gpt pro 5.4 attempt on Math-5547

Round 2 continuation for Erdős Problems #278, #279, #281

We write $d(S)$ for natural density when it exists, $\overline d(S)$ for upper density, and $\underline d(S)$ for lower density.

Problem 279

1) ROUND-2 OBJECTIVE

We pursue (C) obstruction/correction. Round 1 only gave heuristic density considerations. The strongest promising direction is to determine whether the unresolved issue is really one of density or one of finiteness. The new theorem below shows that density is not the obstruction at all.

2) Round-1 FOUNDATION USED

We use two Round-1 ideas.

1. The reformulation: an integer $m$ is covered once there is a modulus $n$ with \[ m\equiv a_n\pmod n \] and with $n$ small enough relative to $m$ so that the quotient $t=(m-a_n)/n$ automatically satisfies $t\ge k$.
2. For pairwise coprime moduli, finite-prefix avoidance densities factor exactly.

3) NEW INSIGHT / TOOL (ROUND-2)

The new insight is that for pairwise coprime moduli, the exceptional set has upper density $0$ for every choice of residues, not merely for some probabilistically chosen one.

In particular, for the primes and any fixed $k\ge 1$, the set of integers not representable as \[ a_p+t p \qquad (t\ge k) \] has upper density $0$ for every residue choice $(a_p \bmod p)_p$.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was this: the heuristic suggested an uncovered proportion about $1/\log x$, but there was no rigorous theorem.

We now prove a sharp partial result:

For any pairwise coprime sequence $(n_i)$ with $\sum_i 1/n_i=\infty$, and for any fixed $k\ge 1$, every choice of residues gives an exceptional set of upper density $0$.

This cleanly isolates the remaining difficulty in Erdős Problem #279: the open problem is not about achieving density $0$, but about upgrading density $0$ to finiteness.

5) WORK (ROUND-2)

Theorem.

Let \[ 1<n_1<n_2<\cdots \] be pairwise coprime positive integers with \[ \sum_{i=1}^\infty \frac{1}{n_i}=\infty. \] Fix $k\ge 1$, and choose arbitrary residues $a_i \pmod{n_i}$. Define \[ E_k(\mathbf a) := \left\{ m\in\mathbf N : \text{there do not exist } i\ge 1 \text{ and } t\ge k \text{ with } m=a_i+t n_i \right\}. \] Then \[ \overline d\big(E_k(\mathbf a)\big)=0. \]

Proof.

Fix $r\ge 1$, and let \[ N_r:=n_1n_2\cdots n_r. \] Define the finite-prefix avoidance set \[ B_r(\mathbf a) := \{\,m\in\mathbf Z : m\not\equiv a_i \pmod{n_i}\ \text{for all } 1\le i\le r\,\}. \] Because the moduli $n_1,\dots,n_r$ are pairwise coprime, the Chinese remainder theorem gives a bijection \[ \mathbf Z/N_r\mathbf Z \longrightarrow \mathbf Z/n_1\mathbf Z \times \cdots \times \mathbf Z/n_r\mathbf Z. \] Hence the proportion of residue classes modulo $N_r$ that avoid all $a_i \pmod{n_i}$ is exactly \[ \prod_{i=1}^r \left(1-\frac{1}{n_i}\right). \] Therefore \[ d\big(B_r(\mathbf a)\big)=\prod_{i=1}^r \left(1-\frac{1}{n_i}\right), \] and this value is independent of the chosen residues.

Now let $m\ge (k+1)n_r$. If $m\notin B_r(\mathbf a)$, then for some $1\le i\le r$ we have \[ m\equiv a_i\pmod{n_i}. \] Write \[ m=a_i+t n_i. \] Since $0\le a_i\le n_i-1$ and $n_i\le n_r$, we get \[ m-a_i\ge (k+1)n_r-(n_i-1)\ge (k+1)n_i-(n_i-1) > k n_i. \] Because $m-a_i$ is divisible by $n_i$, it follows that $t\ge k$.

So every integer $m\ge (k+1)n_r$ outside $B_r(\mathbf a)$ is already covered by one of the first $r$ progressions with quotient at least $k$. Equivalently, \[ E_k(\mathbf a)\cap[(k+1)n_r,\infty)\subseteq B_r(\mathbf a). \] Taking upper densities, \[ \overline d\big(E_k(\mathbf a)\big)\le d\big(B_r(\mathbf a)\big) = \prod_{i=1}^r \left(1-\frac{1}{n_i}\right). \] This holds for every $r$. Since $\sum_i 1/n_i=\infty$, the infinite product on the right tends to $0$. Therefore \[ \overline d\big(E_k(\mathbf a)\big)=0. \] This proves the theorem. $\square$

Corollary (primes).

Take $n_i=p_i$ to be the primes. Then for every fixed $k\ge 1$ and every choice of residues $a_p \pmod p$, \[ \overline d\!\left( \left\{ m : m\neq a_p+t p \text{ for all primes } p \text{ and all } t\ge k \right\} \right) =0. \]

Thus, in Problem #279, the unresolved issue is not density zero but finite exceptionality.

6) ADVERSARIAL VERIFICATION

1. Threshold check. The proof only uses the implication \[ m\ge (k+1)n_r \ \text{ and }\ m\equiv a_i\pmod{n_i} \Longrightarrow m=a_i+t n_i \text{ with } t\ge k. \] This is valid because $a_i\le n_i-1$.

2. Why only upper density? From \[ E_k(\mathbf a)\cap[(k+1)n_r,\infty)\subseteq B_r(\mathbf a) \] we get an upper-density bound. That is exactly what is needed to conclude $\overline d(E_k(\mathbf a))=0$.

3. Does this prove the original problem? No. A set can have upper density $0$ and still be infinite. So the argument does not settle finiteness.

4. Is pairwise coprimality essential? Yes. The exact factorisation \[ d(B_r(\mathbf a))=\prod_{i\le r}(1-1/n_i) \] comes from the Chinese remainder theorem and can fail without pairwise coprimality.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 proves a substantial weakening of the target statement:

For the primes (indeed, for any pairwise coprime sequence with divergent reciprocal sum), the exceptional set has upper density $0$ for every choice of residues and every fixed $k\ge 1$.

So Erdős Problem #279 is now sharply reduced to the problem of upgrading "upper density $0$" to "finite".

8) COMPLETION ESTIMATE

COMPLETION: 45%.

9) REFERENCES

1. P. Erdős and R. L. Graham, Old and New Problems and Results in Combinatorial Number Theory, Monographies de L'Enseignement Mathématique 28, Geneva, 1980.