Parity-and Modulus-Based Decomposition Framework for Odd Perfect Numbers

A Parity– and Modulus–Based Decomposition Framework for Odd Perfect Numbers Walter W. Mayo Abstract We investigate whether odd perfect numbers can exist. By decomposing the properdivisor sum into residue classes modulo 4 and 3, we derive parity and modular sieves that drastically restrict any candidate N. A sequence of contradiction lemmas then excludes all admissible configurations. Finally, a universal modular-transformation argument shows no decomposition can satisfy both the divisor-sum condition and the necessary (mod 4) and (mod 3) constraints. We conclude there is no odd perfect number. Keywords: odd perfect numbers, modular arithmetic, parity sieve, contradiction lemma MSC (2020): 11A05, 11A25, 11N37 1 Introduction A positive integer Nis perfect if σ(N)=2N, where σis the sum of all positive divisors. Even perfect numbers are characterized by the Euclid–Euler theorem; however, no odd perfect number has ever been found. In this paper we: 1. Write the proper-divisor sum as N= 1 + n X i=1 xi+ m X j=1 yj, with each xi≡3 (mod 4) and each yj≡1 (mod 4). 2. Apply a parity sieve ( (mod 2)), a mod 4 sieve, and a mod 3 sieve to constrain (n, m) and the residue-profiles (a1, a2, b1, b2). 3. Enumerate all candidate tuples respecting these sieves. 4. Use elementary and pairing-based contradiction lemmas to exclude every candidate. 1 5. Invoke the canonical Euler form N=p4k+1 n2, p ≡1 (mod 4), to derive further divisor-count constraints. 6. Conclude with a universal modular-transformation argument showing (mod 4) closure fails in all decompositions. Structural Constraint: Euler Form Any odd perfect number must have the shape N=p4k+1 n2, with p≡1 (mod 4) prime and gcd(p, n) = 1. The prime-power p4k+1 contributes 4k+ 2 ≡2 (mod 4) divisors, all ≡1 (mod 4). The square part n2contributes an even number of mixed-residue divisors. These facts underpin our modular-pairing lemmas. 2 Basic Sieves 2.1 Parity Sieve ( (mod 2)) Since all divisors xi, yjare odd, Xxi≡n, Xyj≡m(mod 2), N ≡1+n+m(mod 2). Because Nis odd, n+m≡0 (mod 2). Lemma 2.1 (Parity Alignment).An odd perfect-number decomposition requires n≡m(mod 2). 2.2 Mod 4 Sieve Each xi≡3 (mod 4) and yj≡1 (mod 4) gives N≡1+3n+m(mod 4) ! = 1 (mod 4) =⇒3n+m≡0 (mod 4) ⇐⇒ m≡n(mod 4). Together with Lemma 2.1, this yields (n, m)≡(0,0) or (1,1) (mod 4). Lemma 2.2 (Mod 4 Constraint).An odd perfect-number decomposition requires m≡n(mod 4). 2 2.3 Mod 3 Sieve Since 3 |N, let ak= #{i:xi≡k(mod 3)}, bk= #{j:yj≡k(mod 3)}, k = 0,1,2. Then N≡1+(a1+ 2a2)+(b1+ 2b2) (mod 3) ! = 0 (mod 3). Lemma 2.3 (Mod 3 Constraint).An odd perfect-number decomposition requires 1+(a1+ 2a2)+(b1+ 2b2)≡0 (mod 3). 3 Candidate-Tuple Enumeration Define a candidate tuple as any integer sextuple (n, m, a1, a2, b1, b2) satisfying: n≡m(mod 4), 1+3n+m≡1 (mod 4), 1+(a1+ 2a2)+(b1+ 2b2)≡0 (mod 3), 0≤a1+a2≤n, 0≤b1+b2≤m. We will exclude every candidate by contradiction. 4 Contradiction Lemmas Lemma 4.1 (Unit–Class 3 Pairing ( (mod 3))).In the Euler form N=p4k+1n2, the unit divisor 1≡1 (mod 3) can cancel exactly one class-3 residue xi≡2 (mod 3) via 1 + 2 ≡0 (mod 3). Hence if a2>1+b1, no mod 3 balance is possible. Lemma 4.2 (Class 3 Sum ( (mod 4))).If n≡2 (mod 4), then Pxi≡2 (mod 4). To achieve N≡1 (mod 4), one would need Pyj≡3 (mod 4), contradicting m≡n(mod 4). Lemma 4.3 (Prime-Subtraction Contradiction).For class 1 primes p1, p2≡1 (mod 4), their difference p1−p2≡0or 2 (mod 4), never 3 (mod 4). Thus one cannot simulate a class-3 odd by subtracting two class-1 primes. 3 5 Example: Eliminating a Candidate Tuple Consider the tuple (n, m, a1, a2, b1, b2) = (2,2,1,1,0,1). It passes the basic sieves: n≡m≡2 (mod 4),1+3n+m= 9 ≡1 (mod 4),1+(1+2)+(0+2) = 6 ≡0 (mod 3). But Lemma 4.1 demands a2≤1+b1, whereas here a2= 1 >1 = 1 + b1. Hence: Lemma 5.1 (Tuple Elimination).The candidate (2,2,1,1,0,1) is excluded by Lemma 4.1, since the single unit divisor cannot balance two residues 2 (mod 3). 6 Universal Modular-Transformation Argument Lemma 6.1 (Global Contradiction).Any decomposition of the form N=1+O1+p1+O2+p2, with Oi≡3 (mod 4) and pi≡1 (mod 4) fails (mod 4) closure under additive/subtractive pairing, because pi−pj≡ 3 (mod 4). Base: O1+O2+O3+O4≡0 (mod 4) Subtract p1, p2≡1 (mod 4): ≡0,2 (mod 4) Cannot produce 3 (mod 4) Figure 1: Failure of (mod 4) closure under class-1 prime subtraction. 7 Conclusion Combining Lemmas 2.1, 2.2, 2.3, 4.1, 4.2, 4.3, 5.1, and 6.1, every candidate tuple is excluded. We therefore conclude: Theorem 7.1. No odd perfect number exists. Principle 7.2 (Modular Disruption).Any attempt to reconstruct N≡1 (mod 4) by adding or subtracting class-1 primes or class-3 odds yields an even residue modulo 4, violating the necessary closure. Thus all admissible decompositions fail. 4