Erdős Problem #287

Round 2 continuation for Erdős Problems #282, #283, #287

Erdős Problem #287

1) ROUND-2 OBJECTIVE

I pursue (C) obstruction/correction: isolate a structural condition that every counterexample must satisfy. Round 1 showed only the weaker no-consecutive-denominators result and an exhaustive search for $n_k\le 47$. The exact Round-1 gap was that the $2$-adic uniqueness argument breaks once isolated holes are allowed. The new Round-2 aim is to replace that fragile argument by a general prime-power uniqueness lemma that applies to arbitrary Egyptian partitions.

2) Round-1 FOUNDATION USED

I use the following Round-1 items.

1. The lcm/parity proof that a full interval of consecutive denominators cannot sum to $1$.
2. The interpretation of the conjecture in terms of sets with all successive gaps at most $2$.
3. The exhaustive Round-1 computation showing no counterexample with $n_k\le 47$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a general valuation obstruction.

If \[ 1=\frac1{n_1}+\cdots+\frac1{n_k} \] and for some prime $p$ the maximum value of $v_p(n_i)$ is attained by exactly one denominator, then the identity is impossible. This immediately implies a strong span-divisibility theorem: if \[ W:=n_k-n_1, \] then every denominator $n_i$ must divide \[ L_W:=\mathrm{lcm}(1,2,\dots,W). \] So a hypothetical counterexample of span $W$ is forced into a very rigid set of allowed integers.

A second new tool is a clean harmonic localisation bound: \[ n_1\le \frac{W+e}{e-1}, \qquad n_k\le \frac{e(W+1)}{e-1}. \] Thus a counterexample of span $W$ must lie in a short initial interval of length about $1.582W$, and all of its denominators must be divisors of $L_W$.

4) ATTACK PLAN (ROUND-2)

I prove three precise claims.

1. A general prime-adic uniqueness lemma for arbitrary Egyptian fraction decompositions of $1$.
2. A corollary excluding every prime power divisor $p^e>W$ from every denominator, hence forcing $n_i\mid L_W$.
3. A harmonic upper bound localising the first and last denominator in terms of the span $W$.

This overcomes the exact Round-1 obstacle: the new argument no longer depends on the denominators being consecutive.

5) WORK (ROUND-2)

5.1. A general prime-adic uniqueness lemma.

Suppose \[ 1=\frac1{n_1}+\cdots+\frac1{n_k} \] with distinct positive integers $n_1,\dots,n_k$. Fix a prime $p$, and let \[ t:=\max_{1\le i\le k} v_p(n_i). \] I claim that if $t\ge 1$, then this maximum cannot be attained by exactly one denominator.

Indeed, let \[ L:=\mathrm{lcm}(n_1,\dots,n_k). \] Multiplying the Egyptian fraction identity by $L$ gives \[ L=\sum_{i=1}^k \frac{L}{n_i}. \] Assume for contradiction that $v_p(n_j)=t$ for exactly one index $j$. Then $v_p(L)=t$, so $L/n_j$ is not divisible by $p$, whereas for every $i\ne j$ we have $v_p(n_i)\le t-1$, hence $L/n_i$ is divisible by $p$. Therefore the right-hand side is congruent to $L/n_j\not\equiv 0\pmod p$, while the left-hand side $L$ is divisible by $p$. This is impossible.

So for each prime $p$, either no denominator is divisible by $p$, or the maximal $p$-adic valuation occurring among the denominators is attained at least twice.

5.2. Span-divisibility theorem.

Now assume additionally that the denominators satisfy \[ 2\le n_1<\cdots<n_k, \qquad W:=n_k-n_1. \] I claim that every denominator $n_i$ divides $L_W:=\mathrm{lcm}(1,2,\dots,W)$.

Fix $i$ and a prime $p$. Write $e:=v_p(n_i)$. If $p^e>W$, then two distinct multiples of $p^e$ differ by at least $p^e>W$, so the interval $[n_1,n_k]$ contains at most one multiple of $p^e$. Hence among $n_1,\dots,n_k$, at most one denominator has $p$-adic valuation at least $e$. In particular the maximal $p$-adic valuation occurring in the set is attained uniquely, contradicting the lemma from 5.1.

Therefore for every $i$ and every prime $p$, \[ p^{v_p(n_i)}\le W. \] Equivalently, \[ v_p(n_i)\le \max\{r\ge 0: p^r\le W\}=v_p(L_W). \] Since this holds for every prime $p$, one gets \[ n_i\mid L_W \qquad (1\le i\le k). \] This proves the theorem.

