Aid#59330/Preprints/2019-09-10/www.mathjobs.org
RFSC 04-01 Revised
The Prime Divisors in the Rationality Condition for Odd Perfect Numbers
Simon Davis
Research Foundation of Southern California 8861 Villa La Jolla Drive #13595 La Jolla, CA 92037
Abstract. It is sufficient to prove that there is an excess of prime factors in the product of repunits with odd prime bases defined by the sum of divisors of the integer N = (4k + 4m+1 ℓ 2αi 1) i=1 qi to establish that there do not exist any odd integers with equality (4k+1)4m+2−1 between σ(N) and 2N. The existence of distinct prime divisors in the repunits 4k , 2α +1 Q q i −1 i , i = 1,...,ℓ, in σ(N) follows from a theorem on the primitive divisors of the Lucas qi−1 sequences and the square root of the product of 2(4k + 1), and the sequence of repunits will not be rational unless the primes are matched. Minimization of the number of prime divisors in σ(N) yields an infinite set of repunits of increasing magnitude or prime equations with no integer solutions.
MSC: 11D61, 11K65
Keywords: prime divisors, rationality condition 1. Introduction
While even perfect numbers were known to be given by 2p−1(2p − 1), for 2p − 1 prime, the universality of this result led to the the problem of characterizing any other possible types of perfect numbers. It was suggested initially by Descartes that it was not likely that odd integers could be perfect numbers [13]. After the work of de Bessy [3], Euler proved σ(N) that the condition = 2, where σ(N) = d|N d is the sum-of-divisors function, N d integer 4m+1 2α1 2αℓ restricted odd integers to have the form (4kP+ 1) q1 ...qℓ , with 4k + 1, q1,...,qℓ prime [18], and further, that there might exist no set of prime bases such that the perfect number condition was satisfied.
Investigations of the equation for the sum of the reciprocals of the divisors of N has led to lower bounds on the number of distinct prime divisors. This number has increased from four to nine [23][38][48] while a minimum of 75 total prime factors [25][26] has been established. When 3 6|N, it was shown that there would be a minimum of twelve different prime divisors [24][33][38]. It was demonstrated also that, if 3, 5, 7 ∤ N, greater than 26 distinct prime factors would be required [8][39]. In decreasing order, the three largest prime divisors were bounded below by 108 + 7 [21][31], 104 + 7 [30] and 102 + 1 [29] respectively, 2n+6 while the least prime divisor had to be less than 3 for n different prime factors [22]. 4m+1 2αi Moreover, either one of the prime powers, (4k +1) or qi for some index i, was found to be larger than 1020 [11]. Through the algorithms defined by the sum over reciprocals of the divisors, it has been demonstrated that odd perfect numbers had to be greater than 10300 [5], while the lower bound was increased to 101500 through sieve methods for the abundancy and size of the larges component [5]. An upper bound for an odd perfect 2 4k 44β +2β+3 number 2 if there are k distinct prime factors and 2 if αi = β, i = 1,...,ℓ [49].
One of the possible methods of proof of the odd perfect number conjecture is based τ(N) on the harmonic mean H(N) = 1 , where τ(N) is the number of integer divisors of d|N d N. It has been conjectured that H(N) is not integer when N is odd [43]. This statement P also would imply the nonexistence of odd perfect numbers as the perfect number condition 1 is d|N d = 2 and τ(N) must be even, since N is not a perfect square. The use of the harmonicP mean leads again to the study of the sum of the reciprocals of the divisors, and the values of this sum have only been approximated. For example, it has been found that σ(N) −12 there are odd integer with five different prime factors such that N − 2 < 10 [32].
The uniqueness of the prime decomposition of an integer allows for the comparison
1 between its magnitude and the sum of the divisors. Since the sum of the divisors of an odd 4m+2 4m+1 ℓ 2αi (4k+1) −1 integer N = (4k + 1) i=1 qi can be represented by the product 4k · 2α +1 ℓ q i −1 i , it is sufficient to determine the properties of the prime factors of the i=1 qi−1 Q repunits to establish that there exist no odd perfect numbers. The irrationality of Q 1 2 4m+1 ℓ 2αi 2(4k + 1) σ((4k + 1) ) · i=1 σ(qi ) would imply that σ(N) cannot equal 2N. pIt has been proven for a largeQ class of primes {4k + 1; qi} and exponents {4m + 1; 2αi} that the rationality condition is not satisfied. The irrationality of the square root of the product of 2(4k + 1) and the sequence of repunits is not valid for all sets of primes and exponents, however, and it is verified in §4 that the rationality condition holds for twelve odd integers. The factorizations of these integers have the property that the repunits have prime divisors which form interlocking rings, whereas, in general, the sequence of prime factors does not close. The presence of a sequence of primes of increasing magnitude σ(N) prevents any finite odd integer from being a perfect number. To prove that N 6= 2, it is necessary also to obtain a lower bound for the number of prime divisors in σ(N). qn Since the number of distinct prime factors of i −1 is minimized in the class of repunits qi−1 with exponents containing the prime divisor p when n = p, the exponent is presumed to be prime throughout the discussion. It is demonstrated in Theorem 1 that the minimum number of prime divisors in the product of repunits that could represent the sum of divisors σ(N) of an odd perfect number is greater than or equal to the number of prime factors of N, Then, either σ(N) has an excess of prime divisors or constraints must be imposed on
{4k + 1; qi} and {4m + 1; 2αi} which have no integer solution. The non-existence of odd perfect numbers also follows from the sequence of prime factors of increasing magnitude in the factorization of σ(N), when one of three specified relations is satisfied, and constraints on the basis and exponents otherwise.
2. The Existence of Different Prime Divisors in the Repunit Factors of the Sum of Divisors
To prove the nonexistence of odd perfect numbers, it shall be demonstrated the product of repunits in the expression for σ(N) contains an excess of prime divisors of σ(N), such that the perfect number condition σ(N) = 2N cannot be satisfied. The existence of 2α +1 4m+2 q i −1 distinct prime divisors in the quotients (4k+1) −1 and i , i = 1,...,ℓ follows from 4k qi−1 2αi+1 qi −1 a theorem on the prime factors of the quotients where qi and 2αi + 1 are odd qi−1 primes. A restriction to the least possible number of distinct prime factors in a product
2 2α +1 2αj +1 q i −1 q −1 i · j yields three equations that complement conditions considered previously qi−1 qj −1 [12].
If the inequality 2α +1 q2αi+1 − 1 q j − 1 i 6= j (2.1) qi − 1 qj − 1
2α +1 2αj +1 q i −1 q −1 holds, then either the sets of primitive divisors of i and j are not identical qi−1 qj −1 2α +1 2αj +1 q i −1 q −1 or the exponents of the prime power divisors are different, given that i 6= j . qi−1 qj −1 Let qi >qj. Since
α q2αi+1 − 1 q2αi+1 − 1 q2 j +1 − 1 i = i i 2αj +1 2αj +1 2αj +1 qj − 1 qi − 1 qj − 1 2α +1 2αj +1 q2αi+1 − 1 q 2αj +1 q i − q − 1 1 = i i 1+ i j + (2.2) 2αj +1 q 2αj +1 2αj +1 4αj +2 qi − 1 j qi qj qj 1 − + ... , 2αj +1 4αj +2 qi qj this equals q2αi+1 − 1 q 2αj +1 i i + ǫ 2αj +1 q qi − 1 " j # 2α +1 2α +1 (2.3) q 2αj +1 q j − q j − 1 1 ǫ ≃ i i j + + ... . q 2αj +1 2αj +1 4αj +2 j qi qj qj !
