gpt 6 astra ultra attempt on Math-582

Is Ten Solitary?

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.

The abundancy index is $I(n)=\sigma(n)/n$. A positive integer is solitary if no different positive integer has the same index. Since $\sigma(10)=18$, the question is whether $5\sigma(m)=9m$ has a positive integer solution other than $m=10$. [S1] [S2]

Statement recorded in the supplied source.

Is 10 a solitary number (no other number shares its abundancy index)? [S1]

Residual task reported by the archive.

The embedded review $2026-08-17$ records: Find a friend of 10 or prove that sigma(n)/n=9/5 forces n=10. [S1]

Short English statement

Is ten the only positive integer whose sum of divisors is nine fifths of the integer?

Research attempt

Scope and source correction.

The equation is $5\sigma(m)=9m$, and $m=10$ must be excluded as the trivial solution. The URL in catalogue source [S2] identifies Henry Robert Thackeray's paper, not the author imported in that citation. The corrected primary reference is [E1]. Its stated ten-prime-factor bound is not reproduced or needed below. The attempt proves a smaller set of structural exclusions directly.

Every other candidate is an odd square divisible by five.

The equation implies $5\mid m$. Abundancy $I(t)=\sigma(t)/t$ strictly increases under a proper divisibility extension: prime by prime, each additional exponent increases the finite sum $1+p^{-1}+\cdots+p^{-e}$, and a new prime factor multiplies it by a factor greater than one. If $m$ is even, then $10\mid m$, so $I(m)=I(10)$ forces $m=10$. Thus every other candidate is odd.

For odd $m$, $\sigma(m)=9m/5$ is odd. The divisor-sum parity rule therefore makes every prime exponent even, so $m$ is a square and $5^2\mid m$. These conclusions cover all even candidates, with no finite bound, and reduce the remaining search to odd squares.

The prime three can also be excluded.

Suppose $3\mid m$. If $3^4\mid m$, then $3^4 5^2\mid m$ and \[ I(m)\ge\frac{121}{81}\frac{31}{25} =\frac{3751}{2025}>\frac95, \] contradiction. Since exponents are even, the only remaining option is $3^2\Vert m$. Then $\sigma(3^2)=13$ divides $\sigma(m)$, and the equation $5\sigma(m)=9m$ forces $13\mid m$. Hence $3^2 5^2 13^2\mid m$, and \[ I(m)\ge\frac{13}{9}\frac{31}{25}\frac{183}{169} >\frac95. \] For the last strict comparison, cross multiplication reduces to $13\cdot31\cdot183>405\cdot169$, namely $73749>68445$. Thus $3\nmid m$. Every friend of ten would be an odd square coprime to six, with five among its prime divisors.

A forced prime congruence and exponent pattern.

Write $5^{2b}\Vert m$ with $b\ge1$. Taking five-adic valuations in the equation gives $v_5(\sigma(m))=2b-1>0$. The factor $\sigma(5^{2b})$ is one modulo five, so some other prime power $q^{2e}\Vert m$ has $5\mid\sigma(q^{2e})$. If $q\not\equiv1\pmod5$, the geometric-series identity implies $q^{2e+1}\equiv1\pmod5$. The order of $q$ modulo five then divides both the odd number $2e+1$ and four, so it is one, a contradiction. Therefore $q\equiv1\pmod5$, and now $\sigma(q^{2e})\equiv2e+1\pmod5$, yielding \[ q\equiv1\pmod5,\qquad 2e\equiv4\pmod{10}. \] Thus at least one non-five prime occurs to exponent at least four. The argument proves existence of such a factor, but does not identify which factor or bound how large it can be.

Prime support must absorb divisor-sum factors.

If $p^{2e}\Vert m$, every prime divisor of $\sigma(p^{2e})$ divides $9m$, by the defining equation. Except for the prime three, such primes must already belong to the support of $m$. At the same time $3\nmid m$ but $v_3(\sigma(m))=2$, so all contributions to the prime three from the several divisor sums must total exactly two. Merely recording that three occurs is insufficient; its valuation budget is fixed.

These conditions give a branching search: choose exact exponents, factor their divisor sums, add forced primes, and prune branches with abundancy exceeding $9/5$ or excessive three-adic valuation. The unproved step is termination across arbitrarily many primes and arbitrarily large even exponents. A contradiction in the all-exponents-two branch does not address the forced higher exponent just derived.

Executed exact check.

The script factors every $1\le m\le200000$, evaluates $\sigma(m)$, and compares the integers $5\sigma(m)$ and $9m$. It finds exactly $m=10$. No rounded abundancy values are compared. Code and result are checks_second.py, results_second.json in _Local/Erdos_problems/UnsolvedMath/attempt_audit_2026_10_01/number_theory/. This bound is a modest local check, not the cited paper's structural computation or a claim that every possible friend has been exhausted.

Remaining gap and outcome.

All nontrivial candidates are reduced to odd squares divisible by five, coprime to six, with a specified higher-exponent prime congruence and exact divisor-sum valuation constraints. None of these conditions is shown globally incompatible. Outcome: UNRESOLVED. No friend of ten is constructed and no proof that ten is solitary is obtained.

Sources

• S1 ULAM.AI, UnsolvedMath, supplied problems.json, record 1260 (NT-053), 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 S. R. L. S. de Araujo, Each friend of 10 has at least 10 nonidentical prime factors, arXiv:2310.15900. https://arxiv.org/abs/2310.15900. Bibliographic information from the supplied record.

• E1 Henry Robert Thackeray, Each friend of 10 has at least 10 nonidentical prime factors, arXiv:2310.15900v2 $2024$. https://arxiv.org/abs/2310.15900v2. Author metadata and abstract inspected 1 October 2026. This corrects the attribution attached to the same URL in [S2]; the imported citation is preserved as provenance.