A Rationality Condition for the Existence of Odd Perfect Numbers

A rationality condition is derived for the existence of odd perfect numbers involving the square root of a product, which consists of a sequence of repunits, multiplied by twice the base of one of the repunits. This constraint also provides an upper bound for density of odd integers which could satisfy ${{\sigma(N)}\over N}=2$, where $N$ belongs to a fixed interval with a lower limit greater than $10^{300}$. Characteristics of prime divisors of repunits are used to establish whether the product containing the repunits can be a perfect square. It is shown that the arithmetic primitive divisors with different prime bases can be equal only when the exponents are different, with the exception of a set of cases derived from solutions of a solution prime equation. The proof of this result requires the demonstration of the non-existence of solutions of a more general prime equation, a problem which is equivalent to Catalan's conjecture. Results concerning the exponents of prime divisors of the repunits are obtained, and they are combined with the method of induction to prove to prove a general theorem on the non-existence of prime divisors satisfying the rationality condition.


Introduction
The algorithm for demonstrating the non-existence of odd perfect numbers with fewer than nine different prime divisors requires the expansion of the ratio σ(N) N and strict inequalities imposed on the sums of powers of the reciprocal of each prime divisor [1] [2].
Although it is possible to establish that σ(N) N = 2 when N is divisible by certain primes, there are odd integers with a given number of prime divisors such that σ(N) N > 2 , while σ(N) N < 2 for other integers with the same number of distinct prime factors. Moreover, the range of the inequality for σ(N) N − 2 can be made very small even when N has a few prime factors. Examples of odd integers with only five distinct prime factors have been found that produce a ratio nearly equal to 2: σ(N) N − 2 < 10 −12 [3]. Since it becomes progressively more difficult to establish the inequalities as the number of prime factors increases, a proof by method of induction based on this algorithm cannot be easily constructed.
In §2, it will be shown that there is a rationality condition for the existence of odd perfect numbers. Setting σ(N) N equal to 2 is equivalent to equating the square root of a product, 2(4k + 1) , which contains a sequence of repunits, with a rational number. This relation provides both an upper bound for the density of odd perfect numbers in any fixed interval in N with a lower limit greater than 10 300 and a direct analytical method for verifying their non-existence, since it is based on the irrationality of the square root of any unmatched prime divisors in the product. This condition is used in §3 to demonstrate the non-existence of a special category of odd perfect numbers.
The properties of prime divisors of Lucas sequences required for the study of the square root of the product of the repunits are described in §4 and §5. An induction argument is constructed in §6, which proves that the square root expression is not rational for generic sets of prime divisors, each containing a large number of elements. This is first established for odd integers with four distinct prime divisors and then by induction using the properties of the divisors of the repunits.
The number of square-full integers up to N is N . With a lower bound of 10 300 for an odd perfect number [10], it follows that 2(4k Given a lower bound of 10 6 for the largest prime factor [11], 10 4 for the second largest prime factor and 10 2 for the third largest prime factor of N [12], the density of prime products (4k + 1)  [14]. Given that the probability of an integer being a square is independent of it being expressible in terms of a product of repunits, the density of square-full numbers having the form 2(4k and N 0 is a fixed number, is bounded above by 3.28 × 10 −159 when there are at least eight different prime factors and 5.13 × 10 −163 when N is relatively prime to 3 and has more than ten different prime factors.
Although neither the numerator or the denominator will be squares of integers when q i = 3 or α i = 2, there still remains the possibility that the terms could be equal multiples of different squares. Since the repunit x n − 1 x − 1 is the Lucas sequence with α = x, β = 1, derived from the second-order recurrence where a = α + β = x + 1 and b = αβ = x, the rationality condition can be applied to the 3. Proof of the non-existence of odd perfect numbers for a special class of integers The even repunit (4k+1) 4m+2 −1 4k contains only a single power of 2 since 1+(4k+1)+(4k+ 1) 2 +...+(4k+1) 4m+1 ≡ 4m+2 ≡ 2 (mod 4). Thus, the rationality condition can be applied to a product of odd numbers (4k + 1) which, in turn, requires that (4k + 1) = σ l v 2 and r = σ l vu, so that σ l u|r and σ l u|t, contrary to the original assumption that r and t are relatively prime unless σ ℓ = u = 1. The rationality condition reduces to the existence of solutions to the equation This relation is equivalent to the two conditions x 2m+1 −1 It can be verified that there are no integer solutions to these simultaneous Diophantine equations, implying that when satisfies the gcd condition given above, the square root of (4k + 1) is not a rational number and there is no odd perfect number of the form with this constraint on the pair (4k +1, 4m+2).