2 −1 2αj +1 (2αj +1) 2α +1 2α +1 1+ 2α +1 i 2αj +1 2αj +1 i i qi −1 When qj < qi < qj · (qj − 1) , ǫ · 2αj +1 < 1. The exact h i qi −1 value of the remainderh term is i
2α +1 2αj +1 q2αi+1 − 1 q2αi+1 − 1 q i − q ǫ · i = i i j (2.4) 2αj +1 2αj +1 2αj +1 2αj +1 qi − 1 qi − 1 qj (qj − 1)
2αj +1 1 − 2α +1−2(2α +1) 2αi+1−2(2αj +1) i j which yields the alternative conditions qi < 2 qj and α 3 2 i+1 1 2αj +1 qi < 2 qj for the remainder term to be less than 1. Since the entire quotient is 4α +2 2α +1 2α +1 2α +1 2αi+1 q j (q j −1)+q j −q j qi −1 j j i j 2α +1 2α +1 2α +1 2α +1 , it can be an integer or a fraction with the q j −1 q j q j (q j −1) i i j j 2αi−1 2αj +1 2αj +1 denominator introducing no new divisors other than factors of qi only if qj −qj 2αj +1 2αj +1 contains all of the divisors of qj − 1, presuming that qj is cancelled. Similarly,
3 −1 2α +1 (2α +1)2 j 1+ j 2αj +1 2αj +1 2αi+1 2αi+1 if qj Example 2.1. Consider the prime divisors of pairs of repunits and the integrality of the n qi −1 quotients n when qi, qj and n are odd primes. The first example of an integer ratio for qj −1 2α +1 i 3 3 qi −1 29 −1 29 −1 odd prime bases and exponents is 2αj +1 = 33−1 = 938, but the quotient 29−1 = 13·67 qj −1 2αi+1 2αi+1 qi −1 qi introduces a new prime divisor. For qi, qj ≫ 1, the ratio 2α +1 → , which i qj qj −1 cannot be integer for odd primes qi, qj, implying that one of the repunits, with αi = αj, 2αi+1 has a new prime divisor or qj − 1 contains a higher power of one of the prime factors. The latter possibility is excluded because the numerator and denominator are raised to the same power in the limit of large qi, qj. 2α +1 n q i −1 Theorem 2.2. There will exists a minimum of n prime divisors of i with i=1 qi−1 + odd prime bases and exponents, qi 6= qj for i 6= j and n ∈ Z , for an oddQ perfect number 4m+1 ℓ 2αi N = (4k + 1) i=1 qi , ℓ ≥ n, if 2α +1 Q 2αi+1 j qi − 1 qj − 1 6= (2αi + 1) qi − 1 qj − 1 2α +1 2αi+1 j qi − 1 qj − 1 (2αj + 1) 6= qi − 1 qi − 1 2αi+1 2αi+1 qi − 1 qj − 1 (2αj + 1) 6= (2αi + 1) qi − 1 qj − 1 are valid. Proof. xm−1 yn−1 There are no equalities of the form x−1 = y−1 that have been established yet for odd prime bases and exponents. Equality of two quotients in the rationality condition would yield an integer multiple of the square root of the product of the remaining repunits. It σ(s) would not affect, therefore, the rationality of the square root expression for s , where 4m+1 2 4m+1 2α1 2αℓ N = (4k + 1) s = (4k + 1) q1 ... qℓ . 4 xm−1 The known positive integer solutions to the exponential Diophantine equation x−1 = n y −1 3 62−1 5 53−1 13 903−1 y−1 , m, n > 1, include 2 − 1 = 6−1 , 31 = 2 − 1 = 5−1 and 8191 = 2 − 1 = 90−1 [7][20][40]. The occurrence of the base 2 in one of the repunits is found to be necessary for the integers given by repunits with cubic powers, which have been proven to be the only solutions having an exponent equal to 3 [28]. y−1 The entire set of solutions to this equation has not been determined. If a = δ , b = x−1 y−x s n1 δ , c = δ , δ = gcd(x − 1,y − 1) and s is the least integer such that x ≡ 1(mod by ), mi ni x −1 y −1 mi ni the equation x−1 = y−1 can be represented as (y − 1)x − (x − 1)y = y − x, mi ni ax − by = c, which may be satisfied trivially for m1 = n1 = 1. It has been proven that, if (x, y, m1,n1) is a presumed solution of the Diophantine equation, then m2 − m1 is a multiple of s when (x, y, m2,n2) is another solution with the same bases x,y [7]. Additionally, it has been established that there are at most two solutions for fixed x and y if y ≥ 7 [7]. As x Given that the first exponents are 2αi +1 and 2αj +1, the constraint on a second exponent qj −1 2αi+1 of qj then would be m2 − (2αj + 1) = ιϕ δ qi where ι ∈ Q may be a fraction qj −1 with a denominator which is a divisor of ϕ δ or qi − 1 or qi, and otherwise, ι ∈ Z when qj ∤ (qi − 1). Based on the trivial solution to these equations for an arbitrary pair of odd primes qi,qj, the first non-trivial solution for the exponent of qj would be required 2αi qj −1 to be greater than or equal to than 1 + ιqi (qi − 1)ϕ δ . This exponent introduces 2α qj −1 1+ιq i (q −1)ϕ i i δ q −1 new prime divisors which are factors of j and therefore does not qj − 1 minimize the number of prime factors in the product of repunits. There is a proof of the existence of a maximum of one solution for given x and y [27]. It has been demonstrated that another equation, defined by the equality of the abundance σ(n) I(n)= n for the product of two prime powers, may be solved by prime pairs that would represent sets of solutions to this exponential Diophantine equation [44]. This result may be verified for the repunits with odd prime bases and exponents. 2αi+1 2αi 2αj 1 qi −1 Consider the abundances of products of two prime powers, I(q q )= 2α i j 2αi j qi−1 qi qj ′ 2α +1 2α′ +1 2α +1 q j −1 ′ 2α′ i q j −1 j 2αi j 1 qi −1 j ′ and I(q q ) = ′ , where q 6= q . Let 2α > 2α and q −1 i j 2α′ 2α q −1 q −1 i j i i j i j i j qi qj 5 ′ 2αj > 2αj and suppose that the abundances are equal or ′ α′ 2α +1 2αj +1 2α +1 2 j +1 ′ i q − 1 α α′ i q − 1 2αi−2αi qi − 1 j 2 j −2 j qi − 1 j qi = qj . (2.5) qi − 1 qj − 1 qi − 1 qj − 1 2α′ +1 ′ j ′ 2αi+1 2α −2α q −1 2αj −2α q −1 It would follow that q i i j and q j i . The repunits do not have i qj −1 j qi−1 imprimitive prime divisors except, possibly, for the exponents that divide qi − 1 or qj − 1. However, if (2αi + 1) (qi − 1), for example, gcd(2αi + 1,qi) = 1, and divisibility conditions satisfied qi and qj as primitive divisors would not be affected. Then the qi must be 2α′ +1−1 q j 2αi+1 j qi −1 a primitive divisor of and qj would be a primitive divisor of . Any qj −1 qi−1 imprimitive divisors, such as 2αi + 1, for example, would be matched with divisors of 2α′ +1 j 2α +1 q −1 2α′ −2α q i −1 the corresponding repunit j . Then the prime factors in q i i i would be qj −1 i qi−1 2α′ +1 2α −2α′ q j −1 identified with the prime divisors of q j j j . The exponents also would match j qj −1 because these primes cannot divide the remaining repunits in the equation. A cancellation 2α′ +1 q i −1 of these factors in Eq.(2.5) would yield an equality between the the larger repunits i qi−1 2α +1 q j −1 and j . With a maximum of one solution to the exponential Diophantine equation qj −1 xm−1 yn−1 x−1 = y−1 for each pair of integers (x,y), no other solutions to this equation with the prime bases qi, qj could exist. A one-to-one correspondence between the equality of the abundances of products of even positive powers of qi and qj and a solution to the equation ′ 2α +1 2αj +1 q i −1 q −1 i = j is established. qi−1 qj −1 Lemma 2.3. The product of two odd prime powers with a given abundance is unique if the exponents are even and equal to one less than an odd prime. There are no solutions 2α +1 2αi+1 q j −1 qi −1 j to the equation = for two different odd primes qi and qj and prime qi−1 qj −1 exponents 2αi + 1 and 2αj + 1. 