gpt pro 5.4 attempt on Math-650

Erdős Problem 32 --- Round 3 continuation and gap-closure

1. ROUND-3 OBJECTIVE

I pursue path (C): obstruction/correction. Round 2 established a real obstruction: any proof of an exact one-shift theorem \[ \mathcal P + A \supseteq [N_0,\infty) \quad\text{with}\quad A(x)=o((\log x)^2)\ \text{or}\ A(x)=O(\log x) \] that uses only coarse size information of the primes is impossible, because prime-like random sets of density $1/\log x$ already force a $(\log x)^2$ barrier.

The most promising strict advance is therefore to identify the nearest exact $O(\log x)$ theorem that is actually true after a minimal correction of the representation problem. The new goal in this round is to prove the following corrected statement:

Corrected exact theorem. There exists a set $T\subseteq \mathbb N$ with \[ T(x)=O(\log x) \] such that every sufficiently large integer $n$ can be written in the form \[ n=p+t_1+t_2 \qquad (p\in\mathcal P,\ t_1,t_2\in T). \]

This is an exact, cofinite-coverage theorem at the $O(\log x)$ scale, one extra thin summand away from the original open problem.

2. Round-2 FOUNDATION USED

I use the following Round 2 result as fixed input.

• Round-2 obstruction. There exist prime-like sets $P'\subseteq \mathbb N$ with \[ P'(x)\sim \frac{x}{\log x} \] for which every additive complement $B$ satisfies \[ \liminf_{x\to\infty}\frac{B(x)}{(\log x)^2}\ge 1. \] Hence any successful proof for the actual primes with $A(x)=o((\log x)^2)$ or $A(x)=O(\log x)$ must use arithmetic structure specific to the primes, not just counting-function information.

I also use the trivial monotonicity fact that if $S(x)=O(\log x)$ then $(S+1)(x)=O(\log x)$, and therefore \[ (S\cup(S+1))(x)=O(\log x). \]

3. NEW INSIGHT / TOOL (ROUND-3)

The new ingredient is a prime-specific theorem of Dai and Pan:

External Theorem (Dai--Pan). There exists a set $A\subseteq \mathcal P$ with \[ A(x)=O(\log x) \] such that every sufficiently large odd integer $n$ can be represented as \[ n=a_1+a_2+p \qquad (a_1,a_2\in A,\ p\in\mathcal P). \]

Round 3 adds a parity-shift observation to convert this odd-integer theorem into a cofinite theorem for all sufficiently large integers, still with only $O(\log x)$ many thin summands available.

4. ATTACK PLAN (ROUND-3)

After Round 2, the exact gap was this:

there was a rigorous barrier against density-only proofs of the original one-shift problem, but no positive theorem showing how close one can nevertheless get to an exact $O(\log x)$ statement.

The precise claim to prove now is:

Claim. There exists $T\subseteq\mathbb N$ with $T(x)=O(\log x)$ and an integer $N_1$ such that every $n\ge N_1$ satisfies \[ n\in \mathcal P+T+T. \]

The plan is to start from the Dai--Pan set $A\subseteq\mathcal P$, define \[ T:=A\cup (A+1), \] and then check separately the odd and even cases. Odd integers are covered directly by the external theorem; even integers are reduced to odd integers by subtracting $1$ and absorbing the shift into one of the two thin summands.

5. WORK (ROUND-3)

Step 1: Invoke the prime-specific theorem.

By the Dai--Pan theorem, there is a set $A\subseteq \mathcal P$ and an integer $N_0$ such that \[ A(x)=O(\log x) \] and every odd integer $m\ge N_0$ can be written as \[ m=a_1+a_2+p \qquad (a_1,a_2\in A,\ p\in\mathcal P). \] No extra hypotheses are needed; this is exactly the stated theorem.

Step 2: Define the corrected thin set.

Set \[ T:=A\cup (A+1). \] Then \[ T(x)\le A(x)+A(x-1)=O(\log x), \] so $T$ still has logarithmic counting function.

Step 3: Cover all sufficiently large odd integers.

