GPT 6 Astra Ultra attempt on Math-5369

Problem 11: reusing two square divisors

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. The target is the representation of every sufficiently large odd integer as a positive squarefree integer plus a power of two. I read the complete supplied gpt_pro_5.4/11.tex. Its Chinese remainder construction excludes small exponents using a different prime for each exponent; that is an obstruction to a restricted search, not a counterexample to the representation conjecture. Here two primes are reused periodically. No claim of literature novelty is made.

2) WORK

Write $e_{\min}(n)=\min\{e\ge0:n-2^e\ge1\text{ is squarefree}\}$, with value $+\infty$ if the set is empty. For each integer $K$, let \[ R_K=\{0\le e\le K:6\nmid e,\ 20\nmid e\},\qquad r=|R_K|. \] In every block of 60 exponents, 10 are divisible by 6, 3 by 20, and 1 by both. Thus $r=4K/5+O(1)$. List $R_K$ as $e_1,\ldots,e_r$ and let $q_1,\ldots,q_r$ be the first $r$ primes exceeding 5. Impose \[ n\equiv1\pmod2,\quad n\equiv1\pmod9,\quad n\equiv1\pmod{25},\quad n\equiv2^{e_j}\pmod{q_j^2}\quad(1\le j\le r). \] The Chinese remainder theorem gives one class modulo $M_K=450\prod_{j\le r}q_j^2$. Since $2^6\equiv1\pmod9$ and $2^{20}\equiv1\pmod{25}$, all exponents $0\le e\le K$ are killed by a specified prime square. Every member of the class exceeding $2^K$ therefore has $e_{\min}(n)>K$.

Using the standard prime bound $q_j=O(j\log(j+2))$ gives \[ \log M_K\le2r\log r+O(r\log\log(r+2)) =\frac85K\log K+O(K\log\log(K+2)). \] For $K=\lfloor c\log X/\log\log X\rfloor$, $0<c<5/8$, the number of admissible progression members in $(2^K,X]$ is at least \[ \frac{X-2^K}{M_K}-1 \ge X^{1-8c/5-o(1)}. \] Here $2^K=X^{o(1)}$ and $M_K\le X^{8c/5+o(1)}$. Consequently at least this many odd $n\le X$ satisfy $e_{\min}(n)>c\log X/\log\log X$. This improves the supplied method's constant $1/2$ to $5/8$ by a concrete modification of its congruences.

3) ADVERSARIAL VERIFICATION

The condition $n>2^K$ is essential: divisibility of a nonpositive difference is not a failed positive representation. The congruences use distinct prime moduli, so sharing the same residue 1 modulo 9 and 25 causes no compatibility problem. Inclusion--exclusion counts multiples of 60 once. The assertion for the maximum of $e_{\min}$ alone would be vacuous if small exceptional integers already have value $+\infty$; the growing counting statement above avoids relying on that issue.

4) FINAL

UNRESOLVED. This forces a longer initial interval of bad exponents, but leaves roughly $\log_2 n$ admissible exponents in the original problem. It proves neither an eventual squarefree representation nor failure of every admissible exponent for any sufficiently large $n$.

5) SOURCES CHECKED

The supplied file was read in full. Rosser--Schoenfeld, Approximate formulas for some functions of prime numbers $1962$, publisher record https://doi.org/10.1215/ijm/1255631807, checked 2026-09-23; the prime-growth estimate is the only external input. The official problem-page request returned HTTP 403; no claim of a new live status check is made.

6) COMPLETION ESTIMATE

Subjective progress toward the original question, not a probability of correctness. COMPLETION: 15%