Erdős Problem #289

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.

Erdős Problem #289: independent continuation

Date: 2026-09-23. Research attempt; no claim of a new published result.

1) TARGET AND SOURCE AUDIT

For every sufficiently large $k$, can one choose $k$ disjoint, pairwise nonadjacent integer intervals, each containing at least two integers, whose reciprocal sums total one?

All supplied versions considered: 289.tex. The source investigates only $k=2$ and a terminal prime-gap obstruction. That subproblem is not the full sufficiently-large-$k$ target. I add a valuation constraint applying to every proposed interval family.

2) WORK

Let $S$ be the union of the selected intervals, and suppose $\sum_{n\in S}1/n$ is an integer. For any prime $p$, the maximum of $v_p(n)$ over $S$ cannot occur at exactly one denominator. Indeed let $L=\operatorname{lcm}(S)$ and let that maximum be $e$. Multiplying the sum by $L$ and reducing modulo $p$, every term $L/n$ vanishes except the unique one with $v_p(n)=e$, which is nonzero. But the right-hand side is divisible by $p$, contradiction.

For $p=2$ each single interval has a unique denominator of maximal valuation. To see this, if two maximizers had value $e$, write them as $2^e u<2^e v$ with $u,v$ odd. An even integer lies between $u$ and $v$, producing an interval member with larger valuation. Therefore the globally largest valuation must be attained in at least two different selected intervals.

For two intervals this forces their individual maxima to be equal. Their maximizing denominators are distinct odd multiples of the same power $2^e$, and at least one intervening multiple of $2^{e+1}$ must lie in the omitted gap. This is an exact restriction on placement, independent of asymptotic prime-gap estimates.

3) ADVERSARIAL VERIFICATION

For odd $p$, an interval can contain several denominators with maximal valuation; the uniqueness argument is specifically binary. Having two maximizers is necessary, not sufficient, since their residues may still fail to cancel. None of these conditions rules out all large numbers of intervals.

4) FINAL

UNRESOLVED. The global construction for every sufficiently large $k$ remains missing. The valuation test provides a rejection criterion for searches, not a proof of existence or nonexistence.

5) SOURCES CHECKED

Full 289.tex read; checked the actual sufficiently-large-$k$ target in ErdosProblems/289.lean on the formal-conjectures repository, 2026-09-23.

6) COMPLETION ESTIMATE

This is a conservative subjective measure of progress toward the original question, not a probability of correctness or a claim that the remaining work is routine. COMPLETION: 8%

Erdős Problem 289

1) Formal Restatement

Literal statement (as in the prompt). Is it true that, for all sufficiently large $k$, there exist finite intervals $I_1,\dots,I_k\subset\mathbb{N}$, distinct, not overlapping or adjacent, with $|I_i|\ge 2$ for all $i$, such that \[ 1=\sum_{i=1}^k \sum_{n\in I_i}\frac{1}{n}\ ? \]

Fully explicit formulation. Write each $I_i$ as an integer interval $I_i=[a_i,b_i]\cap\mathbb{N}$ with $1\le a_i\le b_i$. The constraints "not overlapping or adjacent" mean that for $i\ne j$, \[ \text{either }\ b_i+1 < a_j\ \text{or}\ b_j+1 < a_i. \] Equivalently, after reindexing we may assume \[ 1\le a_1\le b_1 < a_2-1\le b_2 < a_3-1\le \dots < a_k-1\le b_k, \] with each length $b_i-a_i+1\ge 2$. The question is whether such configurations exist for all $k$ beyond some absolute threshold.

2) Quick Literature / Context Check (web)

The Erdős Problems site lists this as open. It notes that Erdős and Graham posed the question in 1980 without the "distinct / non-overlapping / non-adjacent" restrictions, and that in this weaker interpretation there is a simple affirmative argument (given in a public comment by V. Kovac). The site also records an explicit representation of $2$ (not $1$) as a sum over $5$ disjoint, non-adjacent intervals found by Hickerson and Montgomery: \[ 2=\sum_{n=2}^{7}\frac1n+\Big(\frac19+\frac1{10}\Big)+\Big(\frac1{17}+\frac1{18}\Big)+\Big(\frac1{34}+\frac1{35}\Big)+\Big(\frac1{84}+\frac1{85}\Big). \]

3) Attack Plan

Two complementary approaches suggest themselves:

1. Constructive "gadget" approach. Find a local identity that replaces one admissible interval (or a small collection of admissible intervals) by a larger collection of admissible intervals separated by gaps, preserving the total sum. If such a gadget increases the interval count by $+1$ while maintaining all constraints, iterating would yield all sufficiently large $k$.

