gpt pro 5.4 attempt on Math-5366

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.