Erdős Problem #7

ROUND-4 OBJECTIVE

I continue along path (C): strengthen the Round-3 obstruction for the minimally corrected statement.

Round 2 already identified the fatal ambiguity in the naive formulation: without distinctness of the moduli, the question is trivial, since \[ \{x \equiv 0 \pmod 3,\ x \equiv 1 \pmod 3,\ x \equiv 2 \pmod 3\} \] is an odd covering system. Thus the genuine problem remains the standard one:

Does there exist a distinct covering system all of whose moduli are odd integers greater than $1$?

Round 3 proved that any such system must contain a modulus at least $127$. In this round I strengthen that finite obstruction substantially:

Main new theorem. Any distinct odd covering system must contain a modulus at least $1001$.

This is still far from a full solution, but it is a genuine theorem strictly stronger than the Round-3 lower bound.

Round-3 FOUNDATION USED

I use the following Round-3 results as fixed input.

1. Corrected formulation. The nontrivial problem is the distinct odd-modulus problem; without distinctness the answer is trivially "yes."
2. Round-3 prime-power assignment obstruction (Lemma 3.1). If a finite set $M$ of odd moduli is assigned to prime divisors $\pi(m)\mid m$ in such a way that, for every prime $p$, \[ \lambda_p:=\sum_{\pi(m)=p} p^{-v_p(m)} < 1, \] then for every choice of residue classes $x\equiv a_m \pmod m$ with $m\in M$, there exists an integer $x$ avoiding all of them. In particular, the chosen congruences do not cover $\mathbb Z$.
3. Round-3 lower bound. Any distinct odd covering system must contain a modulus at least $127$.

I do not re-prove these items here. The new work below uses item 2 and strictly strengthens item 3.

NEW INSIGHT / TOOL (ROUND-4)

The new idea is that the Round-3 prime-assignment lemma can be pushed much further by using the following assignment rule.

• Default rule: assign each odd modulus $m\le 999$ to its largest prime divisor.
• Then make only $23$ explicit exceptions, all involving moduli whose prime divisors are at most $19$.

This assignment was found by an exact mixed-integer feasibility search, but the proof below does not depend on the search. Once the assignment is written down, the verification is completely explicit: one checks that every local load $\lambda_p$ is strictly less than $1$, and then Round-3 Lemma 3.1 applies.

So the genuinely new content of this round is an explicit certificate showing that the Round-3 obstruction already rules out all odd moduli up to $999$.

ATTACK PLAN (ROUND-4)

After Round 3, the exact remaining gap was the following: the best unconditional finite obstruction only showed that a hypothetical distinct odd covering system must contain some modulus at least $127$.

To improve this, I will:

1. use Round-3 Lemma 3.1 as the main engine;
2. define a prime-divisor assignment $\pi(m)$ for every odd $m$ with $3\le m\le 999$;
3. verify that every local load $\lambda_p$ is $<1$;
4. conclude that no cover can use only odd moduli at most $999$, hence any distinct odd covering system must contain a modulus at least $1001$.

This overcomes the Round-3 limitation because the new assignment is much more efficient than the previous explicit partitions: almost every modulus is handled by its largest prime divisor, and only a small exceptional family has to be rerouted.

WORK (ROUND-4)

Theorem 4.1

Let \[ M_{999}:=\{m\in \mathbb N : 3\le m\le 999,\ m\ \text{odd}\}. \] Then for every choice of residue classes $a_m \pmod m$ with $m\in M_{999}$, there exists an integer $x$ such that \[ x \not\equiv a_m \pmod m \qquad (m\in M_{999}). \] Consequently, no distinct odd covering system can have all of its moduli at most $999$. In particular, any distinct odd covering system, if one exists, must contain a modulus at least $1001$.

Proof. By Round-3 Lemma 3.1, it is enough to define a map \[ \pi:M_{999}\to \{\text{odd primes}\} \] such that $\pi(m)\mid m$ for every $m\in M_{999}$ and \[ \lambda_p:=\sum_{\pi(m)=p} p^{-v_p(m)} < 1 \qquad\text{for every odd prime }p. \]

For an odd integer $m$, let $P(m)$ denote its largest prime divisor. Define $\pi(m)$ by the default rule \[ \pi(m)=P(m), \] except for the following $23$ moduli: \[ \begin{aligned} &45,63,135,189,297,351,405,567,675,891,945 \mapsto 3,\\ &165,175,275,325,525,825,875,975 \mapsto 5,\\ &539,637,833 \mapsto 7,\\ &429 \mapsto 11. \end{aligned} \] Each displayed reassignment is to a genuine prime divisor of the modulus, so $\pi(m)\mid m$ always holds.