Prime power divisors of Lucas sequences and Catalan's conjecture
The number of distinct prime divisors of q n −1 q−1 is bounded below by τ (n) − 1 if q > 2, where τ (n) is the number of natural divisors of n [1] [15]. The characteristics of these prime divisors can be deduced from the properties of Lucas sequences. Since the repunits have only odd prime divisors, the proofs in the following sections will have general validity, circumventing any exceptions corresponding to prime q = 2.
For a primary recurrence relation, defined by the initial values U 0 = 0 and U 1 = 1, denoting the least positive integer n such that U n (a, b) ≡ 0 (mod p), the rank of apparition, by α(a, b, p), it is known that α(x + 1, x, p) = ord p (x) [16].
The extent to which the arguments a and b determine the divisibility of U n (a, b) [17] [18] can be summarized as follows: Let p be an odd prime. If p|a, p|b, then p|U n (a, b) for all n > 1.
If p ∤ a, p|b, then either p|U n (a, b), n ≥ 1 or p ∤ U n (a, b) for any n ≥ 1. If p|a and p ∤ b, then p|U n (a, b) for all even n or all odd n or p ∤ U n (a, b) for any n ≥ 1.
If p ∤ a, p ∤ b, p|D = a 2 − 4b, then p|U n (a, b) when p|n. If p ∤ abD, then p|U p−( D p ) (a, b).
For the Lucas sequence U n (q+1, q), there is no prime which divides both q and q+1, and since only q is a divisor of the second parameter, there are no prime divisors of U n (q + 1, q) from this category because q n −1 q−1 ≡ 1 (mod q). If p|(q+1), then q n −1 when n is even. However, p ∤ q n −1 q−1 with n odd, and therefore, prime divisors from this class are not relevant for the study of the product of repunits with odd exponents.
Since q n − 1 = d|n Φ d (q) where Φ n (q) is the n th cyclotomic polynomial, it can be shown that the largest arithmetic primitive factor [21]- [23] of q n − 1 when q ≥ 2 and n ≥ 3 if Φ n (q) and n are relatively prime Φ n (q) p if a common prime f actor p of Φ n (q) and n exists (13) In the latter case, if n = p f p ′f ′ p ′′f ′′ ... is the prime factorization of n, then Φ n (q) is divisible by p if and only if e = n p f = ord p (q) when p ∤ (q − 1), and moreover, p Φ ep f (q) when f > 0 [1].
Division by q − 1 does not alter the arithmetic primitive factor, since it is the product of the primitive divisors of q n − 1, which are also the primitive divisors of q n −1 q−1 . For all primitive divisors, p ′ ∤ (q −1), so that (p ′ ) h q n −1 q−1 if (p ′ ) h |q n −1 and the arithmetic primitive factor again would include (p ′ ) h . The imprimitive divisors would be similarly unaffected because the form of the index n = ep f prevents q − 1 from being a divisor of Φ n (q) when p ∤ (q − 1). If p|(q − 1), the rank of apparition for the Lucas sequence {U n (q + 1, q)} is p, so that it is consistent to set n = p f +1 . Then, p Φ p f +1 (q) and the arithmetic primitive factor is in the expression (5) is where the indices are odd numbers n i = 2α i + 1, p i , i = 1, ..., l, represents the common factor of n i and Φ n i (q i ), and p ℓ+1 is a common factor of 4m + 2 and Φ 4m+2 (4k + 1).
