GPT 6 Astra Ultra attempt on Math-746
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Erdős Problem #288: independent continuation
Date: 23 September 2026. Source versions read in full:288.tex in gpt_pro_5.4.1) OBJECTIVE AND SOURCE AUDIT
The question is whether only finitely many pairs of finite intervals have an integral sum of reciprocals, counting overlaps twice. The supplied singleton reduction is valid, and its prime-tail obstruction is explicitly restricted to disjoint intervals. I add a prime-power condition applicable also when the intervals overlap, keeping the occurrence multiplicities.2) WORK: MAXIMAL VALUATIONS MUST CANCEL
For any finite multiset $D$ of denominators, let $e=\max_{d\in D}v_p(d)$. If $e>0$, integrality of $\sum_{d\in D}1/d$ requires \[ \sum_{\substack{d\in D\\v_p(d)=e}}(d/p^e)^{-1}\equiv0\pmod p. \tag{1} \] To prove this, take $L=\operatorname{lcm}(D)$ and write $L=p^eL_0$. Modulo $p$, the integer numerator $\sum L/d$ receives contributions only from denominators of valuation $e$. Divide the resulting congruence by the nonzero residue $L_0$. In particular a unique maximal valuation is incompatible with integrality, regardless of its exponent.In a nonempty interval $I$, the largest $2$-adic valuation is attained exactly once. Indeed, if two numbers $2^eu<2^ev$ with odd $u,v$ both attained the maximum, the intervening even multiple of $2^e$ would belong to $I$ and have larger valuation. If intervals $I_1,I_2$ have maximal valuations $e_1\ne e_2$, the larger maximum is positive and unique in the combined multiset, contradicting $1$. Therefore every solution satisfies \[ \max_{n\in I_1}v_2(n)=\max_{n\in I_2}v_2(n). \tag{2} \] This remains valid for overlaps: a shared maximum occurs twice, whereas unequal maxima cannot be shared. Equality of maxima is necessary, not sufficient, since further prime-power cancellations remain.
For a singleton $I_2=\{m\}$ and $\sum_{n=a}^b1/n=u/v$ reduced, the source correctly forces $m=v$ and $u\equiv-1\pmod v$. Equation $2$ then gives the exact extra relation $v_2(v)=\max_{a\le n\le b}v_2(n)$. This also follows directly from the unique maximal valuation in one interval. The source's sentence $k=\lceil u/v\rceil$ has a minor exception when $v=1$: then the added unit fraction must be one, so $k=u+1$.