6 Proof. Let 2α +1 2αi+1 j q − 1 qj − 1 I(q2αi q2αj )= i i 2αi 2αj qi (qi − 1) qj (qj − 1) ′ ′ (2.6) 2α +1 2αj +1 ′ α′ i q − 1 2αi 2 j qi − 1 j I(qi qj )= ′ ′ 2αi 2αj qi (qi − 1) qj (qj − 1) 2αi 2αj 2αi 2αj ′ ′ If I(qi qj )= I(qi qj ), with αi ≥ αi and αj ≥ αj, ′ ′ 2αj +1 2αj +1 2α +1 ′ 2αi+1 ′ i α α q − 1 q − 1 2αj −2α q − 1 q − 1 q2 i−2 i i j = q j j i (2.7) i q − 1 q − 1 j q − 1 q − 1 i j i which can be valid only if ′ 2αj +1 ′ 2αi+1 ′ 2αj −2α q − 1 α α q − 1 q j i q2 i−2 i j . (2.8) j q − 1 i q − 1 i j ′ 2αj +1 2α +1 q −1 q i −1 The primitive divisors of j and i must divide the other repunit in this pair q−1 qi−1 2α′ +1 q j ′ ′ j −1 by Eq.(2.7) when αi 6= α and αj 6= α [4]. Similarly, the primitive divisors of i j qj −1 2α +1 q i −1 and i are required by this equation to divide only the other repunit in this pair. qi−1 2αj +1 qj −1 An imprimitive divisor of must divide qj − 1 and 2αj + 1, and an imprimitive qj −1 2α′ +1 i qr qi −1 ′ j −1 divisor of must divide qi − 1 and 2α + 1, there exist divisors , r|(2αj + 1), qi−1 i q−1 t qi −1 ′ or , t|(2α + 1) or the prime exponent divides one less than the base. Then qi and qi−1 i qj cannot be imprimitive divisors of the first kind simultaneously of repunits with bases qj and qi respectively, since the divisibility of qj − 1 by qi and qi − 1 by qi would yield a contradiction. Proper divisors of the second kind do not exist when the exponents in the ′ ′ repunits are prime. When 2αi + 1, 2αj + 1, 2αi + 1 or 2αj + 1 is an imprimitive prime ′ ′ factor, with αi >αi and αj >αj, the divisibility condition ′ 2αj +1 2α −2α′ q − 1 (2α + 1) q j j j , i j q − 1 2α′ +1 q i − 1 (2α + 1) i , j q − 1 i (2.9) 2αj +1 q − 1 (2α′ + 1) j or i q − 1 j ′ 2αi+1 ′ 2αi−2αi qi − 1 (2α + 1) q j i q − 1 i 7 must be satisfied. Therefore, the matching of the divisors with the repunits will be defined by the grouping of prime powers and repunits in Eq.(2.7). A contradiction arises when qi and qj are odd prime primitive divisors of the repunits 2α −2α′ 2αi+1 j j qi −1 with bases qj and qi with odd prime exponents respectively. Let q = and j U(qi−1) 2α′ +1 j 2αi+1 2α −2α′ α′ α q − 1 j j q2 i−2 i i − 1 . (2.10) i U(q − 1) i Since 2α′ +1 2α′ +1 j j 2α −2α′ 2α −2α′ 2αi+1 j j j j (qi − 1) − (U(qi − 1)) 2α′ +1 j (2αi+1) ′ 2α −2α′ j j 2αj + 1 1 = qi 1 − ′ 2α +1 + ... (2.11) 2α − 2αj q i j i α′ 2α′ +1 2 +1 j j ′ α α′ 2α −2α′ 2 j −2 2αj + 1 1 − U j j q j 1 − + ... , i 2α − 2α′ q j j i ′ 2αj +1 ′ the series would terminate for integer ′ , which cannot occur since 2αj + 1 is an odd 2αj −2αj prime, or the infinite series introduces fractional terms that are not divisible by qi. ′ 2α′ ′ ′ 2αi 2αj 2αi j Therefore, Eq.(2.7) is valid only if αi −αi = 0 and αj −αj = 0 and qi qj = qi qj . 2αi+1 2αj +1 It follows that there is only one product of prime powers qi qj with the value 2αi 2αj I(qi qj ), where qi, qj > 2 and αi, αj ≥ 1 such that 2αi + 1 and 2αj + 1 are primes. xm−1 yn−1 Given the one-to-one correspondence with solutions to the equation x−1 = y−1 , there 2α +1 2αj +1 q i −1 q −1 will be no examples of an equality of i and j for different odd prime bases qi−1 qj −1 qi and qj and different prime exponents 2αi + 1 and 2αj + 1. The known solutions to the exponential Diophantine equation are characterized by x = 2. The prime factor 2 may be an imprimitive divisor of a repunit with another base, since 2|q − 1 and n if q is an odd prime and the exponent is even. The equality 4 1 5 52−1 1 4 53−1 3 2 I(2 · 5) = 24·5 (2 − 1) 5−1 = 2352 (2 − 1) 5−1 = I(2 · 5 ), for example, is equivalent to 5 52−1 4 53−1 52−1 5 · (2 − 1) 5−1 = 2 · (2 − 1) 5−1 , and it is evident that 2 is an imprimitive divisor of 5−1 , whereas 5 is a primitive divisor of 24 − 1. After cancellation, the equality of the remaining 5 53−1 factors gives 2 − 1 = 5−1 . Therefore, an exception to the theorem arises when the one of the bases is two and the product of prime powers includes odd exponents. 8 2α +1 2αj +1 q i −1 q −1 If two repunits satisfied i = j = q′h, the product would yield only a qi−1 qj −1 single prime power q′2h and reduce the total number of prime factors of σ(N). There is p q −1 ′h ′ one known solution to q−1 = q , for prime q, q , p [6], and there are no known solutions n xm−1 y −1 ′h to the equation x−1 = y−1 = q for x>y ≥ 2. The introduction of only one prime divisor through the product of two repunits could occur if q2αi+1 − 1 i = q′hi qi − 1 2αj +1 (2.12) qj − 1 = q′hj . qj − 1 There are no imprimitive divisors if (2αi + 1) 6| (qi − 1) and (2αj + 1) 6| (qj − 1), whereas the primitive divisors of the two repunits could be equal if these have the formc ˆ(2αi + bi bj n n 1) (2αj + 1) . The existence of primitive divisors of qi − 1 and qj − 1 for n ≥ 3 implies ′ ′ ′ that q must belong to this category, and q 6|(qi − 1) and q 6|(qj − 1). xn−1 m All solutions to the equation x−1 = y , except (x, y, m, n) = (18, 7, 3, 3), is characterized by the existence of a prime p|x such that m|(p − 1) [6]. Then hi must divide qi − 1 and hj must divide qj − 1 for any integer solution to either relation in Eq.(2.12). This condition is satisfied when the exponents equal 2 and the bases are odd primes, and 35−1 2 there are no solutions other than 3−1 = 11 [36][40], while it is nontrivial for hi, hj ≥ 3. Lemma 2.4. The equations q2αi+1 − 1 i = q′hi qi − 1 2αj +1 qj − 1 = q′hj . qj − 1 ′ ′ ′ cannot be solved by odd primes qi, qj and q , 2αi + 1 and 2αj +1 with hi, hj ≥ 3. Proof. Let τihi = τjhj = l.c.m.(qi − 1,qj − 1). Then it follows from the equalities with the powers of the prime q′ that τ 2α +1 τj q2αi+1 − 1 i q j − 1 i = j (2.13) q − 1 q − 1 i j ! and 2αi+1 qi − 1 ≡ 1 (mod qi) qi − 1 τ 2αj +1 j (2.14) qj − 1 ≡ 1 (mod qi). qj − 1 ! 9 It is known that ordq(a) = q − 1 if a 6≡ 1 (mod q) and gcd(a,q − 1) = 1. Since 2αj +1 q2αi+1 q −1 i −1 j ′hj ≡ 2αi + 1 ≡ 0 (mod 2αi) and gcd ,qi − 1 = gcd(q , 2αj + 1) = 1 or qi−1 qj −1 2α +1 q i −1 a contradiction arises because there must be a primitive divisor of i of the form qi−1 2α +1 q j j −1 2αi ci(2αi +1)+1. Furthermore, ≡ 1(mod qi) only if q ≡ 1(mod qi) or 2αj|(qi −1). qj −1 j Let qi − 1= ηij(2αj) . Then q i−1 +1 ηij qj −1 qj − 1 ′ = q κj . (2.15) qj − 1 κj q −1 i qj −1 η +1 q ij 1 j −1 q −1 Consider the integrality of . The factor (qj − 1) j is irrational for qj −1 κj 1 1 q −1 q −1 κ qj ≥ 3, although (qj − 1) j may be rational if (qj − 1) j = w j for w ∈ Q. The expansion of the numerator is κj qi−1 +1 qj −1 ηij qj − 1 κ κ . j qi−1 j η q −1 + q −1 κ 1 1 κ κ 1 ij j j j j j = qj 1 − q −1 + 1 − + ... , i +1 qi−1 2 qj − 1 ηij 2 qj − 1 qj − 1 +1 q q ηij j j (2.16) 2αj +1 hj which is a product of an irrational number qj , if hj 6|(2αj + 1) and a series of rational 2αj +1 hj terms. Given that 2αj +1 is prime, hj must equal an integer multiple of 2αj + 1 for qj to be an integer, and then the inequality 2αj +1 qj − 1 6= q′c(2αj +1) (2.17) qj − 1 following from 1 1 1 2αj +1 2αj +1 2αj +1 2αj 1 − 2α +1 q q j j − 1 2αj +1 j = qj 1 κj qj −1 1 because 1 − q −1 < 1 and the factor will not compensate for the irrationality i +1 ηij qj of (2α +1 2α +1 j −⌊ j ⌋ 2αj +1 hj hj qj −1 q . Then, if 2αi 6=(qi − 1), 6≡ 1 (mod qi) and τj must equal qi − 1. j qj −1 Similarly, τi must equal qj − 1 and q −1 2α +1 qi−1 q2αi+1 − 1 j q j − 1 i = j . (2.19) q − 1 q − 1 i j ! Then q −q qi−qj i j 2α +1 q −1 2αi+1 q −1 j j q − 1 i qj − 1 q′(hi−hj ) = i = , (2.20) q − 1 q − 1 i j ! which could be valid only if (qj − 1)|(qi − 1). ′ With qi − 1= c (qj − 1), q2αi+1 − 1 [c′(q − 1) + 1]2αi+1 − 1 i j α mod q = ′ ≡ 2 i +1 ( j − 1) qi − 1 c (qj − 1) . 