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Distribution of RSA Number's Divisor on T3 Tree

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Distribution of RSA Number's Divisor on T3 Tree International Journal of Information and Electronics Engineering, Vol. 9, No. 1, March 2019 Distribution of RSA Number’s Divisor on T3 Tree Xingbo Wang and Hongqiang Guo modulus should be in a T3 tree, and thereby provides Abstract —The article investigates the detail distribution of theoretically foundations to the results in paper [13]. RSA modulus’ small divisor in the T3 tree in terms of the divisor-ratio. It proves that, the distribution of the small II. HELPFUL HINTS divisor in T3 tree is completely determined by the divisor-ratio and the parity of the level where the RSA modulus lies, the A. Definitions and Notations small divisor of a RSA modulus lying on an even level lies on Let S be a set of finite positive integers with s and the same level as where the square root of the modulus is 0 clamped, whereas that of a modulus on an odd level possibly sn being the smallest and the biggest terms respectively; an lies on the same level or the higher adjacent level of the square integer x is said to be clamped in S if s x s . Symbol root. Through mathematical induction, it shows that a smaller 0 n divisor-ratio results in a closer position of the small divisor to xS indicates that x is clamped in S . Symbol x is the the square root of a RSA modulus. floor function, an integer function of real number x that Index Terms—Cryptography, RSA modulus, divisor ratio, satisfies inequality x1 x x , or equivalently binary tree. x x x 1 . Let N pq be an odd integer with q 1pq; then k is called the divisor-ratio of N. I. INTRODUCTION p The RSA modulus, which is also called a RSA number, is In this whole paper, symbol T3 is the T3 tree that was a big semiprime composed of two distinct prime divisors, introduced in [3], [7] and symbol N(,)kj is by default the say p and q with 3 pqsuch that 1qp / 2 , node at position j on level k of T3 , where k 0 and according to the American Digital Signature Standard (DSS) 0j 2k 1 . By using the asterisk wildcard *, [1]. As stated in [2], the RSA numbers have been essentially important in cryptography ever since the RSA public symbol N(k ,*) means a node lying on level k . An integer X cryptosystem was established. It is believed that, a is said to be clamped on level k of T if 2kk12X 2 1 systematic theory or method that can factorize the RSA 3 numbers means the failure of the RSA public cryptosystem. and symbol Xk indicates X is clamped on level k . An kk12 Thus factorization of the RSA numbers has been a dream odd integer O satisfying 2 1 O 2 1 is said to be on filled with fantasies of researchers and engineers working on level k of T3 , and use symbol Okto express it. Symbol information security. Nevertheless, the list of unfactroized ()p q k means integers p and q are on the same level RSA numbers gets longer and longer on the bulletin. k or clamped on the same level k . Symbol AB means Recently, a T -tree approach has revealed activities in 3 A holds and simultaneously B holds, symbol AB means studying the integers. The approach originates from paper [3], followed by a series of papers, as list in the references A or B holds. Symbol ()a b c means a takes the value from [4]-[13]. of b and ac . Symbol AB means conclusion B can be In the paper [13], a theorem, which was marked with derived from condition A. Proposition 1 in the paper, was proved to show the scopes of B. Lemmas the divisors p and q in accordance with the variation of the divisor-ratio by 1qp / 2 , 1qp / 1.5 and Lemma 1 (See in [3] & [7]). T3 Tree has the following fundamental properties. 