Division of Φ n i (q i ) by the prime p i is necessary only when gcd(n i , Φ n i (q i )) = 1, and  Proof. Consider the following four cases: I. The arithmetic primitive factors of q n i i − 1 and q n j j − 1 are Φ n i (q i ) and Φ n j (q j ).
Since Φ n (x) is a strictly increasing function for x ≥ 1 [28], Φ n (q j ) > Φ n (q i ) when q j is the larger prime, and equality of Φ n i (q i ) and Φ n j (q j ) could only be achieved, if at all feasible, when n i = n j .
II. The arithmetic primitive factors of q n i i − 1 and q Comparing Φ n (q i ) and Φ n (q j ) p , p = p j is a common factor of n and Φ n (q j ) but it does not . The prime decomposition of e as ρ r 1 1 ...ρ r s s , gcd(ρ t , p) = 1, t = 1, ..., s, leads to the following expressions for Φ n (q i ) and Φ n (q j ), Since e = ord p (q j ), it follows that p|(q e j − 1), and if q e j = 1 + pk j (mod p), then are not divisible by p, consider a primitive prime factor p ′ of Φ n (q i ). It must divide some factor q n ρ t 1 ...ρ t k i − 1 in the expression for Φ n (q i ), and thus, it will also divide q there will be 2 k−1 factors in the numerator and 2 k−1 factors in the denominator divisible by p ′ . When k ≥ 1, the factors of p ′ exactly cancel because each term q [29]. Equivalence of Φ n (q i ) and requires that the prime power divisors of these quantities are equal, so that p ′f a a Φ n (q j ) p for all primes {p ′ a }. However, if p ′f a a q n j −1, then q n i −1 and q n j −1 have the same primitive prime power divisors. The imprimitive prime divisor p which divides q n j − 1 might also divide q n i − 1, although overall cancellation of p in Φ n (q i ) requires that p r q Integer solutions of w = y m , y ≥ 2, m ≥ 2 can be written as w = x n , x ≥ 2 with m|n. Since y|x n , y ∤ (x n − 1) because y ≥ 2. The nearest integers to x n having a similar form, so that none of these integers will have the form y m ± 1. The exception occuring when x = Since the odd primes q i , q j and the exponent n in the prime decomposition of N must be greater than or equal to 3, this restriction is consistent with Catalan's conjecture.
IV. The arithmetic primitive factors of q n i i − 1 and q factor if n i = n j . Thus, the primes p i and p j must be equal, and a comparison can be made between Φ n (q i ) p and Φ n (q j ) p . Again, by the monotonicity of Φ n (x), it follows that these quantities are not equal when q i and q j are different primes. Equality of the arithmetic prime factors could only occur if n i = n j .

The exponent of prime divisors of repunit factors in the rationality condition
Since all primitive divisors of U n (a, b) have the form p = nk + 1, it follows that is the residue index, the exponent p−1 ι(p) will be even for all odd primes p, whereas if ι(p) is even, the exponent p−1 ι(p) may be even or odd.
Given that p|U 2α i +1 (q i + 1, q i ), ι(p) is even and p where N q can be defined to be q . By the logarithmic rule for Fermat quotients, [32]. Since µ i = 0 in general, except when i = q, it follows that q p−1 −1 = 0 (mod p 2 ) except for p − 1 values of q between 1 and p 2 − 1.
By Hensel's lemma [33] [34], each of the integers between 1 and p − 1, which satisfy through the formula Since ϕ(p 2 ) = p(p − 1), a set of p − 1 solutions to equation (17) can also be labelled as for any c 3 between 1 and p − 1.
Theorems concerning the Fermat quotient q r −1 q−1 can be extended to quotients of the type q nr −1 q n −1 . It has been proven, for example, that p q nr −1 q n −1 , p ∤ r, p ∤ q n − 1, then [35], and more generally, if p h When p|(q r − 1), the following lemma is obtained.
for any non-trivial divisor s of 2α i + 1.