2αj +1 (2 21) qj − 1 ≡ 2αj +1 (mod qj − 1). qj − 1 It is evident that c′ 2αi+1 ′hi [− + 1] − 1 ′ 2αi+1 ′−1 q ≡ = [c − 1] + 1 · c (mod qj) (−c′) . (2 22) ′hj q ≡ 1 (mod qj) as 2αi +1 is odd. If hi is a multiple of hj, ′ 2αi+1 ′ [c − 1] + 1 ≡ c (mod qj) (2.23) ′ 2αi+1 ′ [c − 1] ≡ c − 1 (mod qj). The last congruence requires (qi − 1)|2αi. Then 2αi + 1 =c ˜i(qj − 1)+1 and the primitive prime divisor is ′ q =c ˆi[˜ci(qj − 1) + 1] + 1 =c ˆic˜i(qj − 1) +c ˆi + 1 (2.24) 11 for some constantc ˆi. Since c˜i(qj −1)+1 qi − 1 ≡ 1 (mod qj − 1) qi − 1 (2.25) hi hi [ˆcic˜i(qj − 1) +c ˆi + 1] ≡ (ˆci + 1) (mod qj − 1), it follows that ϕ(qj − 1)|hi, where ϕ is the totient function. Let hi = µiϕ(qj − 1). From qc˜i(qj −1)+1 i − 1 µiϕ(qj −1) = [ˆcic˜i(qj − 1) +c ˆi + 1] (2.26) qi − 1 the constraint 1 2αi ϕ(qj −1) ′µi (1 + qi + ... + qi ) = q (2.27) is derived. However, 1 2αi ϕ(qj −1) ϕ(qj −1) 1 1 qi 1+ + ... + 2α qi q i i (2.28) 2αi ∞ n 2αi n ϕ(q −1) (1 − (m − 1)ϕ(q − 1))(q − 1) = q j 1+ m=1 j i . i 2nαi n n n=1 n!qi ϕ(qj − 1) (qi − 1) X Q For sufficiently large qi, this series approximately equals 2αi 2αi ∞ n+1 ϕ(qj −1) (qi − 1) (−1) qi 1+ 2α + n . (2.29) i nϕ(qj − 1)(qj − 1) qi ϕ(qj − 1)(qi − 1) n=2 X 2αi ϕ(q −1) If 2αi = qj − 1, then ϕ(qj − 1) 6| 2αi and q j is irrational. If ϕ(qj − 1)|2αi, the powers of qi will not cancel the denominators in the sum while the summation does not yield an integer and there will be no odd prime solution to Eq.(2.19). It follows that both repunits cannot be powers of the same odd prime. The possibility of equating the product of two repunits in σ(N) with a factor q′h is circumvented. 2αi+1 qi −1 b The primitive prime divisors of are a(2αi +1) +1 while the primitive divisors qi−1 2αj +1 q ′ j −1 ′ b of are a (2αj +1) +1 [4][47]. Since the exponents in the repunits are primes, this qj −1 b′ ′ b th equality is valid only if a = c(2αj +1) and a = c(2αi +1) . For the k common primitive 2α +1 2αj +1 i q −1 ′ qi −1 j b prime divisor of and , the coefficients would be ak = ck(2αj + 1) k and qi−1 qj −1 12 ′ bk bk bk ak = ck(2αi +1) yielding the corresponding factor ak(2αi +1) +1 = ck(2αi +1) (2αj + b′ 1) k +1. If no new primitive divisors are introduced by each repunit, it would follow that 2αi+1 q ′ i − 1 bk b hk =(product of imprimitive divisors) · [ck(2αi + 1) (2αj + 1) k + 1] qi − 1 k Y 2αj +1 q − 1 ′ ′ j bk b h =(product of imprimitive divisors) · [ck(2αi + 1) (2αj + 1) k + 1] k . qj − 1 k Y (2.30) hi1 his hj1 hjs Let qi − 1= p1 ...ps and qj − 1= p1 ...ps . Then phi1 ...phis 2αi+1 ( 1 s + 1) − 1 hi1 his hi1 his 2αi = 2αi +1+ αi(2αi + 1)(p ...p )+ ... +(p ...p ) h1s his 1 s 1 s p1 ...ps h h p j1 ...p js 2αj +1 ( 1 s + 1) − 1 hj1 his hj1 hjs 2αj = 2αj +1+ αj(2αj + 1)(p1 ...ps )+ ... +(p1 ...ps ) . h1s hjs p1 ...ps (2.31) 2αi+1 Let P be a primitive divisor of the qi − 1. Consider the congruence c0 + c1x + ... + 2αi c2αi x ≡ 0 (mod P ), ct > 0, t = 0, 1, 2,..., 2αi. There is a unique solution x0 to the n congruence relation f(x) ≡ c0 + c1x + ... + cnx ≡ 0 (mod P ) within [x0 − ǫ,x0 + ǫ] [35], where ǫ2 ǫf ′(x )+ f ′′(x )+ ... < P. (2.32) 0 2! 0 2αi+1 Setting ct = t+1 , 2αi + 1 2αi + 1 f ′(q − 1) = + 2 (q − 1) + ... + 2α (q − 1)2αi−1, (2.33) i 2 3 i i i the constraint (2.33) implies a bound on ǫ of 1 − 2 qi−1 . Consequently, if 2αi+1 2αi 2 q − 1 |q − q | < 1 − i , (2.34) i j 2α + 1 2α i i the prime P is not a common factor of both repunits. The number of repunits with primitive divisor P will be bounded by the number of integer solutions to the congruence relation. If the same powers of the common primitive prime divisors occurred in the repunits 2α +1 2α +1 q i −1 q i −1 i and j , an inequality between the two integers necessarily would imply the qi−1 qj −1 existence of a distinct primitive factor in one of the repunits. Then, the congruences 2 2αi r 1+ qi + qi + ... + qi ≡ 0 (mod P ) r ≥ 2 (2.35) 2 2αi 1+ qj + qj + ... + qj ≡ 0 (mod P ) 13 could be considered. By Hensel’s lemma, solutions to these constraints would satisfy 2 r−1 qi = qj +c ˆ1P +c ˆ2P + ... +c ˆr−1P , and therefore, qi − qj must be greater than P . 2α +1 2αi+1 q j −1 qi −1 j If P is a prime divisor of and , and αi = αj, qi−1 qj −1 2αi 2αi−1 2αi−1 2αi P |(qi − qj)(qi + (2αi + 1)qi qj + ... + (2αi + 1)qiqj + qj ). (2.36) 2αi 2αi−1 2αi−1 2αi Since gcd(qi−qj,qi +(2αi+1)qi qj +...+(2αi+1)qiqj +qj ) = 1, either P |(qi−qj) 2αi 2αi−1 2αi−1 2αi or P |(qi + (2αi + 1)qi qj + ... + (2αi + 1)qiqj + qj ). When P |(qi − qj), it may satisfy the divisibility condition for the congruence relation. However, there also would 2αi 2αi−1 2αi−1 2αi be a primitive divisor which is a factor of qi + qi qj + ... + qj qi + qj , and not qi −qj. While there may exist primes which divide the difference between the repunits and 2αj +1 q2αi+1 q −1 not each quotient, the common prime factors of i 1 and j must be divisors of qi−1 qj −1 qi − qj for the congruences to hold modulo different powers. The common prime factors 2αi+1 ′ qi −1 {P } which do not divide qi − qj must be equal to the primes that exactly divide qi−1 2α +1 q j −1 and j . qj −1 2α +1 q i −1 The distinct exponents of the primitive prime divisors dividing j must be less qj −1 2α +1 q i −1 than those of i since the P-adic expansion for the reverse inequality would produce a qi−1 contradiction with qi >qj. A precise coincidence of the set of primitive prime divisors then 2α +1 2α +1 q i −1 q i −1 would produce the divisibility condition j i , which is not valid generically qj −1 qi−1 for arbitrarily large pairs (qi,qj), qi 6= qj. It follows that there should be a prime divisor 2α +1 2αi+1 q i −1 qi −1 j of distinct from the factors of . Setting qi = 5 and qj = 3, for instance, it qi−1 qj −1 52αi+1−1 32αj +1−1 can be proven by induction that one of the repunits 4 and 2 has a distinct 53−1 55−1 33−1 35−1 prime divisor, since 5−1 = 31, 4 = 11·71, 2 = 13, 2 = 11·11, and only primitive divisors of the integers occur in these sequences when 2αi +1 is prime. 