2. CRT / large modulus approach. Encode rational pieces of $1$ using unit fractions supported on widely separated blocks of consecutive integers. The blocks must themselves be decomposable into admissible intervals (length $\ge2$) without overlaps. This resembles Kovac's construction for the weaker variant, but the distinctness and disjointness constraints are the sticking point.

4) Work

4.1 A small impossibility check for $k=2$ (bounded search)

As a sanity check, I searched exhaustively for two admissible intervals $I_1,I_2\subset\{1,2,\dots,300\}$, with $|I_i|\ge 2$ and $I_1,I_2$ disjoint and non-adjacent, such that $\sum_{n\in I_1}1/n+\sum_{n\in I_2}1/n = 1$. No such pair exists within this search window.

This does not prove nonexistence for $k=2$ in general; it only indicates that any such solution (if it exists) likely involves quite large endpoints.

4.2 A proved weaker variant (overlaps/repetitions allowed)

If one drops the requirements that the intervals be distinct and non-overlapping/non-adjacent, the problem becomes much easier. A detailed construction (using large moduli and the Chinese Remainder Theorem, plus an explicit decomposition lemma for multisets supported on four consecutive integers) is given by V. Kovac in the public discussion thread for this problem on the Erdős Problems site. That argument yields:

Weaker statement. For all sufficiently large $k$ there exist intervals $I_1,\dots,I_k\subset\mathbb{N}$ (allowed to overlap and repeat) with $|I_i|\ge 2$ such that $\sum_{i=1}^k\sum_{n\in I_i}1/n=1$.

I do not reproduce the full CRT-based construction here (it is several pages even in streamlined form), but the key point is that allowing overlap/repetition lets one realize rational coefficients as multiplicities of unit fractions in a short consecutive block, and then split that multiset into many length-$\ge2$ sub-intervals. This mechanism fundamentally breaks once one forbids repeats and overlaps.

4.3 Why the "distinct + disjoint + non-adjacent" constraints are hard

The disjointness constraints mean the total sum is the sum over a sparse subset of $\mathbb{N}$ made of $k$ blocks. As $k$ grows, each block must typically contribute a very small amount (since the total is fixed at $1$), hence the blocks must lie far out on the number line. But far-out blocks have sums that are arithmetically rigid when required to be exact rationals that fit together to $1$. The usual flexibility of Egyptian fractions (arbitrary denominators) is much reduced by the requirement that denominators come in consecutive blocks.

5) Verification

• The bounded search in § 4.1 is a factual computational claim for endpoints $\le 300$.
• The weaker-variant statement in § 4.2 is explicitly labeled as such and attributed to the discussion-thread construction.
• No unsupported claim is made about the original restricted problem.

6) Final

UNRESOLVED.

1. Farthest point reached: verified computationally that no $k=2$ solution exists with endpoints $\le 300$ under the full restrictions; documented that removing the restrictions yields an affirmative theorem via a CRT/multiplicity method.
2. Best partial results: the weaker-variant affirmative solution (overlaps/repetitions allowed) exists in the literature/discussion; and the explicit $k=5$ representation of $2$ by Hickerson--Montgomery gives evidence that the "disjoint blocks" world is nonempty.
3. Most plausible next moves: find a genuine "interval gadget" that increases the number of disjoint, separated intervals while preserving the total sum; or prove an unconditional infinite family of $k$ for which solutions exist (even a subsequence $k\to\infty$ would be progress).
4. What I would try next with more time: search computationally (with modular pruning and rational reconstruction) for a first explicit representation of $1$ by disjoint non-adjacent intervals of length $\ge2$, even for moderate $k$ (say $k=3,4,5$), as a seed for further "gadget" amplifications.

7) Completion Estimate

Estimate: 10%. (The full restricted problem is open; I only established small-case evidence and recorded a solution to a weaker interpretation.)