We now verify that every local load is $<1$.

Large primes $p\ge 23$

If $p\ge 23$, then no exception is assigned to $p$, so $\pi(m)=p$ means exactly that $p=P(m)$, i.e. $m$ is an odd multiple of $p$ whose prime factors are all at most $p$. In particular, every such $m$ is an odd multiple of $p$ not exceeding $999$.

The odd multiples of $p$ up to $999$ are \[ p,\, 3p,\, 5p,\, \dots,\, (2k-1)p \] with $(2k-1)p\le 999$. Since $p\ge 23$, \[ 2k-1 \le \frac{999}{p} \le \frac{999}{23} < 44, \] hence $k\le 22$. Therefore there are at most $22$ odd multiples of $p$ in $M_{999}$.

Every modulus assigned to $p$ contributes $p^{-v_p(m)}\le 1/p$. Hence \[ \lambda_p \le \frac{22}{p} < 1 \qquad (p\ge 23). \]

Small primes $p\in\{3,5,7,11,13,17,19\}$

Under the above assignment rule, the corresponding fibers $S_p:=\pi^{-1}(p)$ are:

\[ \begin{aligned} S_3&=\{3,9,27,45,63,81,135,189,243,297,351,405,567,675,729,891,945\},\\ S_5&=\{5,15,25,75,125,165,175,225,275,325,375,525,625,825,875,975\},\\ S_7&=\{7,21,35,49,105,147,245,315,343,441,539,637,735,833\},\\ S_{11}&=\{11,33,55,77,99,121,231,363,385,429,495,605,693,847\},\\ S_{13}&=\{13,39,65,91,117,143,169,195,273,455,507,585,715,819,845\},\\ S_{17}&=\{17,51,85,119,153,187,221,255,289,357,425,459,561,595,663,765,867,935\},\\ S_{19}&=\{19,57,95,133,171,209,247,285,323,361,399,475,513,627,665,741,855,931,969\}. \end{aligned} \]

From these explicit sets we read off the $p$-adic exponent counts and hence the local loads:

\[ \lambda_3 = \frac{1}{3}+\frac{3}{3^2}+\frac{7}{3^3}+\frac{4}{3^4}+\frac{1}{3^5}+\frac{1}{3^6} = \frac{715}{729}<1, \] \[ \lambda_5 = \frac{3}{5}+\frac{9}{5^2}+\frac{3}{5^3}+\frac{1}{5^4} = \frac{616}{625}<1, \] \[ \lambda_7 = \frac{5}{7}+\frac{8}{7^2}+\frac{1}{7^3} = \frac{302}{343}<1, \] \[ \lambda_{11} = \frac{10}{11}+\frac{4}{11^2} = \frac{114}{121}<1, \] \[ \lambda_{13} = \frac{12}{13}+\frac{3}{13^2} = \frac{159}{169}<1, \] \[ \lambda_{17} = \frac{16}{17}+\frac{2}{17^2} = \frac{274}{289}<1, \] \[ \lambda_{19} = \frac{18}{19}+\frac{1}{19^2} = \frac{343}{361}<1. \]

Thus $\lambda_p<1$ for every odd prime $p$.

Round-3 Lemma 3.1 now applies, so for every choice of congruences \[ x\equiv a_m \pmod m \qquad (m\in M_{999}), \] there is an integer $x$ avoiding all of them. Therefore the full family $M_{999}$ cannot support a cover.

Finally, if some distinct odd covering system had all of its moduli at most $999$, then its moduli would form a subset $M\subseteq M_{999}$. Restricting $\pi$ from $M_{999}$ to $M$ only decreases each local load $\lambda_p$, so the same argument shows that $M$ also cannot support a cover. This contradiction proves that no distinct odd covering system can have all moduli at most $999$. Hence any such system must contain a modulus at least $1001$. $\square$

Remark. This strictly strengthens the Round-3 bound \[ \max m_i \ge 127 \] to the new unconditional bound \[ \max m_i \ge 1001. \]

ADVERSARIAL VERIFICATION

I check the new argument against the main failure modes.

1. Use of Round-3 results. The only imported result is Round-3 Lemma 3.1, used exactly as intended: once a prime-divisor assignment with all local loads $<1$ is produced, noncoverage follows.