2αj +1 qj −1 Furthermore, there would be a new prime divisor of if αj > αi. This prime qj −1 divisor is congruent to 1 modulo 2αj + 1 and would differ from the other prime factor distinguishing the two repunits with the exponent 2αi + 1 that is congruent to 1 modulo 2αi + 1 unless both equal the same integer having the form a(2αi + 1)(2αj + 1) + 1. Under 2α +1 2αj +1 q i −1 q −1 the first condition, i j has a minimum of one new prime divisor other han qi−1 qj −1 the set of prime factors of each repunit. If the second condition is valid, it remains to be determined if any new prime divisors arise in the product. 14 When the prime exponents are different, the existence of primitive divisors of the αn−βn Lucas numbers α−β for n> 12, when α, β are real, is a consequence of the factorization 2πk F α,β 1≤k< n α cos β into cyclotomic polynomials n( )= 2 − 2 n [4][49]. The Lucas gcd(k,n)=1 sequence has different primitive divisorsQ as n changes because it is not possible to match 2πk all of the factors with cos n . If αi 6= αj, 2αi 2αi+1 k qi − 1= (qi − ω2αi+1) k Y=0 2αj ′ (2.37) 2αj +1 k qj − 1= (qj − ω2αj +1) k′=0 Y2πi ωn = e n . 2αi+1 2αj +1 Quotients of products of linear factors in qi −1 and qj −1 generally will not involve cancellation of integer divisors because products of powers of the different units of unity cannot be equal unless the exponents are a multiples of the primes 2αi + 1 or 2αj + 1. The sumsets which give rise to equal exponents can be enumerated by determining the integers s1,...,st such that the sums are either congruent to 0 modulo 2αi + 1 and 2αj + 1. It is apparent that the entire integer sets {1, 2,..., 2αi} and {1, 2,..., 2αj} cannot be used for unequal repunits. The number of sequences satisfying the congruence relations for each ′ tk tl repunit must equal (2 − 1) and ′ (2 − 1), with sequences of tk=2αi+1 tl=2αj +1 integers which sum to a non-zero value giving rise to cancellation of complex numbers in PP PP the product being included. ′ s′ tk q ωsm t k q ω n Consider two products m=1( i − 2αi+1) and n=1( j − 2αj +1). Upper and lower bounds for these products areQ Q tk q tk < q ωsm < q tk ( i − 1) ( i − 2αi+1) ( i + 1) m=1 Y′ (2.38) tl ′ t′ s t′ q l < q ω n < q l . ( j − 1) ( j − 2αj +1) ( j + 1) n=1 Y t Since the minimum difference between primes is 2, one of the inequalities (qi + 1) k ≤ t′ t′ t t t′ t′ tk (qj − 1) l , (qj + 1) l ≤ (qi − 1) k , (qi + 1) k > (qj − 1) l or (qj + 1) l > (qi − 1) will be ′ satisfied when tl 6= tk. The first two inequalities imply that the factors of the repunits t′ s′ tk q ωsm k q ω n defined by the products m=1( i − 2αi+1) and n=1( j − 2αj +1) are distinct. Q Q 15 If either of the next two inequalities hold, the factors possibly could be equal when ′ ′ tk ti tk tl ′ (qi − 1) = (qj − 1) or (qi + 1) = (qj + 1) . In the first case, tl = nijtk, qi − 1 = nij ′ ′ (qj − 1) , nij > 2. This relation holds for all tk and tl such that k tk = nij l tl. However, 2αi + 1 and 2αj + 1 are prime, so that this equation is not valid.P When qi +1=P nij ′ (qj + 1) , nij ≥ 2, the relation k tk = nij l tl again leads to a contradiction. P P The identification of the products in Eq.(2.37) for each corresponding k,l is necessary, since any additional product would give rise either to a new prime factor after appropriate rescaling, or a different power of the same number, which would would prevent a cancel- t′ lation of an extra prime, described later in the proof. The inequalities (qj − 1) l < (qi − t t′ 1) k < (qj + 1) l imply that ′ ln(qj − 1) ′ ln(qj + 1) tl If these inequalities hold for two pairs of exponents (t ,t′ ), (t ,t′ ), the interval k1 l1 k2 l2 t′ t′ ′ ln(qj −1) ′ ln(qj +1) l2 tkm lm t ,t contains ′ tk . The fractions cannot be equal to ′ for all l2 ln(qi−1) l2 ln(qi−1) t 1 tk t l1 n ln tk 2αi+1 2αi+1 hm, n as this equality wouldi imply that ′ = , which is not possible since is an tl 2αj +1 2αj +1 irreducible prime fraction when αi 6= αj. Then tk1 tk2 1 ′ − ′ ≥ ′ ′ . (2.40) tl tl tl tl 1 2 1 2 ′ ln(qj −1) ′ ln(qj +1) and tk 6∈ t ,t if 2 l2 ln(qi−1) l2 ln(qi−1) h i qj + 1 ln(qi − 1) ln < ′ ′ . (2.41) qj − 1 t t l1 l2 ′ As tl ≤ 2αj + 1, this inequality is valid when 9 q ln(q − 1) > (2α + 1)2. (2.42) j i 4 j ′ Therefore, the products of linear factors with tk < 2αi + 1 and tℓ < 2αj +1 will not be equal given the Eqs.(2.41) and (2.42), When these products cannot be equated, there does 2αi+1 2αj +1 not exist sets of linear terms in the decomposition of qi − 1 and qj − 1 that give the same prime divisor a(2αi + 1)(2αj + 1) + 1 within multiplication by a prime power. qn−1 2πik Since the arithmetic primitive factor of , n 6=6isΦn(q)= 1≤k≤n (qi − e n ) q−1 gcd(k,n)=1 Φn(q) n or p , where p is the largest prime factor of both gcd(n,3) and Φn(Qq) [1][2][4][47][50], any 16 2αi+1 qi −1 prime, which is a primitive divisor of , divides Φ α (qi). It follows that these qi−1 2 i+1 tk q ωsm primes can be obtained by appropriate multiplication of the products m=1( i − 2αi+1) ′ ′ tk sn and (qj − ω ). Although these products may not be the primitive divisors, they n=1 2αj +1 Q t1 q ωsm κ are real, and multiplication of m=1( i − 2αi+1) by 1 ∈ R can be compensated by Q ′ t2 sn −1 multiplication of ′ (qi − ω ) by κ to obtain the integer factors. By Lemma m =1 2Qαj +1 1 2 +1 q i −1 2, the occurrence of the minimal set of prime divisors in the two repunits i and Q qi−1 2α +1 q j −1 j introduces two prime factors in the products of the linear terms. qj −1 Since the prime divisors of the repunits can be obtained by rescaling of products of the form (2.38), t′ and t′ can be set equal, provided that the factor of a power of a l1 l2 prime is included to match the intervals. Suppose, for example, that t′ = t′ = 2. Then, ℓ1 ℓ2 h 2 t h 2 2 ln(qj −1)+h ln P 2 ln(qj +1)+h ln P P (qj − 1) < (qi − 1) k1 < P (qj + 1) and tk ∈ , . 1 ln(qi−1) ln(qi−1) t t k1 k2 h h i Again, for some k1, k2, t′ 6= t′ if αi 6= αj, and |tk1 − tk2 |≥ 1. The inequalities P (qj − l1 l2 2πikl 2πikl 2 t h 2 2α +1 2α +1 1) < (qi−1) k2 < P (qj +1) arise because each of the terms yl =(qj −e j )(qj −e j ), which is matched with a corresponding product of linear factors tk (q − ωsm ), sm=1 i 2αi+1 2 ln(qj −1)+h lnP 2 ln(qj +1)+h lnP has the same upper and lower bound. However, tk 6∈ , 2 ln(qi−1)Q ln(qi−1) q +1 2 ln j h i qj −1 if < 1 which implies qi > 5. The equality is also valid because tk = ln(qi−1) 2 tk ′ 2ln(qj +1)+h lnP h (qi−1) 2 h ′ ′ ′ only if P = 2 and qi −1=(qj +1) , h ≥ 1. Then t = h tk, and if ln(qi−1) (qj +1) l ′ ′ h > 1, summation over k,l gives 2αj +1= h (2αi +1) which is not possible as 2αi +1 and ′ ′ 2αj +1 are prime. If h = 1, tk = tl and 2αi + 1 must be set equal to 2αj + 1, and a new prime divisor will occur. When the limits tk and tℓ′ are set equal to 2 or a bounded integer, ℓ which is feasible since there would exist only 4m +2+ i=1(2αi + 1) prime factors for the k 4m+1 ℓ q2αi tk q ωsm odd integer (4 +1) i=1 i to be a perfect number, the products m=1( i − 2αi+1) ′ ′ P t k sn and (qj − ω ) would equal a real number that must be rescaled by to give an n=1 2αj +1 Q Q integer.Q The rescaling will be given by a prime power in the following. A minimum set of prime divisors will be derived for a product of a given set of repunits. 