gpt pro 5.4 attempt on Math-746

Erdős Problem 288

1) ROUND-2 OBJECTIVE

Path (C): isolate a sharp obstruction and give an exact reformulation of the singleton subcase.

Why this direction is the right continuation of Round 1: Round 1 already had a clean one-interval non-integrality lemma and a general "uniquely occurring prime" obstruction, but it did not rigidly reformulate the still-open singleton case $|I_2|=1$, nor did it prove any structural theorem for ordered disjoint solutions. Those are the two gaps I close here.

2) Round-1 FOUNDATION USED

I use the following Round 1 results without reproving them.

• $F1$ If $[a,b]$ is a nontrivial interval, then $\sum_{n=a}^b \frac1n \notin \mathbb Z$.
• $F2$ If a prime $p$ divides the common denominator to exponent exactly $1$, and among the denominators in the relevant union of intervals there is exactly one multiple of $p$, then the total reciprocal sum cannot be an integer.
• $F3$ The exact small solutions found in Round 1 up to endpoint $300$.

3) NEW INSIGHT / TOOL (ROUND-2)

There are two genuinely new ingredients.

• $N1$ Singleton forcing. In the singleton variant \[ H(a,b)+\frac1m\in\mathbb Z,\qquad H(a,b):=\sum_{n=a}^b \frac1n, \] there is at most one possible $m$ for a fixed interval $[a,b]$: namely, $m$ must equal the reduced denominator of $H(a,b)$.
• $N2$ Terminal composite tail obstruction. In any ordered disjoint two-interval solution \[ \sum_{n=a}^b\frac1n+\sum_{n=c}^d\frac1n\in\mathbb Z,\qquad b<c, \] the later interval $[c,d]$ must lie entirely inside the terminal prime-free tail below $d$. In particular its length is $o(d)$.

4) ATTACK PLAN (ROUND-2)

The exact unresolved gap after Round 1 was this:

there was no rigid reduction for the singleton case, and no theorem showing that the interval containing the maximal denominator must be extremely short in any ordered disjoint solution.

I therefore prove two claims.

• $C1$ If $H(a,b)=u/v$ in lowest terms and $H(a,b)+1/m\in\mathbb Z$, then necessarily $m=v$.
• $C2$ If $b<c$ and $\sum_{n=a}^b 1/n+\sum_{n=c}^d 1/n\in\mathbb Z$, then $[c,d]$ contains no prime in $(d/2,d]$, hence in fact no prime at all, and so $d-c+1=o(d)$.

These two claims are enough to give strict progress beyond Round 1.

5) WORK (ROUND-2)

Proposition 288.1 (singleton forcing).

Let \[ H(a,b)=\sum_{n=a}^b\frac1n=\frac uv \] with $\gcd(u,v)=1$ and $v>0$. Then \[ H(a,b)+\frac1m\in\mathbb Z \] for some positive integer $m$ if and only if \[ u+1\equiv 0\pmod v, \] and in that case necessarily $m=v$.

Proof.

Assume \[ \frac uv+\frac1m=k\in\mathbb Z. \] Then \[ \frac1m=k-\frac uv=\frac{kv-u}{v}. \] Now \[ \gcd(kv-u,v)=\gcd(u,v)=1, \] so the fraction $(kv-u)/v$ is already in lowest terms. But $1/m$ in lowest terms has numerator $1$. Therefore \[ kv-u=1, \] and hence necessarily \[ m=v. \] Conversely, if $u+1\equiv 0\pmod v$, say $u+1=kv$, then \[ \frac uv+\frac1v=\frac{u+1}{v}=k\in\mathbb Z, \] so $m=v$ works. $\square$

So the singleton subcase is equivalent to the exact congruence condition \[ H(a,b)=\frac uv\text{ in lowest terms and }u\equiv -1\pmod v. \] For each interval $[a,b]$ there is at most one candidate singleton.

Proposition 288.2 (terminal composite tail obstruction).