2. Assignment validity. Every exceptional reassignment is to an actual prime divisor: \[ \begin{aligned} &45,135,405,675,975 \text{ are divisible by }5\text{ and }3;\\ &63,189,567,945 \text{ are divisible by }7\text{ and }3;\\ &297,891 \text{ are divisible by }11\text{ and }3;\\ &351 \text{ is divisible by }13\text{ and }3;\\ &165,825 \text{ are divisible by }11\text{ and }5;\\ &175,525,875 \text{ are divisible by }7\text{ and }5;\\ &275 \text{ is divisible by }11\text{ and }5;\\ &325,975 \text{ are divisible by }13\text{ and }5;\\ &539,637,833 \text{ are divisible by }7;\\ &429 \text{ is divisible by }11. \end{aligned} \]

3. Large-prime estimate. For $p\ge 23$, the proof uses only the crude bound "at most $22$ odd multiples of $p$ up to $999$", together with the trivial inequality $p^{-v_p(m)}\le 1/p$. This is independent of any hidden structure and gives a strict inequality $\lambda_p<1$.

4. Small-prime loads. The sets $S_3,S_5,S_7,S_{11},S_{13},S_{17},S_{19}$ are explicit; the exponent counts used in the displayed formulas are read directly from those lists. Every resulting rational number is strictly below $1$.

5. Subset monotonicity. The conclusion for arbitrary distinct odd modulus sets with $\max m_i\le 999$ is legitimate, because restricting the assignment $\pi$ from $M_{999}$ to a smaller set can only decrease the local loads. So if the full family cannot cover, then no subfamily can cover either.

6. Interaction with Round 3. There is no conflict: Round 3 gave the weaker lower bound $127\le \max m_i$, and Theorem 4.1 sharpens it to $1001\le \max m_i$.

No gap remains in the proof of Theorem 4.1.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

The corrected odd-covering problem remains open, but this round proves a substantially stronger finite obstruction than Round 3:

If a distinct odd covering system exists, then it must contain a modulus at least $1001$.

The new ingredient is an explicit prime-divisor assignment certificate on all odd moduli $3\le m\le 999$, verified using the Round-3 prime-power obstruction lemma.

COMPLETION ESTIMATE

COMPLETION: 58%

REFERENCES

1. T. F. Bloom, Erdős Problem #7, https://www.erdosproblems.com/7, accessed 2026-03-07.
2. C. Bispels, M. Cohen, J. Harrington, J. Lowrance, K. Pontes, L. Schaumann, and T. W. H. Wong, A further investigation on covering systems with odd moduli, Discrete Mathematics 349 $2026$, no. 7, 115013.

OBJECTIVE AND SOURCE REVIEW

Erdős #7; independent attempt, 2026-09-23. Does a finite covering of the integers by congruences with distinct odd moduli greater than one exist? I read the complete supplied 7.tex. Its explicit assignment excludes moduli at most $999$, but imports its crucial lemma from a missing earlier round. Here I supply that proof and strengthen its conclusion to a positive lower bound for the density of uncovered integers. This is a verification and quantitative consequence, not a claim of a new literature record.

WORK

Let $M$ be a finite collection of moduli, let $L=\operatorname{lcm}(M)$, and assign to each $m$ a prime $\pi(m)\mid m$. Write $L=\prod p^{e_p}$ and \[ \lambda_p=\sum_{\pi(m)=p}p^{-v_p(m)}. \] For each assigned congruence $x=a_m\pmod m$, forbid its projection $x=a_m\pmod {p^{v_p(m)}}$ in $\mathbb Z/p^{e_p}\mathbb Z$. This removes at most $p^{e_p}\lambda_p$ residues, by the union bound. Consequently at least $p^{e_p}(1-\lambda_p)$ local residues survive. Independently choose a survivor at every prime and apply the Chinese remainder theorem. Every resulting residue modulo $L$ avoids every original congruence, since it fails at its assigned prime. Thus the uncovered set has natural density at least \[ \prod_{p\mid L}(1-\lambda_p)>0 \quad\text{if every }\lambda_p<1. \] No independence between the original congruences is assumed; independence is only between the CRT coordinates.

For the supplied assignment on odd $3\le m\le999$, I recalculated the exact loads by trial division and rational arithmetic. They are $715/729,616/625,302/343,114/121,159/169,274/289,343/361$ at $p=3,5,7,11,13,17,19$, respectively. At $p\ge23$ the supplied argument gives $\lambda_p\le22/p<1$. Therefore a fully explicit lower density bound, uniformly over all residues, is \[ \frac{14}{729}\frac9{625}\frac{41}{343}\frac7{121} \frac{10}{169}\frac{15}{289}\frac{18}{361} \prod_{\substack{23\le p\le999\\p\text{ prime}}}(1-22/p)>0. \] The finite product is small but strictly positive. Restricting to a subset of moduli improves the load bounds.

