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RFSC 04-02 Revised a METHOD for GENERATING MERSENNE PRIMES and the EXTENT of the SEQUENCE of EVEN PERFECT NUMBERS Simon Davis Re

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RFSC 04-02 Revised a METHOD for GENERATING MERSENNE PRIMES and the EXTENT of the SEQUENCE of EVEN PERFECT NUMBERS Simon Davis Re RFSC 04-02 Revised A METHOD FOR GENERATING MERSENNE PRIMES AND THE EXTENT OF THE SEQUENCE OF EVEN PERFECT NUMBERS Simon Davis Research Foundation of Southern California 8837 Villa La Jolla Drive #13595 La Jolla, CA 92039 Abstract. A condition is obtained for the generation of new Mersenne primes from a com- bination of Mersenne numbers with prime indices. It is veri¯ed that all known Mersenne p1 p2 prime indices greater than 19 have the form p1 + p2 ¡ 1, where 2 ¡ 1 is prime and 2 ¡ 1 is composite, and that all Mersenne prime indices less than or equal to 19 have the form 0 0 0 0 p1 p2 p1 + p2 ¡ 1, with 2 ¡ 1 and 2 ¡ 1 both being primes. Arithmetical sequences for the exponents of composite Mersenne numbers are obtained from partitions into consecutive integers, and congruence relations for products of two Mersenne numbers suggest the exis- tence of in¯nitely many composite integers of the form 2p ¡1 with p prime. The congruence properties of di®erences of powers with polynomial exponents are used to prove the in¯nite extent of the sequence of Mersenne primes. MSC: 11N13, 11N32, 11P83 1. Introduction The Lucas-Lehmer test [11][12], together with several characteristics of prime divisors of Mersenne numbers, which are valid for all Lucas sequences can be used to determine theoretically whether the integer 2n ¡1 is prime. The congruence modulo 2n ¡1 is satis¯ed by the known Mersenne primes, although the di±culty of the computation increases for larger values of n. The conjecture of the existence of in¯nitely many Mersenne primes and the problem of establishing the in¯nite extent of the sequence of composite Mersenne numbers with prime indices may be solved without the consideration of speci¯c values of the exponent. The generality of these statements allows for the proofs to be based on conditions imposed on integers of the same order as the exponents. It is shown in x2that since every prime exponent p has the form p1 + p2 ¡ 1, where p1 p1+p2¡1 and p2 are prime, with several conditions being given for 2 ¡ 1 to be composite, all known Mersenne primes greater than 219 ¡ 1 are shown to have exponents of the form p1 +p2 ¡1 where p1 is a Mersenne prime index and p2 is the index of a composite Mersenne number. The Mersenne numbers have a geometrical representation which may be used to derive congruence relations for compositeness based on the partition of the array representing 2n ¡ 1. The solutions to the congruence relations yield arithmetical sequences for the exponent. However, the greatest common denominator of the initial term and the di®erence can be equated with ord2m¡k(k) for some m, k, so that none of the integers in the sequence are prime. Nevertheless, this allows a characterization of the set of exponents greater than 6 of composite Mersenne numbers. The congruence relations for 2p1+p2¡1 ¡ 1 provide a further indication of the existence of in¯nitely many composite Mersenne numbers with prime exponents. A proof of the existence of in¯nitely many Mersenne primes is given in x4. The existence of a ¯nite number of prime solutions to af(n) ¡ bf(n) ´ 0 (mod n) when f(x) does not have a zero at x = 1, is used to demonstrate that there are an in¯nite number of values of n for which 2n ¡ 1 does not have a proper prime divisor. 2. The Exponents of Mersenne Numbers and Arithemetical Progressions There are two in¯nite sequences, of Mersenne numbers with odd index, and primes, in the arithmetical progression 6n + 1, n 2 Z, and the coincidences of these two sequences determine whether the set of even perfect numbers continues inde¯nitely. 