a traditional approach and a fresh method to find the law of factorization of big odd numbers according to the numbers' tails together with their multipliable relations. In contrast to classical methods of sieves, the approaches introduced in this article are elementary, easily-understood and easily-applicable. Deductions and experiments show that they are valid and could be even faster ones. In addition, their being parallelable traits will make them a wider range of applications. Hope the approaches work better in the future. ACKNOWLEDGEMENTS The research work is supported by the national Ministry of science and technology under project 3GA785, Department of Guangdong Science and Technology under projects 5A345 and 5A4, Foshan Bureau of Science and Technology under projects 6AG3, Special Innovative Projects 4KTSCX56, 4SFKC3 and 4QTLXXM4 from Guangdong Education Department. The authors sincerely present thanks to them all. REFERENCES [] Luke Valenta, Shaanan Cohney, Alex Liao,Joshua Fried, Satya Bodduluri, Nadia Heninger,"Factoring as a Service", IACR Cryptology eprint Archive,5,-9 [] X X Liu,X X Zou and J L Tan, "Survey of large integer factorization algorithms, Application Research of Computers",4,3():3-37 [3] Sonal Sarnaik, Dinesh Gadekar, Umesh Gaikwad, "An overview to Integer factorization and RSA in Cryptography", International Journal For Advance Research In Engineering and [4] technology(ijaret), 4,(IX):-7 [4] D R Brown, "Breaking RSA May Be As Difficult As Factoring", Journal of Cryptology archive,6,9():-4 [5] Jongsoo Park,Mathology Sys, "Prime Sieve and Factorization Using Multiplication Table", Journal of Mathematics Research,,4(3):7- [6] Xingbo WANG,"Seed and Sieve of Odd Composite Numbers with Applications in Factorization of Integers", IOSR Journal of Mathematics,6,(5):-7

18 574 Xingbo Wang