Assume \[ 1\le a\le b<c\le d \] and \[ \sum_{n=a}^b\frac1n+\sum_{n=c}^d\frac1n\in\mathbb Z. \] Let $p^\star(d)$ denote the largest prime in $(d/2,d]$. Then \[ c>p^\star(d). \] Consequently $[c,d]$ contains no primes at all, and \[ d-c+1\le d-p^\star(d)=o(d). \]

Proof.

By Bertrand's postulate there exists at least one prime in $(d/2,d]$, so $p^\star(d)$ is well-defined for $d\ge 2$.

Suppose for contradiction that $c\le p^\star(d)\le d$. Then $p^\star(d)\in[c,d]$. Since $p^\star(d)>d/2$, there is no other multiple of $p^\star(d)$ in $[1,d]$; indeed $2p^\star(d)>d$. Because the first interval lies entirely below $c$, the denominator $p^\star(d)$ occurs exactly once in the union of denominators from $[a,b]\cup[c,d]$. By Round 1 foundation $F2$, this forces \[ \sum_{n=a}^b\frac1n+\sum_{n=c}^d\frac1n\notin\mathbb Z, \] a contradiction. Hence $c>p^\star(d)$.

Now every prime in $[c,d]$ would be $>p^\star(d)$, contradicting maximality of $p^\star(d)$ among primes $\le d$. So $[c,d]$ is prime-free. Also \[ d-c+1\le d-p^\star(d). \] Finally, by the prime number theorem, for every $\varepsilon>0$ and all sufficiently large $d$ there is a prime in $((1-\varepsilon)d,d]$. Thus $p^\star(d)>(1-\varepsilon)d$ for large $d$, giving \[ 0\le d-p^\star(d)<\varepsilon d, \] so indeed $d-p^\star(d)=o(d)$ and hence $d-c+1=o(d)$. $\square$

Corollary 288.3.

In any ordered disjoint solution, the interval containing the maximal denominator is a terminal block of consecutive composite numbers, of sublinear length.

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

Using exact rational arithmetic, I checked the singleton subcase up to $b\le 1600$. The only intervals $[a,b]$ for which Proposition 288.1 succeeds are \[ [1,1],\ [1,2],\ [1,3],\ [2,2],\ [2,3],\ [3,6], \] with corresponding singleton values \[ m=1,\ 2,\ 6,\ 2,\ 6,\ 20. \] Thus up to this range the only singleton solutions are the already-known ones.

6) ADVERSARIAL VERIFICATION

I now try to break the new work.

• Quantifier issue in Proposition 288.1. Could there be more than one integer $k$? No. Once $u/v$ is fixed, any solution must satisfy $1/m=k-u/v>0$, so necessarily $k=\lceil u/v\rceil$. The proof did not require naming this $k$, but it is unique.
• Hidden coprimality issue in Proposition 288.1. The step \[ \gcd(kv-u,v)=\gcd(u,v)=1 \] is correct because $kv-u\equiv -u\pmod v$.
• Scope of Proposition 288.2. The proof uses the ordering $b<c$. It does not apply to overlapping solutions such as $([1,2],[1,2])$ from Round 1; this is intended.
• Small-$d$ issue in Proposition 288.2. Bertrand gives a prime in $(d/2,d]$ for every integer $d\ge 2$, so there is no exceptional range to patch.
• Does the proof really imply prime-free tail? Yes. Once $c>p^\star(d)$ and $p^\star(d)$ is the largest prime in $(d/2,d]$, any prime in $[c,d]$ would be $>p^\star(d)$ and still $\le d$, impossible.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

What is new in Round 2 is rigorous, not heuristic:

• the singleton case is reduced to a single exact candidate $m=v(a,b)$ with a congruence condition $u(a,b)\equiv -1\pmod{v(a,b)}$;
• in every ordered disjoint solution, the later interval must be a terminal composite block of length $o(d)$.

These statements are strictly stronger than the Round 1 output, but they do not settle the full problem.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 40%

9) REFERENCES

• [Er288] T. F. Bloom, Erdős Problem #288, 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.