ADVERSARIAL VERIFICATION AND REMAINING GAP

The projection condition is stronger than avoiding the original congruence, which is the direction needed. A positive density bound proves noncoverage for these finitely many moduli, but cannot be passed to an infinite product or to arbitrary maximum modulus. Repeated moduli invalidate the use of this particular certificate. No larger cutoff or solution of the unrestricted question has been established.

SOURCES CHECKED

Complete local source: gpt_pro_5.4/7.tex. The indexed official statement at https://www.erdosproblems.com/7 was checked on 2026-09-23. Its literature remarks are not needed for the self-contained CRT proof. The exact loads are reproduced by python3 verification/batch_3_final_checks.py.

FINAL

UNRESOLVED. The finite obstruction is verified and made quantitative; the unrestricted existence problem remains.

COMPLETION ESTIMATE

Conservative subjective progress toward the original problem, not a probability. COMPLETION: 10%

1) ROUND-2 OBJECTIVE

(C) Obstruction / correction. Round‑1’s “formal restatement” did not explicitly require the moduli to be distinct, while the subsequent investigation (and the classic Erdős–Selfridge “odd covering problem”) implicitly does. Without distinctness, the posed question has an immediate trivial “yes.” The minimal corrected statement consistent with Round‑1’s intent and the literature is:

Odd covering problem (corrected): Does there exist a covering system with distinct moduli, all odd integers > 1?

This is the standard “odd covering system” question and is still listed as open in current references. ([Erdős Problems][1])

I will (i) close the definitional gap and (ii) strictly advance the distinct‑moduli investigation by proving a new general obstruction lemma (union bound + CRT) that replaces Round‑1’s brute‑force Proposition 2 and extends it to a larger finite range.

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2) ROUND-1 FOUNDATION USED

I rely on the following Round‑1 results/ideas (without re‑proving them):

1. Lemma 1 (Round‑1): If ${x\equiv a_i\pmod{m_i}}$ covers $\mathbb Z$, then $\sum_i 1/m_i \ge 1$. 2. Reduction mod $L=\mathrm{lcm}(m_i)$: coverage of $\mathbb Z$ ⇔ coverage of all residues mod (L). 3. Proposition 2 (Round‑1): computationally, no cover using exactly one class for each modulus in ${3,5,7,9,11,13,15}$.

I will strengthen Proposition 2 by giving a short non‑computational proof, and then extend the method.

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3) NEW INSIGHT / TOOL (ROUND-2)

New Lemma (Coprime‑block obstruction via union bound + CRT). If the moduli split into blocks whose lcms are pairwise coprime, and in each block the reciprocal sum is $<1$, then no choice of residues can cover: there is always an integer simultaneously avoiding every congruence.

This is strictly new relative to Round‑1: it converts certain “candidate lists” into deterministic impossibility results without enumeration, and it scales to bigger modulus lists.

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4) ATTACK PLAN (ROUND-2)

Gaps after Round‑1:

* (Gap 0) The problem statement ambiguity: distinct moduli were not enforced, yet Round‑1 proceeded as if they were. * (Gap 1) Proposition 2 was proved only by exhaustive computation; no structure was extracted. * (Gap 2) No extension beyond the $ \le 15$ modulus universe.

Plan to overcome:

1. Close Gap 0: Note the trivial non‑distinct odd covering; then state the minimal corrected “distinct odd moduli” version (the actual open problem). ([Erdős Problems][1]) 2. Close Gap 1: Prove the coprime‑block lemma and use it to re‑prove Proposition 2 uniformly for all residues. 3. Close Gap 2: Apply the lemma to all odd moduli $\le 31$ to get a new unconditional lower bound: any distinct odd covering system must use some modulus $\ge 33$.

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5) WORK (ROUND-2)

5.1 The definitional correction

* If moduli need not be distinct, the original question “all moduli odd?” is trivially YES: \[ {x\equiv 0!!\pmod 3,;;x\equiv 1!!\pmod 3,;;x\equiv 2!!\pmod 3} \] covers $\mathbb Z$ because every integer has a residue $0,1,$ or $2$ modulo $3$.

