Quasiperfect Numbers

Existence of Quasiperfect Numbers

Definitions and mathematical statement

Shared notation.

Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

Write $\sigma(n)=\sum_{d\mid n}d$, summing over positive divisors. A quasiperfect number is a positive integer with $\sigma(n)=2n+1$, equivalently with the sum of its proper positive divisors equal to $n+1$. The problem asks whether any such integer exists. [S1] [S2]

Statement recorded in the supplied source.

Do quasiperfect numbers exist? [S1]

Residual task reported by the archive.

The embedded review $2026-08-17$ records: Construct a quasiperfect number or prove none exist; independently audit and peer-review the new eight-prime-factor computation. [S1]

Short English statement

Can the proper divisors of a positive integer add up to exactly one more than the integer itself?

Research attempt

Scope and source status.

The equation is $\sigma(n)=2n+1$, with the plus sign essential. The 2026 Toyohara--Tao--Yao preprint [S2] claims an eight-distinct-prime lower bound using extensive computation. Its arXiv abstract was inspected on 1 October 2026, but neither its full proof nor its code and formal artifacts were audited here. That claimed bound is not used as a premise. The attempt instead derives an elementary odd-square reduction and a precise obstruction for a prescribed prime support.

Odd divisor sum forces an odd square part.

For an odd prime $p$, $\sigma(p^e)$ is odd exactly when $e$ is even, because it is a sum of $e+1$ odd terms. The factor $\sigma(2^a)=2^{a+1}-1$ is always odd. Since $2n+1$ is odd, a candidate must therefore have the shape \[ n=2^a m^2, \qquad a\ge0,\quad m\text{ odd}. \] This parity argument alone does not imply that $n$ is odd. The even case requires a separate obstruction, provided next.

The even case is impossible.

Assume $a\ge1$ and set $A=2^{a+1}$, so $A\equiv0\pmod4$. Multiplicativity gives \[ (A-1)\sigma(m^2)=Am^2+1, \qquad (A-1)\left(\sigma(m^2)-m^2\right)=m^2+1. \] Thus $A-1$ divides $m^2+1$. But $A-1\equiv3\pmod4$, so its prime factorization contains a prime $r\equiv3\pmod4$ to an odd exponent. In particular $r\mid m^2+1$, which is impossible: it would give $m^2\equiv-1\pmod r$, while raising to the odd power $(r-1)/2$ contradicts Fermat's theorem. Here $r\nmid m$ follows immediately from $m^2\equiv-1$. Therefore $a=0$ and every quasiperfect candidate is an odd square. The case $m=1$ also fails directly, since $\sigma(1)=1\ne3$. This proof tracks the even case instead of assuming a parity conclusion stronger than the parity calculation provides.

Coprimality and support constraints.

The defining equation implies \[ \gcd(n,\sigma(n))=\gcd(n,2n+1)=1. \] Consequently for any exact prime power $p^{2e}\Vert n$, every prime divisor of $\sigma(p^{2e})$ is absent from $n$. This is the opposite support behavior to odd perfect numbers, where divisor sums feed back into the prime support. Mixing those two propagation rules would make a factor-chain argument incorrect.

For instance, if $3^2\Vert n$ then $13\nmid n$, because $\sigma(3^2)=13$. If $13^2\Vert n$, then neither $3$ nor $61$ can divide $n$, since $\sigma(13^2)=3\cdot61$. Thus the exact exponent-two branch cannot contain both three and thirteen. The word "exact" matters: changing the exponent changes the divisor-sum polynomial and its forbidden primes.

Abundancy gives a complementary lower bound on the support size. Since $n$ is odd, \[ 2<\frac{\sigma(n)}n=\prod_{p^{2e}\Vert n} \left(1+p^{-1}+\cdots+p^{-2e}\right) <\prod_{p\mid n}\frac p{p-1}. \] With at most two distinct odd primes the last product is at most $(3/2)(5/4)=15/8$, impossible. Hence at least three distinct primes are necessary by this elementary argument. This is deliberately weaker than the literature's much deeper reported bounds.

A strict reciprocal identity and the last-exponent problem.

Dividing $\sigma(n)=2n+1$ by $n$ and pairing divisors gives \[ \sum_{d\mid n}\frac1d=2+\frac1n. \] The excess over two is exactly $1/n$, not an arbitrary small positive number. A proposed prime support with limiting abundancy at most two is therefore impossible. If its limiting product exceeds two, however, increasing the exponents may move the finite product across two, and the exact excess $1/n$ still needs to be matched. Approximate closeness of abundancy to two does not settle that equality.

There is an exact way to expose the remaining exponent on a fixed branch. Write $n=Mp^{2e}$ with $(M,p)=1$, and put $S=\sigma(M)$. Using the geometric series and multiplying by $p-1$ gives \[ S(p^{2e+1}-1)=(p-1)(2Mp^{2e}+1), \] or \[ \left(pS-2M(p-1)\right)p^{2e}=S+p-1. \] For fixed $M,p$, the coefficient on the left must be positive and the positive rational quotient \[ \frac{S+p-1}{pS-2M(p-1)} \] must be an even power of $p$ with exponent at least two. If the coefficient is zero or negative, the branch is impossible. This is a complete test for one prescribed cofactor and last prime, not a finite test for all possible $M$ and $p$.

Executed finite check.

The script calculates $\sigma(n)$ exactly for every positive $n\le200000$ using its prime factorization. None satisfies $\sigma(n)=2n+1$. This independent check is small and is not a reproduction of the 2026 preprint's claimed computation. It also provides the exact small divisor-sum factorizations used in the support examples. The files are checks.py and results.json under _Local/Erdos_problems/UnsolvedMath/attempt_audit_2026_10_01/number_theory/. No claim of proof-assistant verification or literature-code auditing is made by this local arithmetic check.

Remaining gap and outcome.

The attempt excludes every even integer and all odd supports of at most two primes, derives the forbidden-support rule, and gives an exact last-exponent equation on fixed branches. It does not limit the number of branches or exclude odd squares with larger support. Outcome: UNRESOLVED. The catalogue's request to independently audit the recent large computation is not completed by the present small search and is explicitly left separate from the proved reductions.

Sources

• S1 ULAM.AI, UnsolvedMath, supplied problems.json, record 1250 (NT-043), statement and background. The embedded literature-review date is 2026-08-17; that is the archive’s review date, not a fresh certification. Dataset landing page: https://huggingface.co/datasets/ulamai/UnsolvedMath.

• S2 Akira Toyohara, Ye Tao, and Siqiong Yao, Every quasiperfect number has at least eight distinct prime factors, arXiv:2608.02066 $2026$. https://arxiv.org/abs/2608.02066. Bibliographic information from the supplied record.

• S3 Peter Hagis Jr. and Graeme L. Cohen, Some results concerning quasiperfect numbers, Journal of the Australian Mathematical Society, Series A 33 $1982$, 275-286. https://doi.org/10.1017/S1446788700018790. Bibliographic information from the supplied record.

• E1 Akira Toyohara, Ye Tao, and Siqiong Yao, Every quasiperfect number has at least eight distinct prime factors, arXiv:2608.02066v1 $2026$. https://arxiv.org/abs/2608.02066v1. Abstract inspected 1 October 2026. Its computational and formal-verification claims are reported as claims; they were not independently reproduced here.