Odd perfect numbers

A perfect number equals the sum of its positive divisors other than itself: 6 = 1 + 2 + 3. Could any perfect number be odd?[1] [2]

A positive divisor divides a whole number without a remainder. A perfect number equals the sum of its positive divisors other than itself. Does this ever happen for an odd positive integer? Let $\sigma(N)$ denote the sum including $N$ itself, so perfection means $\sigma(N)=2N$. For six, choosing divisors of its prime factors gives $\sigma(6)=(1+2)(1+3)=12$. For nine, $\sigma(9)=1+3+9=13$, which is not eighteen.

For a factorization $N=\prod_i p_i^{a_i}$, each $p_i$ is a different prime and its exponent $a_i$ counts repeated copies in the product. Independent choices of powers give $\sigma(N)=\prod_i(1+p_i+\cdots+p_i^{a_i})$. Thus a perfect number requires the product of the ratios $(1+p_i+\cdots+p_i^{a_i})/p_i^{a_i}$ to equal two. An odd candidate has only odd prime factors. Euler’s necessary form is $N=p^e m^2$, with $p$ prime, $p$ not dividing $m$, and $p\equiv e\equiv1\pmod4$. In particular, exactly one prime exponent is odd. These conditions restrict a candidate; they do not construct one.[1] [2]

All known examples are even. Restrictions on possible odd examples have not proved that one exists or that none can exist.[1] [2]

The July 2026 preprint assumes that exactly one of the even exponents in the prime factorization is greater than two, while the other even exponents equal two. In that restricted case, the authors prove that the prime with the larger even exponent must be three, and that exponent must be at least 23 billion. Other exponent patterns remain outside the assumption, so this does not prove general nonexistence.[1]

Divisors are tied to how a number is built by multiplication, yet perfection asks their sum to balance the number exactly. Whether that balance can occur for an odd number would tell us whether removing the factor two changes what is possible. It is a meeting point of multiplication and addition in an elementary definition.[2]

The divisor-sum equation also creates dependencies between different primes: a prime dividing one factor’s divisor sum may force another factor into the candidate. Such restrictions can be followed through chains, while the total sum must still balance exactly. The question tests whether all these local factor constraints can coexist in an odd number, connecting prime factorization to a global arithmetic identity.[2]