Sobre la estructura de los números primos. About the Structure of Prime Numbers
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1 Sobre la estructura de los números primos. About the Structure of Prime Numbers Rafael A. Valls Hidalgo-Gato Instituto de Cibernética, Matemática y Física (ICIMAF), Cuba, valls@icimaf.cu Resumen: Después de una introducción fijando el tema de la teoría elemental de números sin excluir posibles aplicaciones en seguridad informática, el tema de primos (absolutos) y relativos es considerado, con el número natural 1 jugando el rol protagónico. Algunos teoremas inéditos sobre enteros consecutivos son incluidos, apuntando a un camino diferente de manejar la relación entre números compuestos y primos. Aparece una forma alternativa de describir la estructura de secuencias infinitas de números, hallando sus diferencias naturales con un carácter periódico infinito. Se logra un proceso recurrente general lo que permite la exclusión de todos los infinitos múltiplos de un primo tras otro, comenzando por el primero. Una breve conclusión puntualiza las posibles aplicaciones y desarrollo futuro. Palabras claves: funciones aritméticas; primos; factorización Página 1 de 11
2 Abstract: After an introduction fixing the topic in the elementary theory of numbers without excluding possible information security applications, the topic of (absolute) primes and relative ones is addressed, with the natural number 1 playing the protagonic role. Some unpublished theorems about consecutive integers are included, pointing to a different road managing the relationship among composites and prime numbers. An alternative way to describe the structure of infinite number sequences appears, findig their natural differences with a finite periodic character. A general recurrent procedure is reached that permit the exclusion of all the infinite multiples of a prime one by one, starting from the first. A brief conclusion points to the possible applications and future development. Keywords : arithmetic functions, primes, factorization. 1. Introduction. The law ruling the distribution of prime numbers continue being and open mistery until today, since the ancient Greek Erathostenes offer us his sieve procedure ([1], p.114) to determine them up to some fixed natural number. This paper continue addressing the essential old sieve approach, with the possible novelty of excluding all the infinite multiples of a given prime, not only the ones up to some fixed natural number. However, the method used never goes out from the Elementary Theory of Página 2 de 11
3 Numbers, the title of a book [1] of the Poland W. Sierpinski, the unique needed reference in the text. Approximation or probabilistic arguments do not appear here, neither a logic different from the classical Aristotle s one. Even if the content is a pure mathematical one, practical applications in the information security topic are not excluded. 2. The qk sets of coprimes to the product of the first k primes. Definition 2.1. Two integer numbers a and b are denoted coprimes (relative primes) one to the other, when their Greatest Common Divisor GCD(a, b)=1 ([1], p.11). Example As all different prime numbers have in common only the divisor 1, any two of them are (trivially) coprimes. Theorem 2.1. If n is the natural difference b-a of two different integer numbers a<b, then both numbers are coprime or not to n. Proof: Let be GCD(a, b)=d, a divisor of both a and b, and then also a divisor of their natural difference n. If d=1, then GCD(a, n)=1 and GCD(b, n)=1, being both a and b coprime to n; if d>1, then GCD(a,n)>1 and GCD(b, n)>1, being both a and b not coprime to n. Then both numbers are coprime or not to n. Q.E.D. Definition 2.2. Pk is the product of the first k prime numbers pk, starting with p1=2. Página 3 de 11
4 Definition 2.3. For any natural k, qk(i) is the infinite (in both senses) ordered sequence of the set qk of coprimes to Pk, with qk(-i) = -qk(i) for any integer i 0. qk(0) is not defined and qk(±1) = 1. It is accepted q0=z, set of integers (negatives, 0 and positives). Example The ordered sequence q1(i) of coprimes to P1=p1=2 (odd integers) is obtained excluding from q0 all the integer multiples of p1=2 (among them the 0 and ±2). Example The ordered sequence qk+1(i) of coprimes to Pk+1 is obtained excluding from qk(i) all the multiples of pk+1. Definition 2.4. φ(n) (Euler s totient natural function) denotes the number of coprimes to n in the first n natural numbers (starting with 1) ([1], p.228). Example φ(1)=1, because GCD(1, 1)=1. Theorem 2.2. If p is any prime and n any natural, then φ(p n )=p n-1 (p-1) ([1], p.229). Example For any prime pk, φ(pk)= pk-1. Theorem 2.3. If m and n are natural coprimes, then φ(mn)= φ(m)φ(n) ([1], p.229). Corolary If n1, n2,, nk are natural numbers, each one coprime to any other, then φ(n1n2 nk)= φ(n1)φ(n2) φ(nk). ([1], p.230). Definition 2.5. Qk= (p1-1)(p2-1) (pk-1). Example From Corolary 2.2.1, Definition 2.2, Example and Definition 2.5, we have Qk= ϕ(pk). Página 4 de 11