2α +1 2αj +1 q i −1 q −1 Since i and j are not powers of the same prime by Lemma 2, there will a qi−1 qj −1 2α +1 2αj +1 q i −1 q −1 minimum of two distinct prime divisors of i j . Suppose that qi−1 qj −1 2α +1 2αi+1 q j −1 qi −1 ′hi j ′hj 1. = q xi = q xj. qi−1 1 qj −1 2 17 2α +1 q k −1 Now consider a third repunit k . Either qk−1 2αk+1 qk −1 ′hk 1a. = q xa qk−1 3 2αk+1 qk −1 ′h1k ′hk 1b. = q q xb qk−1 1 3 2αk+1 qk −1 ′hjk ′hk 1c. = q q xc. qk−1 2 3 2αk+1 qk −1 ′hik ′hjk 1d. = q q xd qk−1 1 2 where xi, xj, xa, xb, xc and xd represent additional factors. Under the first three conditions, there are three distinct prime divisors of the product of the three repunits 2α +1 2αj +1 2α +1 q i −1 q −1 q k −1 i j k . qi−1 qj −1 qk−1 ′ Given that the prime divisors are primitive divisors in Condition (1d), q1 must have the ′ 9 form a(2αi +1)(2αk +1)+1, while q2 would equal b(2αj +1)(2αk +1)+1. If ln(qi −1) ≥ 4 2 or qi ≥ 11, then Eq.(2.42) will be satisfied if qℓ > (2αℓ + 1) . Provided that the odd prime 2αℓ+1 ′ qℓ −1 bases are greater than or equal to 11, and q is labelled qℓ, the repunit will have 1 qℓ−1 2αi+1 2αk+1 qi −1 qk −1 a prime divisor that is not a factor of or if αℓ ≤ αi, αk, and qi−1 qk−1 q2αℓ+1 ℓ − 1 ′h2ℓ ′h3ℓ = q2 q3 xℓ. (2.43) qℓ − 1 Therefore, three prime divisors occur in the product 2α +1 q2αi+1 − 1 q j − 1 q2αℓ+1 − 1 i j ℓ . (2.44) qi − 1 qj − 1 qℓ − 1 If αℓ >αi or αℓ >αk, a primitive prime divisor must be introduced that is different from ′ the repunits with the basis qℓ and exponents less than or equal to αi, αk. It can equal q2 2αj +1 qj −1 only if h ℓ = 0 for an exponent less than or equal to αi, αk. Since the repunits 2 qj −1 2α +1 q ℓ −1 and ℓ both cannot be powers of the same prime q′ by Lemma 2, a third prime qℓ−1 2 ′ factor must occur. The new factor may be labelled q3, and again this product of three repunits is divisible by three distinct primes. ′ Similarly, if qm is set equal to q2, 2αm+1 qm − 1 ′h1m ′h3m = q1 q4 xm. (2.44) qm − 1 18 when αm ≤ αj, αk. and the product 2αj +1 q2αi+1 − 1 q − 1 q2αm+1 − 1 i j m (2.45) qi − 1 qj − 1 qm − 1 would have a minimum of three distinct prime divisors. If αm > αj or αk, a new prime ′ divisor q4 will arise and there would be a minimum of three different prime factors of this product. ′ ′ If q3 6= q4, there would be four different primes divisors of 2αj +1 2α +1 q2αi+1 − 1 q − 1 q j − 1 q2αm+1 − 1 i j ℓ m . (2.46) qi − 1 qj − 1 qℓ − 1 qm − 1 ′ ′ When q3 = q4. it must have the form c(2αℓ + 1)(2αm +1)+1 to be a primitive prime 2αr +1 ′ ′ qr −1 divisor. Setting qr = q = q , has a prime divisor distinct from the factors of 3 4 qr −1 2α +1 ℓ 2αm+1 qℓ −1 qm −1 or if αr ≤ αℓ,αm. If αr > αℓ or αr > αm, a new primitive prime qℓ−1 qm−1 ′ ′ divisor is introduced unless it coincides with q1 or q2. Then another prime factor must occur because the pairs of repunits cannot be powers of the same prime. It follows that 2αj +1 2αj +1 α q − 1 q − 1 q2 ℓ+1 − 1 q2αr +1 − 1 i j ℓ r (2.47) qi − 1 qj − 1 qℓ − 1 qr − 1 or 2α +1 2αj +1 q j − 1 q − 1 q2αm+1 − 1 q2αr +1 − 1 i j m r (2.48) qi − 1 qj − 1 qm − 1 qr − 1 would be divisible by a minimum of four distinct prime factors. Now consider the condition 2α +1 2αj +1 q i −1 q −1 ′h ′h 2. i = q′hi j = q 1j q 2j . qi−1 1 qj −1 1 2 2α +1 2αj +1 q i −1 q −1 It follows that that the differentiation between the prime divisors of i and j qi−1 qj −1 depends on a factor of the form a(2αi +1)(2αj +1)+1, given that exponents do not divide 2α +1 q k −1 either repunit. Since it must be a prime, it will the basis for a new repunit k , where qk−1 qk = a(2αi + 1)(2αj + 1) + 1 for the integer N to be an odd perfect number. The prime satisfies the inequality Eq.(2.42) with respect to the repunit with the lesser exponent if 2αk+1 2αi+1 qk −1 qi −1 αk ≤ αi,αj. Then there will be a prime factor of that does not divide qk−1 qi−1 2α +1 q j −1 or j and there will be a minimum of three different prime divisors of the product qj −1 19 of the three repunits. When αk > αi or αk > αj, there is a new primitive divisor that is ′ ′ a third prime factor unless it coincides with q2, since it cannot be divisible by q1 = qk. Given that q2αk+1 k − 1 ′h2k = q2 xk, (2.49) qk − 1 2αℓ+1 ′ qℓ −1 q would have the form b(2αj + 1)(2αk + 1) + 1. Labelling this prime to be qℓ, 2 qℓ−1 2αj +1 qj −1 would have a new prime factor not dividing if αℓ ≤ αj, αk. If αℓ >αj or αℓ >αk, qj −1 ′ there would be a new primitive prime divisor unless it coincided with q1. Since the repunits 2α +1 2α +1 q i −1 q ℓ −1 i and ℓ both cannot be powers of the prime q′ , it would follow then that qi−1 qℓ−1 1 2α +1 q2αi+1 − 1 q j − 1 q2αℓ+1 − 1 i j ℓ (2.50) qi − 1 qj − 1 qℓ − 1 is divisible by three distinct prime divisors. 2αk+1 2αℓ+1 qk −1 qℓ −1 If αk, αℓ ≤ αi, αj, the new prime divisors in and may be labelled qk−1 qℓ−1 ′ ′ q3 and q4 respectively, and there would be a minimum of four different prime divisors of 2α +1 q2αi+1 − 1 q j − 1 q2αk+1 − 1 q2αℓ+1 − 1 i j k ℓ (2.51) qi − 1 qj − 1 qk − 1 qℓ − 1 ′ ′ ′ ′ if q3 6= q4. When q3 = q4, q2αk+1 − 1 k = q′h3k x q − 1 3 k k (2.52) q2αℓ+1 ℓ − 1 ′h4ℓ = q3 xℓ. qℓ − 1 ′ Either xk 6= 1 or xℓ 6= 1 or xk 6= 1, xℓ 6= 1. A fourth prime factor q5 is introduced, and the product of the four reupunits is divisible by four different primes when qi,...,qℓ ≥ 11. For αk, αℓ >αi or αk, αℓ >αj, it is feasible for the repunits to have the form q2αk+1 − 1 k = q′h2k x q − 1 2 k k (2.53) q2αℓ+1 ℓ − 1 ′h1ℓ = q1 xℓ. qℓ − 1 ′ ′ ′ if the one of the primitive divisors coincides with q2 and q1 respectively. Then q1 = ′ ′ a (2αi +1)(2αj +1)(2αℓ +1)+1 and q2 = b(2αj +1)(2αk +1)+1. When 2αi +1, 2αj +1 < ′ ′ 2αk + 1 ≤ a (2αi + 1)(2αj + 1)(2αk + 1) + 1, there is a prime factor q3 of xk that does p 20 2α +1 2αj +1 2α +1 q i −1 q −1 q ℓ −1 not divide i , j or ℓ . By Lemma 2, there must be a prime factor of qi−1 qj −1 qℓ−1 2α +1 2αi+1 q j −1 2αk+1 2αℓ+1 ′ ′ ′ qi −1 j qk −1 qℓ −1 xℓ that does not equal q , q or q . Then the product 1 2 3 qi−1 qj −1 qk−1 qℓ−1 is divisible by a minimum of four different primes. ′ If αk > αi or αk > αj and 2αk + 1 > a (2αi + 1)(2αj + 1)(2αℓ + 1) + 1, there is a 2αk+1 qk −1 ′ primitive divisor occurring in , different from q and the prime factors of xk, that qk−1 p 2 ′ ′ may be labelled q3. It is possible that h2k = 0 in Eq.