GPT 6 Astra Ultra attempt on Math-650

Erdős Problem #32: independent continuation

Date: 23 September 2026. Source versions read in full: 32.tex in gpt_pro_5.4.

1) OBJECTIVE AND SOURCE AUDIT

The target is a cofinite additive complement $A$ to the primes with $A(x)=o((\log x)^2)$, ideally $O(\log x)$. The supplied file proves a different, two-thin-summand conclusion using an external theorem. Its parity-shift step is valid, but replacing $A$ by $T+T$ restores the quadratic logarithmic bound. I do not count that as closing the original gap, and do not use its unverified random-model assertion as a theorem about primes.

2) WORK: A MULTIPLICITY OBSTRUCTION

Here is a precise condition that any near-minimal complement must evade. Let $A\subset\mathbb N_0$ be fixed and \[ r(n)=\#\{a\in A:n-a\text{ is prime}\},\qquad C(x)=\sum_{n\le x}\binom{r(n)}2. \] Suppose $r(n)\ge1$ for $n\ge n_0$, and suppose also that $r(n)\le K$ for these $n$, for a fixed integer $K\ge1$. Then \[ A(x)\pi(x)\ge x-O(1)+\frac{2C(x)}K-O(1). \] Indeed, interchange the order of summation to obtain $\sum_{n\le x}r(n)=\sum_{a\le x}\pi(x-a)\le A(x)\pi(x)$. For $1\le r\le K$, $\binom r2\le K(r-1)/2$. Summing this inequality over $n_0\le n\le x$ proves the assertion; the finitely many earlier values contribute constants. Consequently, if $C(x)\ge(\delta+o(1))x$ along a subsequence, then the prime number theorem gives \[ A(x)\ge\left(1+\frac{2\delta}{K}+o(1)\right)\log x \] along that subsequence. The unconditional elementary first-order bound is $A(x)\ge(1-o(1))\log x$, obtained without the multiplicity assumption.

The collision term has the exact form \[ C(x)=\sum_{\substack{a<b\\a,b\in A\cap[0,x]}} \#\{n\le x:n-a\text{ and }n-b\text{ are prime}\}. \] Thus an attempted construction close to the counting lower bound must control actual shifted-prime correlations. Cardinality estimates alone do not control this sum.

3) ADVERSARIAL VERIFICATION

The bounded multiplicity hypothesis is additional and has not been proved for an arbitrary thin complement. Nor have I proved a positive linear lower bound for $C(x)$. The displayed obstruction is therefore conditional on these two explicit assumptions, whereas the logarithmic lower bound is unconditional. It is not a disproof of an $O(\log x)$ complement.

7) FINAL AND REFERENCES

UNRESOLVED. This attempt provides a concrete overlap calculation for the original one-shift problem, but neither a sufficiently thin set nor a general obstruction to one. The supplied two-summand result remains a different problem. Literature check: Dai--Pan, Note on the additive complements of primes, https://arxiv.org/abs/1101.1653; the search record was checked, but no stronger theorem from it is used here. The direct official problem-page fetch was unavailable.

8) COMPLETION ESTIMATE

COMPLETION: 8% This is a subjective estimate of progress toward the entire stated problem, not a probability that the conjecture or an unverified proof is correct.