5 Theorem 2.4. ϕ(n) is the number of coprimes to n in any n consecutive integers. Proof: Changing one position to the right any sequence of n consecutuve integers, we obtain a new one excluding the first integer at the left and including a new one at the right. We notice that the natural difference between those two different integers is n, being for that cause both coprime or not to n by Theorem 2.1, remaining the same number of coprimes to n in the new sequence (if one coprime goes out, another one goes into; if no one out, no one into). We obtain the same result changing one position to the left, and repeating the procedure in both senses indefinitely we cover all the integer numbers. As all possible sequences of n consecutive integers have the same number of coprimes to n, that number is equal by Definition 2.4 to the φ(n) one corresponding to the sequence from 1 to n. Q.E.D. 3. The ordered sequences of consecutive coprimes to Pk natural differences. In Definition 2.3 we introduce the ordered sequences qk(i) of coprimes to the product Pk of the first k primes, showing how the members of qk+1(i) can be obtained from the previous qk(i) ones excluding all the multiples of pk+1 (Example 2.3.2). In the road to find an efficient procedure to do it, we consider convenient to indroduce the natural dk(i) ordered sequences of consecutive coprimes to Pk natural differences. Definition 3.1. For i>0, dk(i)= qk(i+1) - qk(i); for i<0, dk(i)= qk(i) - qk(i-1); for i=0, dk(0)=2. Example For k=1 (Example 2.3.1), the set qk are the odd integers, being dk(i)=2 for all i (including i=0). Página 5 de 11
6 Example As both -1 and 1 are coprimes to Pk for any natural k (they are coprimes to any integer), we have for their natural difference always dk(0)=2, as appears in Definition 3.1. We know that the integer divisors of any integer do not depend on its sign, being them the same for both the positive and the negative one ([1], p.7). Definition 3.1 determines always a natural (positive) value for all the dk(i). Example For all natural k and integer i, dk(i)=dk(-i) is always natural. As we can see, the case k=1 corresponds to a periodic sequence dk(i) with a period of only the single element {2}. To obtain the qk(i) elements from the dk(i) ones we only need to apply the formula qk(i)= dk(0)/2+ dk(1)+ dk(2)+ dk(3)+ +dk(i), (3.1) that applies for any natural k. For k=1 reduces to only add 2 indefinitely. Results natural the pass to the following k=2 case. Having already determined d1(i) and q1(i), we know from Example that to find the k=2 elements we need only to exclude from q1(i) the multiples of the following prime p2. But which is it? We can determine it very easily with the following general theorem. Theorem 3.1. For any natural k, pk+1 is the least natural >1 of the set qk. Proof: We know trivially (Example 2.1) that pk+1 is coprime to pk, and also to Pk. It is then a natural >1 belonging to the set qk. Let us suppose that it is not there that least one. Exists Página 6 de 11
7 then in qk a natural < pk+1 with a prime divisor pk+1. The contradiction proves the Theorem. Q.E.D. For any natural k, according to Theorem 3.1 and (3.1) we have then: pk+1= 1+ dk(1), (3.2) Once we have from (3.2) the next prime pk+1=3, the next step is to multiply Pk by pk+1 to obtain Pk+1. If with k=1 we have P1=2, with k=2 we have P2=2.3=6. We notice that in any P1=2 consecutive integers, 1 of them is coprime to P1, according to Theorem 2.3 and φ(p1)=p1-1=1. Having now the first natural P2=6 consecutive integers, 2 of them are coprime to P2 (1 and 5) and φ(p1p2)= φ(p1) φ(p2)= 1.2= 2. Generalizing, in the first Pk+1 natural consecutive integers, we have Qk+1= φ(pk+1) coprimes to Pk+1 (supported by Theorem 2.3, Corolary and Definition 2.5). When multiplying P1=2 by p2=3, we are extending the period {2} of d1(i) to {2, 2, 2}, that is not yet the period {4, 2} of d2(i) when the multiples of pk+1=p2=3 are excluded. The first 2 marked ( ) represents the multiples of pk+1 that need to be excluded, an action that is done substituting the marked item by the adding of it with the next one, obtaining already the mentioned right period of d2(i). Applying (3.1) to d2(i) we obtain then the sequences: q2(i) = 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, (3.3) d2(i) = {4, 2}, {4, 2}, {4, 2}, {4, 2}, 4, (3.4) 4. General procedure to obtain dk+1(i) from dk(i). Página 7 de 11