Suppose that p| Then, by equation (19), By a theorem on congruences, if q e ≡ 1 (mod p), where e|(p − 1) and q p−1 ≡ 1 (mod p 2 ), then q e ≡ 1 (mod p 2 ) The theorem on congruences can be extended to larger prime powers: q e ≡ 1 (mod p n ) and q p−1 ≡ 1 (mod p n+1 ), then q e ≡ 1 (mod p n+1 ). From the first congruence relation, q e = 1 + k ′ p n for some integer k ′ . Raising this quantity to the power p−1 e , it follows that Since p−1 e < p, the integer k ′ must be a multiple of p. Thus, q e = 1 + k ′′ p n+1 ≡ 1 (mod p n+1 ). By the generalized congruence theorem, and equation (20) in turn implies that p 4 q p−1 i −1 q i −1 . Since this process can be continued indefinitely to arbitrarily high powers of the prime p, a contradiction is obtained once the maximum exponent Similarly, If s is a non-trivial divisor of 2α i + 1, then p , because it is a primitive divisor of U 2α i +1 (q i + 1, q i ). Given that p h U 2α i +1 (q i + 1, q i ), by equation (21), Imprimitive prime divisors of U n (a, b) are characterized by the property that p|U d (a, b) for some d|n. The exponent of the imprimitive prime power divisor exactly dividing q n −1 q−1 can be determined by a further lemma: if p h q n −1 q−1 , then either gcd(n, p − 1) = 1, q ≡ 1 (mod p), p h |n (mod p) or e = gcd(n, p−1) > 1, p k |Φ e (q), p h−k n [15]. Since v p (Φ e (q)) = v p (q e − 1) if p ∤ q − 1, the general formula [12] for the exponent of a prime divisor of a repunit is The exponent also can be deduced from the congruence properties of q-numbers [n] = q n −1 q−1 and q-binomial coefficients [36], as it equals s = ǫ 0 h + ǫ 1 + ... + ǫ k−1 where p h q e − 1 and n − 1 = a 0 + e(a 1 + a 2 p + ... + a k p k−1 ) with ǫ i equal to 0 or 1, which is consistent with equation (21) because ǫ 0 = 1 and v p (n) = Specializing to the case of h = 2, it follows that if the quotient q n −1 q−1 is exactly divisible by p 2 , then (i) gcd(n, p − 1) = 1, p|(q − 1) or p ∤ q n − 1, p 2 n (ii) p Φ e (q), where e = α(q + 1, q, p) is the rank of apparition of p, p n (iii) p 2 Φ e (q), p ∤ n and the only indices n i which allow for exact divisibility of q n i i −1 q i −1 by p 2 are n i = µp 2 , when p|(q i − 1) or e i ∤ n i , n i = µe i p when p Φ e i (q i ) and n i = µe i when p 2 Φ e i (q i ). Since n i is odd, the three categories can be defined by the conditions: (i) n i = µp 2 , (ii) n i = µe i p, p is a primitive divisor of The equation is known to have finitely many integer solutions for m, n, x, y, given a and b such that gcd(a, b) = 1, a(y − 1) = b(x − 1), and max(m, n, x, y) < C where C is an effectively computable number depending on a, b and F where |x − y| < F z (log z) 2 (log log z) 3 with z = max(x, y) [37] [38]. Using this relation to re-express , it can be established that there are unmatched primes in the product of the repunits (5) and that the square root of this expression is irrational for several different categories of prime divisors {q i , i = 1, ..., ℓ; 4k + 1}.
Theorem 2. The square-root expressions 2(4k + 1) Similarly, if p ∤ (q j − 1) but p| (q ℓ − 1), then an odd power of p divides the product of the two repunits if Q q j ≡ 0 (mod p) or , with h ′ j odd, and p remains an unmatched prime divisor.
(iii) When n j = 2α j + 1 is set equal to n ℓ = 2α ℓ + 1, the primitive prime divisors of (iv) Additional prime divisors are introduced when the exponents are adjusted, so that, in general, there will be unmatched prime divisors in the products of the repunits

Proof.