(2.53) and the prime q2 rather than 2α +1 q k −1 q′ is introduced into the factorization of k . Then the proofs of the existence of a 3 qk−1 fourth prime divisor will be interchanged and conclusions remain valid. To minimize the ′ number of primes, suppose that xℓ is divisible only by q3. Then q2αk+1 − 1 k = q′h3k x′ q − 1 3 k k (2.54) q2αℓ+1 ℓ − 1 ′h1ℓ ′h3ℓ = q1 q3 . qℓ − 1 ′ It follows that q3 = c(2αk + 1)(2αℓ + 1) + 1 ≡ qr. If αr ≤ αk, αℓ, there exists a prime 2α +1 2α +1 q2αr +1 q k −1 q ℓ −1 factor of r −1 that does not divide k or ℓ and may be labelled q′ . If qr −1 qk−1 qℓ−1 4 ′ ′ q4 6= q2, there is a minimum of four distinct prime divisors of 2αj +1 α q2αi+1 − 1 q − 1 q2 k+1 − 1 q2αr +1 − 1 i j k r (2.55) qi − 1 qj − 1 qk − 1 qr − 1 or 2αj +1 α q2αi+1 − 1 q − 1 q2 ℓ+1 − 1 q2αr +1 − 1 i j ℓ r (2.56) qi − 1 qj − 1 qℓ − 1 qr − 1 for qi, qj, qk, qℓ, qr ≥ 11. q2αr +1 If q′ coincides with q′ , the number of primes will be minimized when r −1 = q′h2r . 4 2 qr −1 2 ′ ′ Then q2 = b (2αj +1)(2αk +1)(2αr +1)+1. A fourth prime factor arises when αℓ <αj, αk, α q2 ℓ+1 ′ ℓ −1 Given that 2αj + 1, 2αk + 1 < b (2αj + 1)(2αk + 1)(2αr + 1) + 1, will have a qℓ−1 2αj +1 2α +1 q −1 q ℓ −1 q2αr +1 prime divisor not dividing j p , ℓ or r −1 . Therefore, the least number of qj −1 qℓ−1 qr −1 prime factors would occur when q2αℓ+1 ℓ − 1 ′h1ℓ ′h3ℓ ′h4ℓ = q1 q3 q4 (2.57) qℓ − 1 Then the product of the four repunits with bases qi, qj, qk, qℓ ≥ 11 would have a minimum ′ of four different prime divisors. If 2αℓ + 1 > b (2αj + 1)(2αk + 1)(2αℓ + 1) + 1, q2αℓ+1 ℓ − 1p ′h5ℓ ′ = q5 xℓ (2.58) qℓ − 1 21 ′ ′ ′ ′ where q5 does not equal q1, q2 or q3. Again, there are four primes in the factorization of the product of the four repunits. Suppose that 2α +1 2αj +1 ′h q i −1 q −1 ′h q 2j . 3. i = q′h1i q′h2i j = q 1j 2 qi−1 1 2 qj −1 1 ′ ′ Let qk = q1 = a(2αi + 1)(2αj + 1) + 1 and qℓ = q2 = b(2αi + 1)(2αj + 1) + 1. If 2α +1 2αk+1 2αi+1 q j −1 qk −1 qi −1 j αk ≤ αi, αj, there is a prime factor of that does not divide or qk−1 qi−1 qj −1 ′ ′ and it may be labelled q3. Similarly, there is prime divisor q4 that is not a factor of both repunits in Condition 3. Consequently, given these inequalities for the exponents, 2α +1 2αj +1 2α +1 2α +1 2αj +1 2α +1 q i −1 q −1 q k −1 q i −1 q −1 q ℓ −1 i j k and i j ℓ have a minimum of three different q−i−1 qj −1 qk−1 qi−1 qj −1 qℓ−1 prime factors. Furthermore, there is a minimum of four distinct prime divisors of 2α +1 q2αi+1 − 1 q j − 1 q2αk+1 − 1 q2αℓ+1 − 1 i j k ℓ (2.59) qi − 1 qj − 1 qk − 1 qℓ − 1 2α +1 2α +1 q k −1 q ℓ −1 when q′ 6= q′ . If q′ = q′ , both k and ℓ cannot be powers of the same prime 3 4 3 4 qk−1 qℓ−1 by Lemma 2, and a fourth prime divisor is introduced into the product of the four repunits. ′ If αk > αi or αk > αj, there would be a new primitive divisor different from q3 that ′ ′ could be q2, but it cannot equal q1. Suppose that q2αk+1 k − 1 ′h2k = q2 xk. (2.60) qk − 1 ′ Similarly, if αℓ >αi or αℓ >αj, the primitive divisor could coincide with q1 such that q2αℓ+1 ℓ − 1 ′h1ℓ = q1 xℓ. (2.61) qℓ − 1 ′ ′ ′ ′ Then qk = q1 = a (2αi + 1)(2αj + 1)(2αℓ + 1) + 1 and qℓ = q2 = b (2αi + 1)(2αj + ′ 1)(2αk + 1) + 1. If 2αi + 1, 2αj + 1 < 2αℓ + 1 < b (2αi + 1)(2αj + 1)(2αℓ + 1) + 1, 2α +1 2αk+1 2αi+1 q j −1 qk −1 qi −1 j must have a prime factor that does not dividep or and xk = qk−1 qi−1 qj −1 ′′ ′ q3 . When 2αℓ + 1 > b (2αi + 1)(2αj + 1)(2αℓ + 1) + 1, it follows that there is a new ′ ′′ ′′ primitive divisor differentp from q2 or q3 , which may be labelled q5 . Similarly, there 2α +1 2α +1 2αj +1 q ℓ −1 q j −1 q −1 would be a prime factor q′′ of ℓ that does not divide i or j if 4 qℓ−1 qj −1 qj −1 ′ 2αi + 1, 2αj + 1 < 2αℓ + 1 < a (2αi + 1)(2αj + 1)(2αk + 1) + 1. When 2αℓ + 1 > p 22 ′ ′ ′′ a (2αi + 1)(2αj + 1)(2αk + 1) + 1, the primitive divisor would be different from q1, q4 ′ pand may be labelled q6. Therefore, 2α +1 2αj +1 2α +1 2α +1 2αj +1 2α +1 q i −1 q −1 q k −1 q i −1 q −1 q ℓ −1 i j k and i j ℓ always has a minimum of three qi−1 qj −1 qk−1 qi−1 qj −1 qℓ−1 different prime divisors if the odd prime bases are greater than or equal to 11. There is a ′ ′ minimum of four distinct prime divisors of the product of the four repunits if q5 6= q6. 2α +1 2α +1 q k −1 q ℓ −1 If q′ = q′ , it is not possible for both k and ℓ to be powers of this prime. 5 6 qk−1 qℓ−1 Therefore, q2αk+1 − 1 k = q′h1k x′ q − 1 5 k k (2.62) q2αℓ+1 ℓ − 1 ′h5ℓ ′ = q5 xℓ qℓ − 1 ′ ′ ′ ′ ′ ′ where xk 6= 1 or xℓ 6= 1. Both do not equal a power of q1 or q2. Then xk or xℓ has a ′ ′ ′ ′ prime factor q7 that does not equal q1, q2 or q5. Consequently, a minimum of four different 2α +1 2αj +1 2α +1 q i −1 q −1 q k −1 q2α+1−1 primes divides i j k ℓ . qi−1 qj −1 qk−1 qℓ−1 The prime divisors of any remaining repunits with odd prime bases less then 11 may be listed. It has been proven that no odd perfect number can have 3, 5 and 7 as factors simultaneously [34]. Consequently, the following sets may occur amongst these three primes: {3}, {5}, {7}, {3, 5}, {3, 7} and {5, 7}. It is evident that the number of distinct prime factors will be greater than or equal to the number of repunits for each of these sets by Lemma 2. There could be a coincidence between these divisors and factors of repunits with larger odd prime bases. Yet the conclusions reached previously are unaffected, because these primes would divide remaining factors xa, xb, xc, xd, xi, xj, xk, xℓ, xm and xr in the repunits. Suppose that the product of n − 1 repunits with odd prime bases and exponents 2αi +1 n q r −1 −1 ir has a minimum of n − 1 different prime divisors, which occurs in the sum r=1 qir −1 2α +1 q in −1 ofQ divisors of an odd perfect number. If there is a prime factor of in that divides qin −1 this product, either it is a power of 2αir +1, r = 1,...,n−1 or it is a primitive prime factor of the other repunits. The first condition will be considered afterwards. Secondly, it has + the form ar(2αir + 1) + 1 for any a ∈ Z and r = 1,...,n − 1, a set of n +3 repunits can be produced with a minimum of n +3 prime divisors by the preceding discussion. Therefore, 2αi +1 n q r −1 a minimum of n distinct prime divisors exist in a product ir , which occurs r=1 qir −1 in the sum of divisors of an odd perfect number, by inductionQ for n = 4z, 1 + 4z, 2 + 4z, z ∈ Z+ when the exponents are not prime divisors. It can be demonstrated, however, that 23 any odd perfect number with three or four prime factors also must have a minimum of σ(N) three or four, and such integers cannot satisfy the condition N = 2 [12]. It is sufficient to prove, for a given prime divisor of N, that there are three other prime factors such that the product of the four repunits with prime exponents is divisible by a minimum of four prime divisors to extend the result to all primes divisors of N. Consequently, if the prime exponents are not divisors of the repunits, there would be a minimum of n different prime n 2αi divisors in the product of n repunits arising from σ( i=1 qi ) for all positive integers n, N k 4m+1 ℓ q2αi if = (4 + 1) i=1 i is an odd perfect number.Q Q xm−1 yn−1 For the known solutions to the equation x−1 = y−1 , one of the prime bases is 2. Since qj − 1 = 1, the theorem is circumvented because the bounds (2.39) are satisfied for a larger set of primes qi and exponents tk. Furthermore, the integers sets {1, 2,..., 2αi} and {1, 2,..., 2αj} are used entirely in the products to give the same integer, which must 2αi 2αi then be a prime. The existence of solutions to the inequalities 1 < (qi − 1) < 3 2α +1 2αj +1 q i −1 implies that these bounds do not exclude the equality of 2 −1 and i for some qj −1 qi−1 2αj + 1,qi, 2αi + 1. 