Let $n\ge N_0$ be odd. By the theorem, \[ n=a_1+a_2+p \qquad (a_1,a_2\in A,\ p\in\mathcal P). \] Since $A\subseteq T$, this is already a representation \[ n=p+t_1+t_2 \qquad (t_1,t_2\in T). \] Thus every sufficiently large odd integer lies in $\mathcal P+T+T$.

Step 4: Cover all sufficiently large even integers.

Let $n\ge N_0+1$ be even. Then $n-1\ge N_0$ is odd, so again by the theorem, \[ n-1=a_1+a_2+p \qquad (a_1,a_2\in A,\ p\in\mathcal P). \] Hence \[ n=(a_1+1)+a_2+p. \] Now $a_1+1\in A+1\subseteq T$ and $a_2\in A\subseteq T$, so \[ n=p+t_1+t_2 \qquad (t_1,t_2\in T). \] Therefore every sufficiently large even integer also lies in $\mathcal P+T+T$.

Conclusion.

If we choose $N_1:=N_0+1$, then every integer $n\ge N_1$ belongs to $\mathcal P+T+T$, with \[ T(x)=O(\log x). \] This proves the corrected exact theorem.

Corollary (exact $O((\log x)^2)$ complement recovered from an $O(\log x)$ set). Let \[ C:=T+T. \] Then every sufficiently large integer belongs to $\mathcal P+C$, and \[ C(x)\le T(x)^2=O((\log x)^2) \] in the crude counting-function sense. So the classical Erdős-scale exact complement bound is recovered from a logarithmic set $T$ by one extra additive layer.

6. ADVERSARIAL VERIFICATION

Parity check.

Odd $n$ are covered directly. Even $n$ are handled by writing $n-1$ as an odd integer covered by the external theorem and shifting one thin summand by $+1$. No parity case is omitted.

Growth check.

Passing from $A$ to $A+1$ does not change the order of magnitude of the counting function, and taking the union preserves the estimate $O(\log x)$. Thus the corrected theorem genuinely remains at logarithmic scale.

Quantifier check.

The external theorem gives a single set $A$ and a single threshold $N_0$ working for all sufficiently large odd integers. From this we constructed a single set $T$ and a single threshold $N_1=N_0+1$ working for all sufficiently large integers. There is no hidden dependence on $n$.

Compatibility with Round 2.

There is no contradiction with the Round 2 barrier. Round 2 ruled out density-only approaches to the one-shift problem $\mathcal P+A$ with $A(x)=o((\log x)^2)$ or $O(\log x)$. The present result uses genuine prime-specific input and proves a different statement, namely coverage by \[ \mathcal P+T+T, \] not by $\mathcal P+T$.

Minimality claim.

I do not claim to have proved that adding one extra thin summand is logically minimal among all possible corrections. What has been proved is that this is the closest exact $O(\log x)$ correction supplied by the checked literature used here.

7. FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

Round 3 adds a rigorous positive theorem beyond the Round 2 obstruction:

There exists a set $T\subseteq\mathbb N$ with $T(x)=O(\log x)$ such that every sufficiently large integer is representable as \[ n=p+t_1+t_2 \qquad (p\in\mathcal P,\ t_1,t_2\in T). \]

Thus the nearest checked exact $O(\log x)$ variant currently available is true after adding one further thin summand. Combined with Round 2, this sharply isolates the frontier:

• exact one-shift coverage $\mathcal P+A$ at size $O(\log x)$ remains open,
• density-only methods cannot settle it,
• but exact two-thin-shift coverage $\mathcal P+T+T$ at size $T(x)=O(\log x)$ is already provable.

8. COMPLETION ESTIMATE

COMPLETION: 58%

9. REFERENCES

• {[DP14 }] Li-Xia Dai and Hao Pan, The additive complements of primes and Goldbach's problem, Acta Arithmetica 162 $2014$, 209--221.
• {[Ko96 }] Mihail N. Kolountzakis, On the additive complements of the primes and sets of similar growth, Acta Arithmetica 77 $1996$, 1--8.
• {[EP32 }] Erdős Problem #32, erdosproblems.com.