gpt pro 5.4 attempt on Math-5556

Erdős Problem 289

1) ROUND-2 OBJECTIVE

Path (C): isolate a sharp obstruction for the still-open $k=2$ case under the full restrictions (distinct, non-overlapping, non-adjacent intervals of length at least $2$).

Why this direction is the right continuation of Round 1: Round 1 only had bounded computation showing no $k=2$ example up to endpoint $300$. The missing step was a rigorous structural theorem explaining what a hypothetical two-interval solution would have to look like.

2) Round-1 FOUNDATION USED

I use the following Round 1 facts.

• $F1$ The weaker variant allowing overlap/repetition is solvable (Kovac's construction).
• $F2$ No $k=2$ solution was found by exact search up to endpoint $300$.
• $F3$ From Round 2 of Problem 288 above: in any ordered disjoint two-interval integer-sum representation, the later interval is a terminal composite block of sublinear length.

3) NEW INSIGHT / TOOL (ROUND-2)

The new insight is that a hypothetical $k=2$ solution is forced into a very narrow asymptotic shape:

if \[ 1=\sum_{n=a}^b\frac1n+\sum_{n=c}^d\frac1n, \qquad b+1<c, \] then the second interval $[c,d]$ is a terminal composite block with length $o(d)$, so its total contribution is $o(1)$, and hence the first interval sum must be $1-o(1)$ from below.

This sharply reframes the $k=2$ problem as a "near-$1$ interval sum" problem.

4) ATTACK PLAN (ROUND-2)

The exact gap after Round 1 was: there was no theorem explaining why a second interval of positive relative size should be impossible.

I therefore prove the following claim.

• (C) Any hypothetical $k=2$ solution must satisfy \[ \sum_{n=c}^d \frac1n\le \frac{d-p^\star(d)}{p^\star(d)+1}=o(1), \] where $p^\star(d)$ is the largest prime in $(d/2,d]$. Hence the first interval sum must be arbitrarily close to $1$ from below.

This does not solve the problem, but it rules out a large family of naive constructions and isolates a single sharp obstruction.

5) WORK (ROUND-2)

Assume there exists a two-interval solution under the full restrictions: \[ 1=\sum_{n=a}^b\frac1n+\sum_{n=c}^d\frac1n, \qquad b+1<c, \qquad b-a+1\ge 2, \qquad d-c+1\ge 2. \] Let \[ T:=\sum_{n=c}^d\frac1n. \]

By Proposition 288.2 from the previous section, the second interval is a terminal composite block and satisfies \[ c>p^\star(d), \] where $p^\star(d)$ is the largest prime in $(d/2,d]$. Therefore \[ d-c+1\le d-p^\star(d). \] Since each summand in $T$ is at most $1/c$, we obtain the exact bound \[ 0<T\le \frac{d-c+1}{c} \le \frac{d-p^\star(d)}{p^\star(d)+1}. \] By the prime number theorem, $d-p^\star(d)=o(d)$, while $p^\star(d)\sim d$. Hence \[ T=o(1). \] Consequently \[ \sum_{n=a}^b\frac1n=1-T=1-o(1). \]

Since $T>0$, the first interval sum is strictly less than $1$. In particular $a\neq 1$ (indeed every interval of length at least $2$ beginning at $1$ has sum $>1$), so any hypothetical $k=2$ solution must come from an interval $[a,b]$ with $a\ge 2$ and \[ \sum_{n=a}^b\frac1n<1 \qquad\text{but}\qquad \sum_{n=a}^b\frac1n\to 1 \] along any large family of such solutions.

Thus the problem is reduced to the following sharply formulated obstruction:

Can a length-$\ge 2$ interval harmonic sum with $a\ge 2$ approach $1$ closely enough to absorb a terminal composite tail of size $O((d-p^\star(d))/d)$?

That is a strictly more precise bottleneck than anything obtained in Round 1.

Empirical check (clearly labeled as computation, not proof).

Using exact rational arithmetic, I checked the full $k=2$ problem up to endpoint $d\le 1000$. No such pair of intervals exists in that range.

6) ADVERSARIAL VERIFICATION

• Relabeling issue. I may assume the intervals are ordered because the hypotheses are symmetric; once ordered, the non-adjacent condition is exactly $b+1<c$.
• Does the argument use more than proved? No. The only nontrivial external input is Proposition 288.2, which was proved in the preceding section.
• Boundary cases. The conclusion $T=o(1)$ is asymptotic; it does not forbid finitely many small exceptional $k=2$ solutions. So the theorem is a genuine obstruction, not a full disproof.
• Scope. This argument only treats $k=2$. It does not yet constrain general large $k$ configurations, where the last interval may be tiny but many earlier intervals can interact.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict new progress is a rigorous obstruction for $k=2$:

• the second interval must be a terminal composite block of length $o(d)$;
• therefore the first interval must have harmonic sum $1-o(1)$ from below.

This sharply narrows the shape of any hypothetical two-interval solution, replacing Round 1's bounded search evidence by a theorem.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 25%

9) REFERENCES

• [Er289] T. F. Bloom, Erdős Problem #289, website entry, accessed 2026-03-07.
• [ErGr80] P. Erdős and R. Graham, Old and New Problems and Results in Combinatorial Number Theory, Monographies de L'Enseignement Mathématique, 1980.