5.3. Immediate corollaries of the span-divisibility theorem.

1. No denominator can contain a prime factor $>W$.
2. More strongly, no denominator can be a prime power $>W$.
3. Hence every denominator $>W$ is necessarily composite and has at least two distinct prime factors.

These are all exact consequences of $n_i\mid L_W$.

5.4. Harmonic localisation.

Let $a:=n_1$ and $b:=n_k=a+W$. Since the denominator set is contained in the interval $[a,b]$, \[ 1=\sum_{i=1}^k \frac1{n_i}\le \sum_{n=a}^{b}\frac1n. \] Using the standard integral bound for a decreasing function, \[ \sum_{n=a}^{b}\frac1n\le \int_{a-1}^{b}\frac{dx}{x}=\log\frac{b}{a-1}. \] Hence \[ 1\le \log\frac{b}{a-1}, \qquad\text{so}\qquad e\le \frac{b}{a-1}=\frac{a+W}{a-1}. \] Rearranging gives \[ (e-1)a\le W+e, \qquad\text{i.e.}\qquad a\le \frac{W+e}{e-1}. \] Substituting $b=a+W$ then yields \[ b\le W+\frac{W+e}{e-1}=\frac{e(W+1)}{e-1}. \] So every counterexample of span $W$ must lie inside the interval \[ \left[2,\ \frac{e(W+1)}{e-1}\right] \] with first term at most $(W+e)/(e-1)$.

5.5. Synthesis.

Any counterexample to Problem #287 must therefore have the following rigid form. There exists an integer $W\ge 2$ and an interval $[a,a+W]$ such that

1. $a\le (W+e)/(e-1)$,
2. every chosen denominator lies in $[a,a+W]\cap \mathrm{Div}(L_W)$,
3. the chosen denominators have successive gaps at most $2$,
4. their reciprocals sum to $1$.

This is a substantial sharpening of the raw statement of the conjecture: the search for a counterexample is reduced to a divisor-theoretic problem tied to $L_W$.

6) ADVERSARIAL VERIFICATION

1. Does the prime-adic lemma require consecutive denominators? No. The proof uses only the lcm identity and works for an arbitrary finite set of denominators.
2. Could there be two denominators with different large powers of the same prime, so that the maximal valuation is not unique? If one denominator contains $p^e>W$, then there is at most one multiple of $p^e$ in the whole interval of length $W$, so there is at most one denominator with $p$-adic valuation at least $e$. Hence the maximal valuation is indeed unique.
3. Is the conclusion $n_i\mid L_W$ too strong? No. It is exactly equivalent to the prime-by-prime inequalities $p^{v_p(n_i)}\le W$ proved above.
4. Could the harmonic bound be off by one? The bound used is \[ \sum_{n=a}^{b}\frac1n\le \int_{a-1}^{b}\frac{dx}{x}, \] which is standard for decreasing $1/x$. The resulting inequalities are therefore valid exactly as written.
5. How does this interact with Round 1? The Round-1 no-consecutive theorem becomes a special case of the new philosophy: in a full interval one can force a unique maximal $2$-adic valuation. Round 2 generalises the uniqueness obstruction to arbitrary prime powers and arbitrary sets.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The main conjecture remains open, but Round 2 proves a theorem that is strictly stronger than any Round-1 result: if a counterexample has span \[ W=n_k-n_1, \] then every denominator must divide $\mathrm{lcm}(1,2,\dots,W)$, and the whole counterexample must lie in a short initial interval with \[ n_1\le \frac{W+e}{e-1}, \qquad n_k\le \frac{e(W+1)}{e-1}. \] So a hypothetical counterexample must be an unusually dense cluster of divisors of $L_W$ with no two consecutive omissions and reciprocal sum exactly $1$. This is a genuine structural reduction of the original problem.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 55%

9) REFERENCES

Checked or used in this round:

1. T. F. Bloom, Erdős Problem #287, webpage accessed 2026-03-07.

Problem 287: a stronger two-adic span obstruction

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. I read all of 287.tex. The target is to rule out an Egyptian partition of one with distinct denominators $2\le n_1<\cdots<n_k$ and every successive gap at most two. The supplied general valuation lemma yields $n_i\mid\operatorname{lcm}(1,\ldots,W)$, where $W=n_k-n_1$. I refine the factor-of-two restriction and record the exact residue condition when an odd prime power has only two representatives.

