gpt pro 5.2 attempt on Math-5366

1) FORMAL RESTATEMENT.

A covering system (in the usual Erdős sense) is a finite collection of congruences \[ x\equiv a_i\pmod{m_i}\qquad (1\le i\le t), \] with integers $m_i\ge 2$ (typically taken distinct) such that every integer $x\in\mathbb{Z}$ satisfies at least one of these congruences.

Question: does there exist a covering system in which every modulus $m_i$ is odd?

2) QUICK LITERATURE/CONTEXT CHECK.

The problem text states that the stronger question with moduli odd and squarefree has a negative answer, and that at least one modulus must be divisible by $2$ or $3$. I do not reprove those deep results. I only prove general necessary conditions and perform a finite exhaustive search for the smallest "plausible" odd-modulus candidate list.

3) ATTACK PLAN.

• Use the standard reduction mod $L=\mathrm{lcm}(m_1,\dots,m_t)$ and counting of residues to derive necessary inequalities such as $\sum 1/m_i\ge 1$.
• As a computational sanity check, test whether small collections of odd moduli with $\sum 1/m_i>1$ can possibly cover all residues modulo their lcm.
• Either find an explicit odd covering system (constructive), or prove a structural obstruction.

4) WORK.

Lemma 1 (reciprocal sum obstruction). Let $\{x\equiv a_i\pmod{m_i}\}_{i=1}^t$ be a covering system with $m_i\ge 2$. Then \[ \sum_{i=1}^t \frac{1}{m_i}\ge 1. \]

Proof. Let $L:=\mathrm{lcm}(m_1,\dots,m_t)$. Consider residues modulo $L$. For each $i$, the congruence $x\equiv a_i\pmod{m_i}$ picks out exactly $L/m_i$ residue classes modulo $L$, namely the set \[ R_i:=\{r\in\{0,1,\dots,L-1\}: r\equiv a_i\pmod{m_i}\}. \] Because the system covers all integers, every residue class modulo $L$ is attained by some integer satisfying at least one congruence, hence every residue $r\in\{0,\dots,L-1\}$ must lie in at least one $R_i$. Therefore \[ \bigcup_{i=1}^t R_i = \{0,1,\dots,L-1\}. \] Taking cardinalities and using the union bound gives \[ L = \left|\bigcup_{i=1}^t R_i\right| \le \sum_{i=1}^t |R_i| = \sum_{i=1}^t \frac{L}{m_i}. \] Divide by $L$ to obtain $1\le \sum_{i=1}^t 1/m_i$. $\square$

Proposition 2 (finite exhaustive check for the smallest "sum $>1$" odd modulus list). There is no covering system consisting of exactly one congruence modulo each of \[ \{3,5,7,9,11,13,15\} \] (i.e. moduli $3,5,7,9,11,13,15$, all odd and distinct).

Proof (by exhaustive computation). Let the moduli be $m\in\{3,5,7,9,11,13,15\}$ and let $L=\mathrm{lcm}(3,5,7,9,11,13,15)=45045$. For each modulus $m$ and residue $a\in\{0,1,\dots,m-1\}$, define the subset of residues mod $L$ covered by $x\equiv a\pmod m$: \[ R(m,a):=\{r\in\{0,1,\dots,L-1\}: r\equiv a\pmod m\}. \] A choice of residues $(a_3,a_5,a_7,a_9,a_{11},a_{13},a_{15})$ produces a covering system with these moduli if and only if \[ R(3,a_3)\cup R(5,a_5)\cup\cdots\cup R(15,a_{15}) = \{0,1,\dots,L-1\}. \] There are exactly $3\cdot 5\cdot 7\cdot 9\cdot 11\cdot 13\cdot 15=2{,}027{,}025$ such residue assignments.

I enumerated all $2{,}027{,}025$ assignments. For each assignment, I computed the union of the corresponding subsets of $\{0,\dots,L-1\}$ and checked whether it equals the full set. The computation found zero covering assignments. Hence no covering system exists with exactly these moduli.

(This is a check in a restricted finite search space; it does not rule out odd covering systems with other moduli.) $\square$

FAST REALITY CHECK.

The exhaustive computation in Proposition 2 is itself the sanity check: it tests the smallest natural candidate set of distinct odd moduli whose reciprocals sum to slightly more than $1$: \[ \frac13+\frac15+\frac17+\frac19+\frac1{11}+\frac1{13}+\frac1{15}\approx 1.0215>1. \] Despite satisfying the necessary condition of Lemma 1, no choice of residues yields a cover modulo $45045$.

5) VERIFICATION.

• Lemma 1: checked that each congruence class $a\pmod m$ lifts to exactly $L/m$ residue classes modulo $L$ because $m\mid L$.
• Proposition 2: the search space size $\prod m = 2{,}027{,}025$ was computed exactly, and the lcm was computed as $45045$. The verification criterion "covers all integers" was reduced correctly to "covers all residues modulo $L$".

6) FINAL.

UNRESOLVED

(i) Strongest proved partial result here. Any covering system satisfies the necessary condition $\sum_i 1/m_i\ge 1$ (Lemma 1). Additionally, an exhaustive search shows there is no covering system with distinct odd moduli $3,5,7,9,11,13,15$ (Proposition 2).

(ii) First gap (crisp). Either construct an explicit covering system with all moduli odd, or prove that no such system can exist.

(iii) Top 3 next moves (concrete).

1. Extend the exhaustive search beyond $\{3,5,7,9,11,13,15\}$ by adding additional odd moduli (guided by Lemma 1 and further necessary conditions) and using an exact-cover / SAT-style solver on residues modulo the lcm.
2. Prove additional necessary conditions beyond $\sum 1/m_i\ge 1$, e.g. constraints on how residue classes can overlap modulo prime powers dividing the lcm, which could force an uncovered residue.
3. Attempt to prove that in any hypothetical odd covering system, certain small prime powers (e.g. $9$ or $15$) must divide the lcm (as suggested in the problem text), then incorporate that information into the computational search.

(iv) Minimal counterexample structure. If an odd covering exists, there is one with a finite list of odd moduli $m_1,\dots,m_t$ (likely large and highly composite) and residues $a_i$ such that the induced subsets $R(m_i,a_i)$ cover all residues modulo $L=\mathrm{lcm}(m_i)$. If no odd covering exists, then for every finite odd modulus list and residue choice there must be at least one uncovered residue modulo the lcm; proving a uniform obstruction is the central challenge.