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Open problem, Arithmetic and number theory, posed 100

Odd perfect numbers

Open

Does there exist an odd positive integer NN that equals the sum of its proper positive divisors — equivalently, an odd integer N≥1N \ge 1 satisfying σ(N)=2N\sigma(N) = 2N, where σ(N)=∑d∣Nd\sigma(N) = \sum_{d \mid N} d is the sum-of-divisors function?

Research frontier as of 2026

As of 2026 no odd perfect number has been found, and whether one can exist remains open. Instead of a direct proof of nonexistence, a web of necessary conditions has been established that any hypothetical odd perfect number NN must satisfy: by Euler's theorem N=qkm2N = q^k m^2 where qq is a prime with q≡k≡1(mod4)q \equiv k \equiv 1 \pmod{4} and gcd⁡(q,m)=1\gcd(q, m) = 1; N>101500N > 10^{1500} and its total number of prime factors counted with multiplicity is Ω(N)≥101\Omega(N) \ge 101 (Ochem–Rao, 2012); NN has at least ω(N)≥10\omega(N) \ge 10 distinct prime factors, and ω(N)≥15\omega(N) \ge 15 if 3∤N3 \nmid N (Nielsen, 2015); its largest prime factor exceeds 10810^8 (Goto–Ohno, 2008), its second largest exceeds 10410^4, and its largest prime-power component exceeds 106210^{62}. However, as Sylvester and later researchers noted, cyclotomic factorizations of σ(pa)=∏d∣(a+1),d>1Φd(p)\sigma(p^a) = \prod_{d \mid (a+1), d>1} \Phi_d(p) apply equally to 'spoof' factorizations where a prime factor is replaced by a composite quasi-prime, and since Descartes's spoof 198585576189198585576189 exists, purely local divisibility chains face a spoof barrier unless they use a deeper global property of primes.

Best known results

  • Euler's structural theorem: any odd perfect number has the form N=qkm2N = q^k m^2 with qq prime, gcd⁡(q,m)=1\gcd(q, m) = 1, and q≡k≡1(mod4)q \equiv k \equiv 1 \pmod{4} (Euler, 1747).
  • Any odd perfect number satisfies N>101500N > 10^{1500}, has Ω(N)≥101\Omega(N) \ge 101 prime factors with multiplicity, and has a prime-power component >1062> 10^{62} (Ochem–Rao, 2012).
  • Any odd perfect number has at least ω(N)≥10\omega(N) \ge 10 distinct prime factors, and ω(N)≥15\omega(N) \ge 15 if 3∤N3 \nmid N (Nielsen, 2015).

Tools and where they stop

ToolAchievedWhere it stops
Cyclotomic factorization chains and branch-and-bound search (Brent–Cohen–Ochem–Rao)Propagates prime divisors of σ(pa)=∏d∣(a+1),d>1Φd(p)\sigma(p^a) = \prod_{d \mid (a+1), d>1} \Phi_d(p) through σ(N)=2N\sigma(N) = 2N using factor tables of cyclotomic polynomials to rule out all candidates up to 10150010^{1500}.Any finite search tree can only push the lower bound on NN higher and encounters unfactored composite numbers of hundreds of digits along deep branches.
Abundancy inequalities and Diophantine bounds on ω(N)\omega(N) (Sylvester, Nielsen)Uses the inequalities ∏i=1ω(N)σ(piai)piai=2<∏i=1ω(N)pipi−1\prod_{i=1}^{\omega(N)} \frac{\sigma(p_i^{a_i})}{p_i^{a_i}} = 2 < \prod_{i=1}^{\omega(N)} \frac{p_i}{p_i - 1} to bound the smallest prime factors when ω(N)\omega(N) is fixed, ruling out ω(N)≤9\omega(N) \le 9.Each increment in ω(N)\omega(N) causes a combinatorial explosion of cases, and purely local multiplicative relations also hold for Descartes-type 'spoof' odd perfect numbers, which do exist.

Open questions

  • Can one prove that no odd perfect number is divisible by 33, 55, or 77 — or more generally establish a lower bound on the smallest prime factor that grows faster than the upper bound forced by ω(N)\omega(N)?
  • Is there an invariant that distinguishes true prime factorizations from Descartes-type spoof factorizations well enough to bypass the spoof barrier in σ(N)=2N\sigma(N) = 2N?

References

  1. Leonard Eugene Dickson (1919). History of the Theory of Numbers, Volume I: Divisibility and Primality
  2. Pascal Ochem, Michaël Rao (2012). Odd perfect numbers are greater than 10150010^{1500} · DOI:10.1090/s0025-5718-2012-02563-4
  3. Pace P. Nielsen (2015). Odd perfect numbers, Diophantine equations, and upper bounds · DOI:10.1090/s0025-5718-2015-02941-x
  4. Takeshi Goto, Yasuo Ohno (2008). Odd perfect numbers have a prime factor exceeding 10810^8 · DOI:10.1090/s0025-5718-08-02050-4