* The nontrivial problem (and the one Round‑1 was actually addressing) is the distinct‑moduli odd covering problem, as standardly formulated. ([Erdős Problems][1])

From here on, “odd covering system” means: distinct odd moduli $>1$. ([Erdős Problems][1])

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5.2 New Lemma: Coprime‑block obstruction

Lemma 2 (Coprime‑block obstruction). Let (M) be a finite set of moduli $m\ge 2$, and for each $m\in M$ fix a residue $a_m\pmod m$. Suppose (M) can be partitioned into blocks \[ M = M_1 \sqcup \cdots \sqcup M_r \] such that the integers \[ L_j := \mathrm{lcm}(M_j) \] are pairwise coprime, and for every (j), \[ \sum_{m\in M_j} \frac{1}{m} < 1. \] Then there exists an integer (x) such that \[ x\not\equiv a_m \pmod m \quad \text{for every } m\in M. \] In particular, no selection of one congruence class for each $m\in M$ can cover $\mathbb Z$.

Proof. Fix (j). Work modulo $L_j$. For each $m\in M_j$, the congruence $x\equiv a_m\pmod m$ corresponds to a set of residues modulo $L_j$ of size exactly $L_j/m$ (because $m\mid L_j$). Let that residue-set be $R_{m}\subseteq \mathbb Z/L_j\mathbb Z$.

By the union bound, \[ \left|\bigcup_{m\in M_j} R_m\right| \le \sum_{m\in M_j} |R_m| = \sum_{m\in M_j} \frac{L_j}{m} = L_j\sum_{m\in M_j}\frac{1}{m} < L_j. \] Hence the union is a proper subset of $\mathbb Z/L_j\mathbb Z$, so we can choose a residue class $b_j \pmod{L_j}$ outside it. That means: \[ b_j \not\equiv a_m \pmod m \quad\text{for all } m\in M_j. \]

Now the $L_j$ are pairwise coprime, so by the Chinese Remainder Theorem there exists an integer (x) satisfying \[ x \equiv b_j \pmod{L_j}\quad (j=1,\dots,r). \] Fix any $m\in M_j$. Since $m\mid L_j$, the congruence $x\equiv b_j\pmod{L_j}$ implies $x\equiv b_j\pmod m$, and therefore $x\not\equiv a_m\pmod m$. This holds for every $m\in M$. ∎

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5.3 Strengthened replacement for Round‑1 Proposition 2 (no computation needed)

Proposition 2′ (uniform nonexistence for ${3,5,7,9,11,13,15}$). For every choice of residues $(a_3,a_5,a_7,a_9,a_{11},a_{13},a_{15})$, there exists an integer (x) avoiding all congruences \[ x\not\equiv a_m \pmod m \quad (m\in{3,5,7,9,11,13,15}). \] Hence no covering system can consist of exactly one class for each modulus in ${3,5,7,9,11,13,15}$.

Proof. Partition the moduli into two blocks: \[ M_1={3,5,9,15},\qquad M_2={7,11,13}. \] Their lcms are $L_1=\mathrm{lcm}(3,5,9,15)=45$ and $L_2=\mathrm{lcm}(7,11,13)=1001$, and $\gcd(45,1001)=1$.

Compute the reciprocal sums (equivalently, the union‑bound counts modulo the block lcms):

* Modulo $45$: sizes are $45/3=15,;45/5=9,;45/9=5,;45/15=3$, so \[ 15+9+5+3=32<45;;\Longrightarrow;;\sum_{m\in M_1}\frac1m=\frac{32}{45}<1. \] * Modulo $1001$: sizes are $1001/7=143,;1001/11=91,;1001/13=77$, so \[ 143+91+77=311<1001;;\Longrightarrow;;\sum_{m\in M_2}\frac1m=\frac{311}{1001}<1. \]

Thus Lemma 2 applies, producing an (x) avoiding all seven congruences. ∎

Strict strengthening over Round‑1: this removes the exhaustive search entirely and proves a stronger, residue‑uniform impossibility statement.

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5.4 New extension: no distinct odd covering system with all moduli $\le 31$

Let \[ M_{\le 31}={m\in\mathbb Z_{\ge2}: m\ \text{odd},\ m\le 31} ={3,5,7,9,11,13,15,17,19,21,23,25,27,29,31}. \]

Theorem 3 (new finite-range obstruction). For every choice of residues ${a_m\pmod m}*{m\in M*{\le 31}}$, there exists an integer (x) such that \[ x\not\equiv a_m \pmod m\quad\text{for all }m\in M_{\le 31}. \] In particular, no covering system can have all distinct odd moduli $\le 31$. Consequently, any distinct odd covering system (if it exists) must include some modulus $\ge 33$.

