GPT 6 Astra Ultra attempt on Math-5555

Problem 287: a stronger two-adic span obstruction

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. I read all of 287.tex. The target is to rule out an Egyptian partition of one with distinct denominators $2\le n_1<\cdots<n_k$ and every successive gap at most two. The supplied general valuation lemma yields $n_i\mid\operatorname{lcm}(1,\ldots,W)$, where $W=n_k-n_1$. I refine the factor-of-two restriction and record the exact residue condition when an odd prime power has only two representatives.

2) WORK

Let $t=\max_i v_2(n_i)$. If $t\ge1$, multiply the identity $\sum_i1/n_i=1$ by the least common multiple $L$. Modulo two, precisely the terms of valuation $t$ contribute one. Their number must therefore be even and at least two. Any two such denominators are $2^t$ times distinct odd integers, so they differ by at least $2^{t+1}$. Consequently \[ 2^{v_2(n_i)}\le W/2\qquad\text{for every even }n_i. \] Combining this with the supplied prime-power bound gives the stronger necessary divisibility \[ n_i\mid 2^{\max\{0,\lfloor\log_2W\rfloor-1\}} \prod_{\substack{p\le W\\p\text{ odd prime}}}p^{\lfloor\log_pW\rfloor} \qquad(W\ge2). \] The greatest allowed power of two has been reduced by a factor two.

For an odd prime $p$, let $t$ be the maximum positive valuation and suppose exactly two denominators have valuation $t$, say $p^tu$ and $p^tv$, where $p\nmid uv$. Reduction of the same cleared identity modulo $p$, followed by cancellation of the common unit $L/p^t$, gives \[ u^{-1}+v^{-1}\equiv0\pmod p, \quad\text{equivalently}\quad u+v\equiv0\pmod p. \] In particular, if $p^t>W/2$, at most two multiples of $p^t$ fit in the interval. Both must occur and they are consecutive multiples: $v=u+1$. Thus \[ 2u+1\equiv0\pmod p. \] This is a residue restriction beyond merely demanding a repeated maximum valuation.

3) ADVERSARIAL VERIFICATION

The two-adic argument concerns denominators at the maximal valuation; counting all multiples of $2^t$ would lose the required odd-coefficient spacing. These conditions hold for every finite distinct Egyptian partition, not only the dense gap-two configuration. For $1=1/2+1/3+1/6$, $W=4$ and the two maximal even denominators are exactly four apart, so the factor $W/2$ cannot be decreased in this general lemma. No inference from these necessary conditions to impossibility of dense configurations has been justified.

4) FINAL

UNRESOLVED. The span sieve is stricter and includes exact cancellation congruences. A global argument showing that all gap-two candidate sets fail is still missing.

5) SOURCES CHECKED

The complete supplied 287.tex, read 2026-09-23. All refinements above follow directly from its cleared-denominator identity and are proved here; no unverified search or literature novelty is claimed.

6) COMPLETION ESTIMATE

Subjective progress toward excluding every proposed dense partition. COMPLETION: 15%