Erdős Problem #277
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Erdős problems openaigpt 6 astra ultra 100% cataloguedSolvedSolvedStatement
For every real $c$, does some positive integer $n$ satisfy $\sigma(n)>cn$, while no covering system has pairwise distinct moduli drawn from the divisors $d>1$ of $n$? Yes. Haight originally proved this. We give the later short deduction from Hough's minimum-modulus theorem, as suggested in the current discussion. The proof below concerns the displayed original question; it does not assert the numerical leading constant in the editorial's additional extremal asymptotic. No novelty is claimed.
Restatement from Math-5546-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
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GPT 6 Astra Ultra attempt on Math-5546
Erdős Problem #277
Statement and attribution
For every real $c$, does some positive integer $n$ satisfy $\sigma(n)>cn$, while no covering system has pairwise distinct moduli drawn from the divisors $d>1$ of $n$? Yes. Haight originally proved this. We give the later short deduction from Hough's minimum-modulus theorem, as suggested in the current discussion. The proof below concerns the displayed original question; it does not assert the numerical leading constant in the editorial's additional extremal asymptotic. No novelty is claimed.External input
Hough proved that an absolute constant $C$ bounds the smallest modulus in every finite covering system with distinct moduli greater than one. This is the substantial external input. We require only the existence of $C$, not its numerical value.Work: remove all small prime factors
For $y>C$, set \[ n_y=\prod_{C<p\le y}p. \] Every divisor $d>1$ of $n_y$ has a prime divisor exceeding $C$, hence $d>C$. A covering system with distinct moduli among these divisors would have smallest modulus greater than $C$, contradicting Hough's theorem.Since $n_y$ is squarefree, multiplicativity gives \[ \frac{\sigma(n_y)}{n_y}=\prod_{C<p\le y}\left(1+\frac1p\right). \] This product is unbounded. Here is the needed elementary argument. If $\sum_p1/p$ converged, then the products $\prod_{p\le y}(1-1/p)^{-1}$ would be bounded, since $-\log(1-1/p)\le1/p+2/p^2$. But expansion of the finite Euler product includes all integers at most $y$, so it is at least $\sum_{m\le y}1/m$, which diverges. Thus $\sum_p1/p=\infty$. Removing the finitely many primes $p\le C$ preserves divergence, and \[ \log(1+1/p)\ge\frac1{2p} \] shows that the product for $\sigma(n_y)/n_y$ tends to infinity. Choose $y$ so that it exceeds $c$. Then $n=n_y$ has both required properties.
Verification and outcome
The obstruction applies to any subset of the allowable divisors, regardless of the chosen residues. Large $\sigma(n)/n$ is only a sum-of-reciprocals condition; it does not by itself supply a cover. Distinctness of moduli is essential to the cited Hough theorem and is present in the question.SOLVED, with Hough's theorem as an explicit external input. The construction and its unbounded divisor-sum ratio are proved here.