$\sigma(p^k) = 2m^2$ forces $p \equiv k \equiv 1 \pmod{16}$

Proved

OddPerfectNumber.prime_and_exp_mod_sixteen_of_sigma_eq_two_mul_sq

by Gabewhigham · Sep 8, 2026 · Mathlib 0df444a (Lean v4.33.1)

diophantine-equationsdivisor-sumsnumber-theoryperfect-numbers

Write $\sigma(n)=\sum_{d\mid n} d$ for the sum-of-divisors function.

Let $p$ be a prime with $p \equiv 1 \pmod 4$ and let $k$ be a natural number with $k \equiv 1 \pmod 4$ . If

\sigma(p^{k}) \;=\; 2m^{2}

for some natural number $m$ , then necessarily

p \equiv 1 \pmod{16} \qquad\text{and}\qquad k \equiv 1 \pmod{16}.

The statement is not vacuous: for $p = 17$ and $k = 1$ one has $\sigma(17) = 18 = 2\cdot 3^{2}$ , and indeed $17 \equiv 1 \pmod{16}$ and $1 \equiv 1 \pmod{16}$ ; likewise $\sigma(97) = 98 = 2 \cdot 7^{2}$ with $97 \equiv 1 \pmod{16}$ .

Role in the theory of odd perfect numbers. If $N = p^{k}m^{2}$ is an odd perfect number in Euler form, so that $p \equiv k \equiv 1 \pmod 4$ and $p \nmid m$ , the Euler equation $\sigma(p^{k})\sigma(m^{2}) = 2p^{k}m^{2}$ together with $p \nmid \sigma(p^{k})$ yields natural numbers $s$ with $2m^{2} = \sigma(p^{k})s$ and $\sigma(m^{2}) = p^{k}s$ . One has $s = 1$ exactly in the extremal situation $m^{2} \le p^{k}$ , and then $\sigma(p^{k}) = 2m^{2}$ . The congruences above therefore restrict that extremal situation severely: unless both the special prime and the special exponent are $\equiv 1 \pmod{16}$ , one must have $p^{k} < m^{2}$ .

Preamble

import Mathlib
open Finset

Formal statement

namespace OddPerfectNumber

theorem prime_and_exp_mod_sixteen_of_sigma_eq_two_mul_sq (p k m : ℕ) (hp : p.Prime)
    (hp4 : p % 4 = 1) (hk : k % 4 = 1)
    (h : (∑ d ∈ (p ^ k).divisors, d) = 2 * m ^ 2) :
    p % 16 = 1 ∧ k % 16 = 1 := by sorry

end OddPerfectNumber

Source

Elementary consequence of the classical splitting of the Euler equation for odd perfect numbers; the parametrisation 2m^2 = sigma(p^k) s, sigma(m^2) = p^k s is due to J. A. B. Dris, 'The abundancy index of divisors of odd perfect numbers', Journal of Integer Sequences 15 (2012), Article 12.4.4, Section 2.

View graph

σ(pk)=2m2\sigma(p^k) = 2m^2σ(pk)=2m2 forces p≡k≡1(mod16)p \equiv k \equiv 1 \pmod{16}p≡k≡1(mod16)

$\sigma(p^k) = 2m^2$ forces $p \equiv k \equiv 1 \pmod{16}$