1qp / 2 . The paper also proposed three interval- (P1). Every node is an odd integer and every odd integer subdivisions that could indicate which subinterval the two bigger than 1 must be on the T3 tree. Odd integer N with divisors lie in. However, that paper was lack of N>1 lies on level log2 N 1 . mathematical analysis to show why the proposed (P2). N is calculated by subdivision should be those ones. It seemed that the (,)kj proposed subdivisions were coming suddenly from the kk1 heaven and accordingly a question whether there is better N(,)kj 2 1 2 j , j 0,1,...,2 1 one is naturally asked by readers. To make it clear, this paper proves in detail where the small divisor of a RSA and thus kk12 Manuscript received December 5, 2018; revised March 14, 2019. 2 1 N(,)kj 2 1 The authors are with Department of Mechatronic Engineering, Foshan University, Foshan, PRC and Guangdong Engineering Center of Information Security for Intelligent Manufacturing System, Foshan, China (P3) The biggest node on level k of the left branch (e-mail: [email protected]; [email protected]). doi: 10.18178/ijiee.2019.9.1.699 23 International Journal of Information and Electronics Engineering, Vol. 9, No. 1, March 2019 is 2kk1 2 1whose position is j 21k 1 , and the smallest yields kk1 node on level k of the right branch is 2 2 1 whose n x nx n( x 1) 1 position is j 2k 1 . On the same level, there is not a node that is a multiple of another one. III. MAIN RESULTS AND PROOFS (P4) Multiplication of arbitrary two nodes of T3, Proposition 1. Let N 3 be an odd integer and say N(,)m and N(,)n , is a third node of T3. Let kNlog2 1 ; then J 2(1mn 2) 2(1 2) 2 ; the NN k1 k 1 k 1 multiplication (,)(,)mnis given by 1 N 22 2 2 2 2 (1) 2 mn2 NNJ(,)(,)mn 2 1 2 when k is even, whereas mn1 If J 2 , then NNN(m , ) ( n , ) ( m n 1, J ) lies on level k1 k 1 k 1 k 1 12 mn1 2 2 N 2 2 mn1 of T3; whereas, if J 2 , 2 2 2 2 (2) mn1 2 NNN(m , ) ( n , ) ( m n 2, ) with J 2 lies on level mn2of T3. when k is odd. Lemma 2(See in [10 ]). Let N 3 be an odd integer and k 1 Proof. Direct calculation yields kNlog2 1 ; then N 1 . Particularly, 2 kk12 k 1 kk12 k 1 2NN 2 222 2 (N 22 2 ) 1 when k is odd, whereas kk12 2 11N 2222 k 1 2 k 1 ( 22 2N ) 1 when k is even. 2 By Lemma 4 (P13) it holds Lemma 3(See in [13 ]). Let N pq be an odd integer kk12 11 N 2222 q with 1pqand 1 ; then 2 p k 1 k 2 1 1 Let 2 2 B ; then 222 B and thus 31 N p N N q N 22 N BB2 (3) where x 0 if x 0 . 2 Particularly, when 2 , it yields kk11 Let ; then 0 when k is odd and N 3 22 p N N q N 22 0.5 when k is even. By Lemma 4 (P14), it holds 3 when , it yields k1 k 1 k 1 2 2 2 1 2 BB2 2 2 (2 1) (4) 35 N p N N q N 44 and and when 2 it holds k1 k 1 k 1 1 BB2 22 2 2 2 2 2 (22 1) (5) 2 1 2 1 (1 ) N p N N q N 22 Now suppose k is even; then (4) becomes Lemma 4 (See in [14]). For real numbers x and y, it holds (P13) x y x y kk11 22 0.5 1 (P14) n x n x B 2 2 (2 1) Lemma 5 (See in [15]). Let and x be a positive real k1 k 1 k 1 21 1 numbers; then it holds 22 ( 1) 2 2 ( 1) 2 2 22 x 1 x ( x 1) Particularly, if is a positive integer, say n , then it and (5) becomes 24 International Journal of Information and Electronics Engineering, Vol. 9, No. 1, March 2019 kk11 11 By Lemma 4 (P13) it holds 22 22 B 2 2 2 (2 1) 0 k1 k 1 k 2 k 2 1 2 1 2 2 23 2 2 2 2 N 2 2 which says an even k yields 4 k1 k 1 k 1 1 kk11 kk22 2 2 N 2 12 12 2 2 2 (6) Let 2222 ; then 222 2 2 and thus 2 3 If k is odd; then (4) becomes N 2 (12) 4 k1 k 1 k 1 1 2 2 01 2 B 2 2 (2 1) 2 (7) Note that, by Lemma 4 (P14), it holds k1 k 1 k 1 and (5) becomes 2 2 2 12 2 2 2 (2 2 1) (13) kk11 1 0 22 2 B 2 2 2 (2 1) and (8) k1 k 1 k 1 k1 k 1 k 1 13 2 23 2 2 2 2 2 2 22 (14) 2 ( 1) 2 ( 1) 2 2 2 2 2 2 (2 2 1) 24 That is, an odd k leads to 3 2 1 By 1 , it can see by Lemma 5 that, when k 0 k1 k 1 k 1 k 1 4 32 12N 22 2 2 2 2 2 2 (9) is even 2 kk11 22 0.5 1 0.5 2 Corollary 1.
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