2αi+1 If the primes, which divide only qi −1, are factors of qi −1 or qj −1, these would be 2α +1 2αi+1 q j −1 qi −1 j partially cancelled in a comparison of and . Any divisor of qi − 1 which qi−1 qj −1 2αi+1 qi −1 is a factor of also must divide 2αi + 1. When 2αi + 1 and 2αj +1 are prime, this qi−1 divisor would have to be 2αi + 1 or 2αj + 1. An exception to the result concerning the occurrence of a distinct prime factor in one of the repunits could occur if equations of the form 2α +1 2αi+1 j qi − 1 ν qj − 1 = (2αi + 1) qi − 1 qj − 1 2α +1 2αi+1 j ν qi − 1 qj − 1 (2αj + 1) = (2.63) qi − 1 qj − 1 2α +1 q2αi+1 q i − 1 ν1 i − 1 ν2 j (2αj + 1) = (2αi + 1) qi − 1 qj − 1 2αj +1 qj −1 are satisfied, as the set of prime divisors of the repunits would be identical if 2αi+1 qj −1 2α +1 2αi+1 q j −1 qi −1 j in the first relation, 2αj + 1 in the second condition and 2αi + 1 , qi−1 qj −1 2αi+1 qi −1 2αj + 1 in the third relation. The conditions with ν = ν1 = ν 2 = 1 follow if qi−1 24 p ν different powers of the other primes do not arise. However, qi − 1 ≡ 0 (mod p ) only if p p qi − 1 ≡ 0 (mod p), and since qi − q ≡ 0 (mod p), this is possible when p|(qi − 1). If p p qi −1 qi −1 ν p|(qi − 1), then ≡ p and 6≡ 0 (mod p ), ν ≥ 2. Setting ν = ν = ν gives qi−1 qi−1 1 2 2α +1 2αi+1 j qi − 1 qj − 1 = (2αi + 1) qi − 1 qj − 1 2α +1 2αi+1 j qi − 1 qj − 1 (2αj + 1) = (2.64) qi − 1 qj − 1 2αi+1 2αi+1 qi − 1 qj − 1 (2αj + 1) = (2αi + 1) qi − 1 qj − 1 2α +1 q j −1 2αi+1 j qi −1 with 2αi + 1 6| and 2αj + 1 6| . It follows that either 2αi + 1 or 2αj + 1 qj −1 qi−1 is a prime which divides only one of the repunits. Furthermore, qi − 1 and qj − 1 are integers which are not rescaled, such that distinct prime divisors arise in the products of the remaining linear factors. If the exponents 2αi +1 and 2αj + 1 are unequal, these factors do not divide both repunits. When the exponents 2αi + 1 and 2αj + 1 are equal, it would be necessary for nij to be equal to one, which is 2α +1 2αi+1 q j −1 qi −1 j not feasible since qi 6= qj. It follows also that and cannot be equal, and qi−1 qj −1 the original assumption of their inequality is valid. However, by the first two relations in Eq.(2.55), there exists a prime which is not a divisor of both repunits. The non-existence of odd integers satisfying σ(N) = 2N is related to the number of prime divisors in the square root expression 2(4k + 1)σ(N). A proof of the excess of prime divisors of σ(N) would begin with the introductionp of a new factor for each repunit. Example 2.5. An example of a congruence with more than one solution less than P is (2α + 1)2α (2α − 1) (2α + 1)2α (2α − 1)(2α − 2) (2α + 1) + α (2α + 1)x + i i i x2 + i i i i x3 i i i 3! 4! (2α + 1)2α (2α − 1)(2α − 2)(2α − 3) + i i i i i x4 =0(mod 11) 5! (2.65) which is solved by x = 2, 4 when 2αi +1 = 5. This is consistent with the inequality (2.32), 35−1 55−1 since ǫ = 0.074897796 when x0 = 2. Consequently, 2 = 11 · 11 and 4 = 11 · 71 both 25 35−1 have the divisor 11 and the distinct prime divisor arises in the larger repunit. Indeed, 2 2α +1 q k −1 would not have a different prime factor from a repunit k that has 11 as a divisor. qk−1 A set of primes {3,qk,...} and exponents {5, 2αk,...} does not represent an exception to 35−1 the Theorem 1 if 2 is chosen to be the first repunit in the product, introducing the prime divisor 11. Each subsequent repunit then would have a distinct prime factor from the previous repunit in the sequence. 3. Formulation of Conditions on the Factors of Integers having the Form of an Odd Perfect Number 4m+1 ℓ 2αi Let N = (4k + 1) i=1 qi [18] and the coefficients {ai} and {bi} be defined by 2αi+1Q 4m+2 qi − 1 (4k + 1) − 1 ai = bi gcd(ai,bi) = 1. (3.1) qi − 1 4k Bounds on the number of solutions to these equations for given a, b, qi and qj have been σ(N) derived [45]. If N 6= 2, 1 q2α1+1 − 1 q2α2+1 − 1 q2αℓ+1 − 1 (4k + 1)4m+2 − 1 2 2(4k + 1) 1 2 ... ℓ " q1 − 1 q2 − 1 qℓ − 1 4k # p (ℓ+1) 1 4m+2 2 (b1...bℓ) 2 (4k + 1) − 1 (3.2) = 2(4k + 1) 1 · a ...a 2 4k ( 1 ℓ) p ℓ 2m+1 αi 6= 2(4k + 1) qi i=1 Y or ℓ 2αi+1 ℓ b1...bℓ qi − 1 4k = · 4m+2 a ...aℓ qi − 1 (4k + 1) − 1 1 i=1 Y (3.3) ℓ 4k ℓ+1 −1 q2αℓ+1 − 1 k 4m+1 q2αi ℓ . 6= 2(4 + 1) 4m+2 i · (4k + 1) − 1 qℓ − 1 i=1 Y When ℓ > 5 is odd, there exists an odd integer ℓo and an even integer ℓe such that ℓ = 3ℓ0 + 2ℓe, so that b ...b b a b a b a b b 1 ℓ = 13 2 46 5 ... 3ℓo−2,3ℓo 3ℓo−1 3ℓo+1 3ℓo+2 a1...aℓ a13 b2 a46 b5 a3ℓ −2,3ℓ b3ℓ −1 a3ℓ +1a3ℓ +2 o o o o o (3.4) b b s2 ... ℓ−1 ℓ · a a t2 ℓ−1 ℓ 26 2 b3¯i−2,3¯i a3¯i−1 s with s, t ∈ Z. It has been proven that 6= 2(4k + 1) · 2 for any choice of a3¯i−2,3¯i b3¯i−1 t b a ¯ ,a ¯ ,a ¯ and b ¯ ,b ¯ ,b ¯ consistent with Eq.(3.1) [12]. Similarly, ℓ 6= 2(4k + 3i−2 3i−1 3i 3i−2 3i−1 3i aℓ s2 1) · t2 so that 2 b ¯ ¯ a ¯ ρ¯ ¯ s 3i−2,3i 3i−1 k 3i−2 ≡ 2(4 + 1) · 2 a ¯i , ¯i b ¯i χ¯ ¯i t 3 −2 3 3 −1 3 −2 . 2 (3 5) b ¯ b ¯ ρˆ ¯ s 3j+1 3j+2 k 3j+2 = 2(4 + 1) · 2 a3¯j+1a3¯j+2 χˆ3¯j+2 t where the fractions are square-free and gcd(¯ρ3¯i−2, χ¯3¯i−2) = 1, gcd(ˆρ3¯j+2, χˆ3¯j+2) = 1. * Then, b ...b ρ¯ ρ¯ ρˆ ρˆ s2 1 ℓ k 1 ... 3ℓo−2 ℓ−2ℓe+2,2 ... ℓ2 . . = 2(4 + 1) · · 2 (3 6) a1...aℓ χ¯1 χ¯3ℓo−2 χˆℓ−2ℓe+2,2 χˆℓ2 t When ℓ = 2ℓo + 3ℓe > 4 for odd ℓ0 and even ℓe, b ...b 4k + 1)4m+2 − 1 ρˆ ρˆ ρ¯ ρ¯ s2 1 ℓ = 2(4k + 1) · 22 ... 2ℓo,2 ℓ−3ℓe+1 ... ℓ−2 · . (3.7) a ...a 4k χˆ χˆ χ¯ χ¯ t2 1 ℓ 22 2ℓo,2 ℓ−3ℓe+1 ℓ−2 If the products ℓo ℓe ρ¯ ¯ ρˆ ¯ 3i−2 3ℓo+2j,2 (3.8) χ¯ ¯ χˆ ¯ ¯ 3i−2 ¯ 3ℓo+2j,2 iY=1 jY=1 for odd ℓ and ℓo ℓe−1 ρˆ ¯ ρ¯ ¯ 2i,2 2ℓo+3j+1 (3.9) χˆ ¯ χ¯ ¯ ¯ 2i,2 j=1 2ℓo+3j+1 iY=1 Y for even ℓ are not the squares of rational numbers, b ...b s2 1 ℓ 6= 2(4k + 1) · ℓ is odd a ...a t2 1 ℓ (3.10) b ...b 4k + 1)4m+2 − 1 s2 1 ℓ 6= 2(4k + 1) · ℓ is even a ...a 4k t2 1 ℓ which would imply the inequality (3.3) and the non-existence of an odd perfect number. Conditions on the factors of N have been given [19]. A bound for the odd exponent of the prime 4k + 1 in the product representation of this integer will be derived. Lemma 3.1. The exponent 4m + 1 in the prime factorization must be greater than or 4m+1 ℓ 2αi equal to 5 for the integer N = (4k + 1) i=1 qi to be an odd perfect number. * The notation has been changed from thatQ of reference [12] with a different choice of index forρ ¯,χ ¯. 27 σ(n) Proof. Let I(n) = n be the abundance of an odd integer. The equality of σ(N) with 2N requires the following limits [14] 1 (4k + 1)4m+2 − 1 4k + 1 I (4k + 1)4m+1 = < (4k + 1)4m+1 4k 4k ℓ 8k I q2αi > i 4k + 1 i=1 ! Y (3.11) ℓ 3(4k + 1)2 − 4(4k + 1) + 2