Suppose {a i } and {b i } are defined by define {a ij } and {b ij } with gcd(a ij , b ij ) = 1, and Since the fraction b 1 a 1 can be expressed in terms of b 2 where ρ r 12 12 , χ s 12 12 denote products of various powers of different primes, with r 12 , s 12 representing the sets of exponents, (r 12 ) 0 , (s 12 ) 0 labelling a collection of exponents consisting of 0 or 1 and ρ 12 , χ 12 being products of these primes with all of the exponents equal to 1. The sets (r 12 ) 0 , (s 12 ) 0 are chosen so that r 12 − (r 12 ) 0 = 2r 12 , s 12 − (s 12 ) 0 = 2s 12 represent even exponents. Since a similar relation exists between a 2 b 2 and a 3 b 3 , and where gcd(p, q) = 1. If a 2 = (4k + 1)ρ 12 ρ 23 * denotes the square of a rational number.
Since gcd(a 3 , b 3 ) = 1, the square-free factors can be separated in the fraction Since a 3 is even, andâ 3 is divisible by a single factor of 2, χ 12 = ρ 23 p 2q , and similarly, becauseb 3 is odd, ρ 12 = 1 4k+1 χ 23 qp. Since ρ 12 ρ 23 χ 12 χ 23 = 2 4k+1 qp pq , rationality of also could be achieved by setting a 2 = (4k + 1)qp p ′2 2 and b 2 = pqq ′2 . Then Either there is an overlap between the prime factors of (4k+1) p 2 andq orâ 2 = (4k+1) p 2q = (4k +1) p ′ 2q ′ , and similarly, there is either an overlap between the prime factors of q andp or b 2 = qp = q ′p′ . Removing any overlap, then the remaining square factors can be separated in a 2 and b 2 obtaining the formâ 2 b 2 for the square-free part of the ratio a 2 b 2 . The equalities containingâ 2 andb 2 imply thatp >p ′ ≥ p ′ > p andq >q ′ ≥ q ′ > q. By interchanging the roles of a 2 , b 2 and a 3 , b 3 in the above argument, the inequalities p >p and q >q can be derived, implying a contradiction. Thus, when ℓ = 3, it should not be possible to find coefficients {a i } and {b i } satisfying equation (25)  A variation of the standard induction argument can be used to show that there cannot be different odd perfect numbers with prime decompositions (4k When ℓ is odd, rationality of square root of the product of repunits with ℓ−1 prime bases Since the values q ℓ = 3 and α ℓ = 2 can be excluded from the product of repunits, ρ ℓ is odd and does not equal 1, so that b 1 ...b ℓ a 1 ...a ℓ = 2(4k + 1) . The square root of the product of repunits with ℓ prime bases {q i , i = 1, ..., ℓ} is therefore not rational.
The proof can be continued for ℓ > 3 by assuming that there do not exist any odd primes q 1 , ..., q ℓ−1 and 4k + 1 such that 2(4k + 1) rational and proving that the same property is valid when ℓ odd primes q 1 , ..., q ℓ arise in the prime decomposition of the integer N .
is integer, and non-existence of odd perfect numbers of the form (4k + 1) 4m+1 q 2α 1 1 ...q ..a ℓ = 2(4k + 1) . Since Since the irrationality of the square root expression is assumed to hold generally for ℓ − 1 odd primes {q i } and any value of 4k + 1, the effect of the inclusion of another prime q ℓ can be deduced. Thus, given an arbitrary set of ℓ odd primes, q 1 , ..., q ℓ and some prime of the form 4k + 1, irrationality of the square root of expression (43) implies that However, by equation (25) ℓ , separating the square-free factors from the factors with even exponents. it follows that The form of the relation (45) is valid for arbitrary values of b ℓ a ℓ , but the choice of ρ ℓ is specific to the repunit q ℓ 2α ℓ +1 −1 is the square of an integer only when q ℓ = 3, α ℓ = 2, it is preferable to represent the rationality condition for ℓ − 1 and ℓ primes when ℓ is odd. Irrationality of the square root expression for ℓ−1 primes {q i , i = 1, ...ℓ−1}, which requires that ω ℓ−1 = 1 is a square-free integer, implies irrationality for ℓ primes {q i , i = 1, ..., ℓ} if ω ℓ−1 ρ ℓ = ω ℓ = 1 is square-free.