1 Since 6n + 1 can be factorized only if n has the form 6xy § (x + y) with x and y, the Mersenne number 2p ¡ 1 is prime only if it equals 6n + 1; n = 6xy § (x + y) + z, z 6= 0 with 6xy § (x + y) + z 6= 6x0y0 § (x0 + y0) for any integers x0; y0. Given the condition p 2 ¡ (6z + 1) = (6x § 1)(6y § 1), consider two primes p1 and p2 such that p1 2 ¡ (6z1 + 1) = (6x1 § 1)(6y1 § 1) = 6h1 + 1 h1 = 6x1y1 § (x1 + y1) (2:1) p2 2 ¡ (6z2 + 1) = (6x2 § 1)(6y2 § 1) = 6h2 + 1 h2 = 6x2y2 § (x2 + y2) Multiplying these two integers gives p1+p2 2 ¡(6z1 +1)(6z2 +1)¡(6z1 +1)(6h2 +1)¡(6z2 +1)(6h1 +1) = (6h1 +1)(6h2 +1) (2:2) or equivalently p1+p2¡1 2 = [3(h1 + z1) + 1][6(h2 + z2) + 2] (2:3) It can be checked that the known Mersenne prime indices greater than 31 cannot be p1 p2 expressed as p1 + p2 ¡ 1 for two primes p1; p2 such that both 2 ¡ 1 and 2 ¡ 1 are prime. p1 p2 If z1 6= 0, and 2 ¡ 1 is prime, while z2 is set equal to zero, so that 2 ¡ 1 is allowed to be composite, p1+p2¡1 2 ¡[6((1¡γ1)z1+(1¡γ2)h1+(1¡γ3)h2)+1] = 6(3(h1+z1)h2+γ1h1+γ2z1+γ3h2)+1 (2:4) 0 0 0 0 0 0 for some fractions γ1; γ2; γ3 with 3(h1 +z1)h2 +γ1h1 +γ2z1 +γ3h2 = 6x y §(x +y ); x ; y 2 Z. The Mersenne number 2p1+p2¡1 ¡ 1 is prime if there is no solution to equation (2.4) with γ1 = γ2 = γ3 = 1. If p1 is a given Mersenne prime index, it can be conjectured that p1 ¡ 1 can expressed as the di®erence between two primes p and p2, since an even integer 2N equals p ¡ p2 if 2(N + p2) is given by the sum of the two primes p; p2. The estimated number of prime pairs (p; p + 2N) with p · x is conjectured to be x Y p ¡ 1 ¼ (x) » 2C (2:5) 2N 2 (log x)2 p ¡ 2 p>2 pjN ³ ´ Q 1 where C2 is the twin-prime constant p>2 1 ¡ (p¡1)2 [16][23], and · µ ¶¸ x log log x ¼ (x) · 6:836 C 1 + O (2:6) 2 2 (log x)2 log x Theorem. Every ¯nite even positive integer 2N can be expressed as the di®erence be- tween two primes if the Goldbach conjecture is valid. 2 Proof. For any even integer 2N, N > 2, there exists two primes q1; q2 such that 2N = q1 + q2 if the Goldbach conjecture [7][8] is correct. This equality implies a rela- tion of the form 2N¹ =q ¹1 ¡ q¹2,q ¹1; q¹2 prime, for some N¹ < N, since 2(N ¡ q2) = q1 ¡ q2. It can be assumed that this property holds for all integers 1 · N~ < N¹ and that the lesser prime in each of the di®erences is bounded above by 2N~ ¡ 2. The existence of a prime pair with a di®erence to an arbitrary even integer shall be demonstrated by induction on N¹. 2N¹ + 2 =q ¹1 ¡ q¹2 + 2 = 2N + 2 ¡ 2¹q2 (2:7) = q3 + q4 ¡ 2¹q2 = (q3 ¡ q¹2 + k) ¡ (¹q2 ¡ q4 + k) Since q3 + q4 =q ¹1 +q ¹2 + 2, the indices can be chosen such that q3 > q¹1; q4 < q¹2 + 2 or q¹1 > q3 > q¹2; q4 > q¹2 + 2 unless q3 =q ¹1, q4 =q ¹2 + 2. If the equalities are valid, then 2N¹ ¡2 = q3 ¡q4. Continuing this process with 2N +2 replaced by 2N +2{, { ¸ 2, it follows that the equalities would imply that all even integers less than 2N¹ can be expressed as di®erences between pairs of primes even when arbitrarily large values of N¹ are chosen by 0 subtracting 2q6 from an arbitrarily large integer N = q5 +q6. Consequently, the equalities would imply that every even integer is equal to the di®erence between two primes. Suppose then that the ¯rst set of inequalities holds. Then q3¡q¹2¡1 > 0 andq ¹2¡q4+1 > 0. Since 0 0 2N¹ + 2 = (q3 ¡ q¹2 ¡ k ¡ k ) ¡ (¹q2 ¡ q4 ¡ k ¡ k ) (2:8) 0 0 0 andq ¹2 ¡ q4 ¡ k ¡ k =q ¹2 ¡ k1 ¡ k ¡ (q4 + k2), k1 + k2 = k, k1; k2; k can be chosen such 0 0 that q4 + k2 =q ¹2 ¡ k1 ¡ k ¡ p¹2 withq ¹2 ¡ k1 ¡ k andp ¹2 prime, since q4 + k2 < 2N¹ when 0 q¹2 ¡ q4 ¡ k ¡ k > 0 andq ¹2 < 2N¹. Suppose that the equality between q4 + k2 and the 0 0 0 di®erence between two primes ¯xes k1 + k . Then, k can be adjusted to ¯x q3 ¡ q¹2 ¡ k ¡ k 0 to be prime. It follows that 2N¹ + 2 = q3 ¡ q¹2 ¡ k ¡ k ¡ p¹2 is a di®erence between two primes, withp ¹2 < 2N¹. If q3 < q¹1; q4 > q¹2 + 2, consider the equality 0 0 2N¹ + 2 =q ¹1 ¡ q¹2 + 2 = (¹q1 + q4 ¡ k ¡ k + 2) ¡ (q4 ¡ q¹2 ¡ k ¡ k ) (2:9) 0 0 As q4 ¡ q¹2 ¡ k ¡ k = (q4 ¡ k1 ¡ k ) ¡ (¹q2 ¡ k2), andq ¹2 ¡ k2 is an even integer less than 0 0 2N¹,q ¹2 ¡ k2 = q4 ¡ k1 ¡ k ¡ p¹3 wherep ¹3 is prime and k1 + k is adjusted to render 0 0 0 q4 ¡ k1 ¡ k to be prime. Then k also can be chosen such thatq ¹1 + q4 ¡ k ¡ k + 2 is 0 prime. The lesser prime in the di®erence satis¯es the inequality q4 ¡ q¹2 ¡ k ¡ k < 2N¹ since q4 ¡ q¹2 < q3 ¡ q¹2 < q¹1 ¡ q¹2 = 2N¹.
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