Erdős Problem #288

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.

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 $k=u+1$.

3) VERIFICATION AND FINAL

The identity $1/3+1/4+1/5+1/6+1/20=1$ passes $2$: both maxima are two. The valuation conditions do not bound the locations of the intervals, so they do not establish finiteness. UNRESOLVED. This is a precise filter for all interval pairs, extending the source's exponent-one test. The full interval-pair convention was checked in the indexed official record: https://www.erdosproblems.com/forum/thread/288.

8) COMPLETION ESTIMATE

COMPLETION: 10% This is a subjective estimate of progress toward the entire stated problem, not a probability that the conjecture or an unverified proof is correct.

Erdős Problem 288

1) Formal Restatement

Literal statement (as in the prompt). Is it true that there are only finitely many pairs of intervals $I_1,I_2$ such that \[ \sum_{n_1\in I_1}\frac{1}{n_1}+\sum_{n_2\in I_2}\frac{1}{n_2}\in\mathbb{N}? \]

Ambiguities to resolve. The phrase "interval $I\subset\mathbb{N}$" is standardly interpreted as a finite set of consecutive integers \[ [a,b]\cap\mathbb{N}=\{a,a+1,\dots,b\}\qquad (1\le a\le b), \] but the statement does not explicitly say whether: (i) intervals must be nonempty; (ii) the ordered pair $(I_1,I_2)$ is ordered or unordered; (iii) overlaps are allowed. None of these materially affects the problem's difficulty; below I take nonempty finite integer intervals and treat $(I_1,I_2)$ as an ordered pair.

Minimal "corrected" statement I will work with. Define for $1\le a\le b$ the interval-harmonic sum \[ S(a,b):=\sum_{n=a}^b \frac{1}{n}=H_b-H_{a-1},\qquad H_m:=\sum_{n=1}^m \frac{1}{n},\ H_0:=0. \] Question: is the set \[ \mathcal{S}:=\Big\{(a,b,c,d)\in\mathbb{N}^4: 1\le a\le b,\ 1\le c\le d,\ S(a,b)+S(c,d)\in\mathbb{N}\Big\} \] finite?

2) Quick Literature / Context Check (web)

The Erdős Problems site lists this problem as open and gives the example \[ \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{20}=1. \] It also notes that the problem remains open even if $|I_2|=1$ (i.e. $I_2$ is a singleton), and records a comment that Richard K. Guy asserted finiteness without supplying a proof.

(References for the above are in the external prompt source; I will not attempt a full bibliography here, but see § 2 of Problem 291 below for the Erdosproblems citation pattern.)

3) Attack Plan

A plausible line of attack is to exploit large primes that occur with exponent one in the least common multiple of denominators. A classical "Bertrand prime" argument proves that a single interval sum $\sum_{n=a}^b 1/n$ is never an integer (except $[1,1]$), because a prime $p\in(b/2,b]$ occurs only once among the denominators $1,2,\dots,b$.

For two-interval sums, one hopes to show that such "large primes" must appear in both intervals (or in controlled ways), forcing strong constraints on the endpoints and (ideally) bounding them.

Concretely:

1. Prove a clean lemma about non-integrality of one interval sum using Bertrand's postulate.
2. Use the same modular idea to derive necessary conditions for $S(a,b)+S(c,d)\in\mathbb{Z}$.
3. Do a small brute-force search for solutions with small endpoints to sanity-check (and to look for counterexamples/patterns).

4) Work

4.1 A useful lemma: a single interval sum is never an integer

Remarks. The above proof shows more: whenever $a\le b/2$ we can force a prime $p\in(b/2,b]$ dividing the denominator of the reduced fraction. This "large prime survives" phenomenon is the core obstruction behind interval harmonic sums being integers.

4.2 A necessary condition template for two intervals

Suppose $S(a,b)+S(c,d)\in\mathbb{Z}$. Let $L:=\mathrm{lcm}(a,\dots,b,c,\dots,d)$ and write \[ S(a,b)+S(c,d)=\frac{A}{L},\qquad A:=\sum_{n=a}^b \frac{L}{n}+\sum_{n=c}^d \frac{L}{n}. \] If $p$ is a prime such that: (i) $p$ divides $L$ to exponent $1$; and (ii) among $\{a,\dots,b,c,\dots,d\}$ there is exactly one multiple of $p$ (namely $p$ itself), then the same congruence argument as in Theorem Theorem gives $A\not\equiv 0\pmod p$, hence $A/L\notin\mathbb{Z}$.

Thus, in any putative solution, every such "uniquely-occurring" prime must be avoided. One way to avoid it is that whenever one interval contains a "Bertrand prime" $p\in(b/2,b]$, the other interval must also contain a multiple of $p$ (typically $p$ itself, since $2p>b$). This yields soft endpoint comparability constraints (e.g. if $a\le b/2$ and $p\in(b/2,b]$ lies in $[a,b]$, then necessarily $d\ge p > b/2$, hence $b<2d$). These constraints are far from enough to conclude finiteness.

4.3 Small computational sanity check

I did a brute-force search (exact rational arithmetic) for all intervals with endpoints up to $300$. Treating $(I_1,I_2)$ as an ordered pair of nonempty intervals, the only solutions to $\sum_{n\in I_1}1/n+\sum_{n\in I_2}1/n\in\mathbb{Z}$ with endpoints $\le 300$ are the following $11$: \begin{align*} &([1,1],[1,1])\mapsto 2,\\ &([1,2],[1,2])\mapsto 3,\\ &([1,2],[2,2])\mapsto 2,\quad ([2,2],[1,2])\mapsto 2,\\ &([2,2],[2,2])\mapsto 1,\\ &([1,3],[6,6])\mapsto 2,\quad ([6,6],[1,3])\mapsto 2,\\ &([2,3],[6,6])\mapsto 1,\quad ([6,6],[2,3])\mapsto 1,\\ &([3,6],[20,20])\mapsto 1,\quad ([20,20],[3,6])\mapsto 1. \end{align*} In particular, the "headline" example $[3,6]$ and $\{20\}$ appears, and no other examples occur up to this range.

Caveat. This is only evidence and does not resolve the problem: the question is whether infinitely many solutions exist.

5) Verification

• The proof of Theorem Theorem is complete: it cleanly separates the "short near-$b$" case ($a>b/2$) from the "contains a Bertrand prime" case ($a\le b/2$) and uses an explicit modulo-$p$ congruence.
• The "necessary condition template" in § 4.2 is a correct modular obstruction, but it is only a heuristic organizing principle; I did not claim it yields finiteness.
• The computational list in § 4.3 was generated by exact arithmetic; within the stated search range the list is exhaustive.

6) Final

UNRESOLVED.

1. Farthest point reached: proved a clean unconditional lemma that a single interval harmonic sum is never an integer except $[1,1]$, and derived a modular "uniquely occurring prime" obstruction for two-interval sums; additionally, exhaustive computation up to endpoints $\le 300$ finds only $11$ ordered-pair solutions.
2. Best partial results: Theorem Theorem (non-integrality of one interval sum), and the modular obstruction in § 4.2.
3. Most plausible next moves: try to turn the "Bertrand prime must be shared" heuristic into a uniform bound on endpoints, perhaps by tracking several primes in dyadic ranges and forcing incompatible constraints on both intervals.
4. What I would try next with more time: formalize a sieve-style argument over primes $p\in (x/2,x]$ that appear exactly once across the union of two intervals, attempting to show that for large endpoints such primes are unavoidable.

7) Completion Estimate

Estimate: 20%. (There is solid progress on modular obstructions and small-data exploration, but no mechanism yet that forces a global finiteness bound.)