For any prime divisor p v p (ω ℓ−1 ) = where e i = ord p (q i ). It follows that Suppose that p is one of the extra prime divisors so that v p (ω ℓ−1 ) = 1. If e ℓ ∤ n ℓ or p ∤ n ℓ , then p q ℓ n ℓ −1 q ℓ −1 and v p (ω ℓ ) = 1.
Since it has been assumed that the square root expression with ℓ − 1 primes {q i , i = 1, ..., ℓ − 1} is irrational, there is either an unmatched primitive divisor or an imprimitive divisor in the product . Suppose that the extra prime divisor p j is a factor of the repunit q 2α j +1 j −1 q j −1 . By equation (27), To proceed further, it is first useful to choose the exponent 2α ℓ +1 to be equal to 2α j +1.
where h ′ ℓ is odd, the prime divisor p in the product of the two repunits cannot be entirely absorbed into the square factors. Similar conclusions hold when p ∤ (q j − 1) and p|(q ℓ − 1).
Let p be an imprimitive prime divisor such that p ∤ (q j − 1) and p ∤ (q ℓ − 1), then . If p h |n ℓ , and n j = n ℓ , then h j = h ℓ = h, again implying that the prime divisor can be absorbed into the square factors.
The arithmetic primitive factors of respectively, are different when n j = n ℓ , except possibly for solutions generated by the prime equation q n ℓ −1 q n j −1 = p required when either p j = gcd(n j , Φ n j (q j )) or p ℓ = gcd(n ℓ , Φ n ℓ (q ℓ )) equals 1. The algebraic primitive factors Φ n j (q j ) and Φ n ℓ (q ℓ ) necessarily will be different if n j = n ℓ . Consider a prime divisor p ′ of the arithmetic primitive factors which is raised to a different power in Since it has been established that square classes of the repunits q n −1 q−1 consist of only one element [40], it follows that (q and there is only one representative from each sequence {q n j j − 1, n j ∈ Z}, {q n ℓ ℓ − 1, n ℓ ∈ Z} which has a specified square-free factor κ. Thus, unless n q j (κ) coincides with n q ℓ (κ).
If q n ℓ ℓ − 1 = κ(y ′ 2 ) 2 and q n ℓ j − 1 = κ(y ′ 1 ) 2 , and . Both inequalities for y 1 cannot be satisfied if q n ℓ 4 ℓ < √ κκ 2 or equivalently q n ℓ 2 ℓ < gcd(q n ℓ j − 1, q n ℓ ℓ − 1). When the pair of primes (q j , q ℓ ) satisfies the last inequality, the prime divisors in ρ j and ρ ℓ do not match and the product of the repunits , it follows that y 2 1 (q n ℓ − 1) ≃ y 2 2 (q n j − 1) leading to consideration of the inequality |y 2 2 q n j − y 2 1 q n ℓ | ≤ |y 2 2 − y 2 1 |. The constraint placed on q j is q ℓ , it is sufficient for q j to satisfy the stronger constraint which is equivalent to an upper bound for y 2 of Any adjustment in n ℓ will introduce additional prime divisors. Either they shall be new prime factors of the exponent or primitive divisors [42]- [45]. If n ℓ is multiplied by a prime factorp r ℓ , wherep|ρ ℓ , then the product ρ ℓp r ℓ will contain the powerp 1+r ℓ ℓ . While the prime power can be absorbed into the product of square factors only when r ℓ is odd, the repunit q n ℓp r ℓ ℓ −1 q ℓ −1 has extra primitive divisors, giving rise to a non-trivial ω ℓ , implying irrationality of the square root expression with ℓ primes {q i , i = 1, ...ℓ}. Moreover, gcd(Φpi(q), Φpj (q)) = 1 when i = j and p ∤ (q − 1), multiplication of the index byp r ℓ will introduce new prime divisors through the decomposition of the repunit The abstract argument given for ℓ = 3 could also be extended to higher values of ℓ. This approach would consist of the demonstration of the property b 1 ...b ℓ a 1 ...a ℓ · = 2(4k + 1) · if ℓ is odd, and b 1 .