8 a) Determine the following prime pk+1= 1+dk(1). b) Start building the period {dk+1} repeating the {dk(i)} one pk+1 times. c) To prepare the exclusion of all pk+1 multiples, start the summation (3.1) with accumulator ac=0 and i=0, marking the corresponding item each time ac=pk+1dk(i) and returning ac=0. Stop after Qk marks. d) To exclude all pk+1 multiples, substitute each marked item by the adding of it with the next one. Stop after Qk substitutions. Can be proved easily that each one of the Qk items on the period of dk(i) in c) and d) is marked one and only one time, having also a symmetric equal value in both different senses from the dk(0)=2 and also from the (Pk/2)=4 one for k>1, and the number of 2 s (and also 4 s) in {dk(i)} can be proved equal to Rk= (p2-2)(p3-2) (pk-2). Using the showed procedure, the infinite sequence of prime numbers stays determined in a recurrent way, revealing a special internal periodic structure close related with their proper multiples, the composite numbers. Being the quantitative aspect of the considered procedure determined by Pk, it suggests the development of the following special numerical system to manage prime numbers. Página 8 de 11
9 5. Positional number system with Pk as a variable base. In the system with a constant base b=10, a figure f= 0, 1, 2,, 9 in position k=0, 1, 2, represents the value f.10 k. We propose to build an analog positional number system where a figure f=0, 1,, Pk in position k=0, 1, 2, represents the value f.pk. Definition 5.1. Let us denote Prime Number System (PNS) the positional number system with Pk as a variable base described in the previous paragraph. For the translation of a decimal number to the binary system (b=2), we use the known algorithm with successive divisions by b (starting with the number) of the found quotients, being the obtained ordered residues the desired figures. We can verify easily that the previous mentioned algorithm can be modified to obtain a successful translation of any number to the PNS, dividing by the prime numbers pk in succession (instead of the constant b). Example 5.1. Let us consider the number = x ; = x ; = x ; = x ; = x ; = x ; = 844 x ; 844 = 44 x ; 44 = 1 x ; 1 = 29 x ; Página 9 de 11
10 translated to PNS is 1(21) with10 PNS figures from k=0 to 9. We are using ( ) to distinguish PNS figures with more than 1 decimal figures. The values represented by each figure are the following: 1 x P0 = 1 ; 0 x P1 = 0 ; 3 x P2 = 18 ; 5 x P3 = 150 ; 9 x P4 = 1890 ; 2 x P5 = 4620 ; 7 x P6 = ; 8 x P7 = ; 21 x P8 = ; 1 x P9 = ; The adding of all them equals the number. Theorem 5.1. Any number in PNS is a multiple of pk if (and only if) the number represented by its last k figures is also a multiple of pk. Proof: In PNS (as in any other positional number system), any number is always equal to the addition of the number represented by its last k figures (positions 0 to k-1) and the one with all those figures considered 0. This last number is always a multiple of Pk, the addition of multiples of all primes up to pk by Definition 2.2 and then also such a multiple. Then, the original number is a multiple of pk if (and only if) the number represented by its last k figures is also a multiple of pk. Q.E.D. Example In the example 5.1, the number 1(21) PNS (= ) is not multiple of p4=7, because 1.P0+0.P1+3.P2+5.P3=5301PNS (16910) is not multiple of p4. It can be verified in this case that none of the 10 first prime numbers involved in the PNS representation of the number are divisors of it. Página 10 de 11
11 6. Conclusions. A possible efficient recurrent procedure to determine the prime numbers is reached, what points to some applications including factorization and primality. A detailed analysis of the algorithmic complexity of the procedure is necessary and still for done. 7. References. [1] Sierpinski, Waclaw (1964), Elementary Number Theory, Poland Academy of Science. Página 11 de 11
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