Proof.

Step 1: Partition into coprime blocks. Consider the block of all ${3,5,7}$-smooth odd moduli up to 31: \[ A:={3,5,7,9,15,21,25,27}. \] Let $L_A=\mathrm{lcm}(A)=3^3\cdot 5^2\cdot 7=27\cdot 25\cdot 7=4725$.

The remaining moduli in $M_{\le 31}$ are the primes \[ P:={11,13,17,19,23,29,31}. \] Each $p\in P$ is coprime to $4725$, and the primes are mutually coprime. So we have a partition into blocks \[ M_{\le 31}=A ;\sqcup; {11};\sqcup;{13};\sqcup;\cdots;\sqcup;{31}, \] whose lcms are $4725,11,13,17,19,23,29,31$, pairwise coprime.

Step 2: Check the “sum < 1” condition on each block.

* For each singleton prime block ({p}), clearly $\sum_{m\in{p}}1/m = 1/p<1$.

* For the block (A), compute sizes modulo $L_A=4725$: \[ \frac{4725}{3}=1575,; \frac{4725}{5}=945,; \frac{4725}{7}=675,; \frac{4725}{9}=525,; \frac{4725}{15}=315,; \frac{4725}{21}=225,; \frac{4725}{25}=189,; \frac{4725}{27}=175. \] Summing, \[ 1575+945+675+525+315+225+189+175 = 4624 < 4725. \] Therefore \[ \sum_{m\in A}\frac{1}{m}=\frac{4624}{4725}<1. \]

Step 3: Apply Lemma 2 and conclude. Lemma 2 applies to this coprime block decomposition, so for any choice of residues ${a_m}$ there is an integer (x) avoiding every congruence $x\equiv a_m\pmod m$ for all $m\in M_{\le 31}$. Hence no selection of residue classes on those moduli can cover $\mathbb Z$.

Finally, if a distinct odd covering system had all moduli $\le 31$, its modulus set would be a subset of $M_{\le 31}$; but any cover for a subset would extend to a cover for the full set $M_{\le 31}$ by adding arbitrary extra congruences (union only grows), contradicting the impossibility just proved. Therefore no distinct odd covering system can have maximum modulus $\le 31$, so any such system must contain some modulus $\ge 33$. ∎

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6) ADVERSARIAL VERIFICATION

* Quantifiers / “for all residues” check: Lemma 2 and its applications do not depend on any specific residue choices; the union bound uses only the sizes $L_j/m$, which are residue‑independent. Thus Proposition 2′ and Theorem 3 truly hold uniformly over all residue assignments.

* CRT hypotheses: In Proposition 2′, $\gcd(45,1001)=1$ (prime factors ${3,5}$ vs ${7,11,13}$). In Theorem 3, $4725=3^3\cdot 5^2\cdot 7$ is coprime to each of $11,13,17,19,23,29,31$, and those primes are mutually coprime; CRT applies.

* Potential loophole: “but a covering might omit some small moduli”: Addressed: if a subset of moduli admits a covering, then adding more congruences on additional moduli cannot destroy coverage (union is monotone). So disproving coverage for the full set $M_{\le 31}$ automatically disproves coverage for every subset.

* Boundary case where $\sum 1/m =1$: Not used; every block we applied Lemma 2 to satisfies a strict inequality, so the union bound yields a strict deficit in residue count, guaranteeing a complement residue exists.

All checks pass; no hidden assumption beyond distinctness of moduli in the “odd covering problem” interpretation.

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7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

* The original question without distinctness is trivial “YES” (e.g. the three classes mod $3$). * Under the minimal corrected and standard interpretation (distinct odd moduli $>1$), the existence problem remains open. ([Erdős Problems][1]) * Strict advances beyond Round‑1:

1. A new general obstruction lemma (Lemma 2) using union bound + CRT. 2. A noncomputational, residue‑uniform proof strengthening Round‑1 Proposition 2. 3. A new finite-range obstruction: no distinct odd covering system can have all moduli $\le 31$ (hence any candidate must include some modulus $\ge 33$).

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8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 35%

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9) REFERENCES

* Erdős Problems page for Problem #7 (odd covering system) — definition/context and current open status. ([Erdős Problems][1])

[1]: https://www.erdosproblems.com/prizes/25/open "https://www.erdosproblems.com/prizes/25/open"