if ℓ is even, given that there are no sets of primes {q i } with less than ℓ elements satisfying the rationality condition. It may be noted that since and b 13 a 13 ] are square-free factors, the quotient will equal It has been established that b ℓ a ℓ = 2(4k + 1) · because there is no odd integer of the form (4k + 1) 4m+1 q 2α ℓ ℓ which satisfies the relation σ(N) · because of the non-existence of odd perfect numbers of the type (4k + 1) 4m+1 q (57) and the coefficients {a i , b i } will not satisfy the rationality condition when the square-free ,χ ℓ1 ,χ ℓ1 have prime divisors other than 2 and 4k + 1 which do not match to produce the square of a rational number.
When ℓ is odd and greater than 5, there always exists an odd integer ℓ o and an even integer ℓ e such that ℓ = 3ℓ o + 2ℓ e , implying the following identity Consequently, is not the square of a rational number.
If ℓ is even and greater than 4, there always exists an odd integer ℓ o and an even integer ℓ e such that ℓ = 2ℓ o + 3ℓ e . From the identity χ ℓ−3j+1 is not the square of a rational number.

Conclusion
The rationality condition provides an analytic method for investigating the existence of odd perfect numbers. The aim of this approach then becomes the proof of the existence of an unmatched prime divisor in the product of the repunits, since the square root of any such divisor would be irrational, contrary to the condition for the existence of an odd perfect number. An upper bound for the density of odd integers greater than 10 300 , in an interval of fixed length, which could satisfy σ(N) N = 2, may be found by considering the square root expression containing the product of repunits, combining the estimate of the density of square-full numbers in this range with the probability of an integer being expressible as the product of repunits with prime bases multiplied by 2(4k + 1). Repunits form a special class of Lucas sequences, and the properties of primitive and imprimitive prime divisors of these sequences can be used to determine the powers of primes dividing the product of repunits. A comparison of the divisibility properties of Lucas sequences U n (q+1, q) with different values of q has been undertaken in §4. Specifically, the arithmetic primitive factors of these repunits, products of the primitive prime power divisors, can be compared for different values of the prime basis, and it has been shown that they could only be equal if the indices of Lucas sequences differ, except possibly for pairs of divisors Φ n (q i ), Φ n (q j ) p j generated by the prime equation q n j −1 q n i −1 = p. In the second theorem, nonexistence of the odd perfect numbers for a large set of primes {q i , i = 1, ..., ℓ; 4k + 1}, exponents {2α i + 1, i = 1, ..., ℓ; 4m + 1} and values of ℓ using the method of induction adapted to the coefficients {a i , b i } in the product of repunits. An abstract argument is given for the non-existence of coefficients satisfying the rationality condition when ℓ = 3 and then various results are proven for ℓ > 3 by using the properties of prime divisors of product of two repunits, q 2α j +1 j −1 q j −1 and q 2α ℓ +1 ℓ −1 q ℓ −1 , belonging to each of the four categories: (i) p|(q j − 1), p|(q ℓ − 1) (ii) p|(q j − 1), p ∤ (q ℓ − 1) (iii) p ∤ (q j − 1), p|(q ℓ − 1) (iv) p ∤ (q j − 1), p ∤ (q ℓ − 1). Irrationality of the square root expression for any set of ℓ − 1 primes {q i , i = 1, ..., ℓ − 1} implies that each unmatched prime divisor in the product of repunits with bases {q i , i = 1, ..., ℓ − 1, 4k + 1} can be associated with a single repunit, because factors of other repunits divisible by this prime contain powers of the prime with the exponent summing up to an even integer. Supposing, for example, that the repunit containing this extra prime divisor is q 2α j +1 j −1 q j −1 . The problem of determining whether this prime divisor remains unmatched, when an additional prime q ℓ in the decomposition of the odd integer N is included, depends on the feasibility of matching the prime divisors of each pair of repunits (U n j (q j + 1, q j ), U n ℓ (q ℓ + 1, q ℓ )) as j takes all values in the range {1, 2, ..., ℓ − 1} such that the repunit U n j (q j + 1, q j ) contains an extra prime divisor.