2) WORK

Let $t=\max_i v_2(n_i)$. If $t\ge1$, multiply the identity $\sum_i1/n_i=1$ by the least common multiple $L$. Modulo two, precisely the terms of valuation $t$ contribute one. Their number must therefore be even and at least two. Any two such denominators are $2^t$ times distinct odd integers, so they differ by at least $2^{t+1}$. Consequently \[ 2^{v_2(n_i)}\le W/2\qquad\text{for every even }n_i. \] Combining this with the supplied prime-power bound gives the stronger necessary divisibility \[ n_i\mid 2^{\max\{0,\lfloor\log_2W\rfloor-1\}} \prod_{\substack{p\le W\\p\text{ odd prime}}}p^{\lfloor\log_pW\rfloor} \qquad(W\ge2). \] The greatest allowed power of two has been reduced by a factor two.

For an odd prime $p$, let $t$ be the maximum positive valuation and suppose exactly two denominators have valuation $t$, say $p^tu$ and $p^tv$, where $p\nmid uv$. Reduction of the same cleared identity modulo $p$, followed by cancellation of the common unit $L/p^t$, gives \[ u^{-1}+v^{-1}\equiv0\pmod p, \quad\text{equivalently}\quad u+v\equiv0\pmod p. \] In particular, if $p^t>W/2$, at most two multiples of $p^t$ fit in the interval. Both must occur and they are consecutive multiples: $v=u+1$. Thus \[ 2u+1\equiv0\pmod p. \] This is a residue restriction beyond merely demanding a repeated maximum valuation.

3) ADVERSARIAL VERIFICATION

The two-adic argument concerns denominators at the maximal valuation; counting all multiples of $2^t$ would lose the required odd-coefficient spacing. These conditions hold for every finite distinct Egyptian partition, not only the dense gap-two configuration. For $1=1/2+1/3+1/6$, $W=4$ and the two maximal even denominators are exactly four apart, so the factor $W/2$ cannot be decreased in this general lemma. No inference from these necessary conditions to impossibility of dense configurations has been justified.

4) FINAL

UNRESOLVED. The span sieve is stricter and includes exact cancellation congruences. A global argument showing that all gap-two candidate sets fail is still missing.

5) SOURCES CHECKED

The complete supplied 287.tex, read 2026-09-23. All refinements above follow directly from its cleared-denominator identity and are proved here; no unverified search or literature novelty is claimed.

6) COMPLETION ESTIMATE

Subjective progress toward excluding every proposed dense partition. COMPLETION: 15%

Erdős Problem #287

1) FORMAL RESTATEMENT

For $k\ge 2$, let $2\le n_1<\cdots<n_k$ be distinct integers with \[1=\frac1{n_1}+\cdots+\frac1{n_k}. \] The conjecture asserts that necessarily \[\max_{1\le i\le k-1} (n_{i+1}-n_i)\ge 3. \] Equivalently: there is no Egyptian-fraction representation of $1$ with all consecutive gaps in the denominator set bounded by $2$.

2) QUICK LITERATURE/CONTEXT CHECK

The bound "$\ge 3$ would be best possible" is witnessed by the classical example \[1=\frac12+\frac13+\frac16,\] whose denominator gaps are $1$ and $3$. It is known (Erdős, 1932) that the weaker bound $\ge 2$ holds, i.e. $1$ is not a sum of reciprocals of consecutive integers $n,n+1,\dots,n+k-1$ with $n>1$.

A heuristic/conditional route mentioned in the literature: the conjecture would follow (up to finitely many exceptions) from the existence, for all large $N$, of a prime $p\in[N,2N]$ such that $(p+1)/2$ is also prime (a "Sophie Germain pair").

3) ATTACK PLAN

1. Strengthen the Erdős $1932$ parity method. The classical "no consecutive denominators" result can be proved via $2$-adic valuation of the lcm. Try to adapt this to the "gaps $\le 2$" setting by looking at a prime (or prime power) that divides exactly one chosen denominator to an extremal exponent.
2. Density/interval viewpoint. If gaps $\le 2$, the denominators occupy an interval with at most isolated missing points. Translate the equation into a near-harmonic sum and attempt to derive a contradiction from congruences or from Diophantine approximation properties of harmonic numbers.
3. Finite counterexample search. Because the conjecture is falsifiable, perform exhaustive search for counterexamples with $n_k$ bounded.

4) WORK

4.1. Proof of the weaker bound $\max(n_{i+1}-n_i)\ge 2$ (no consecutive denominators)

4.2. Exhaustive search for a counterexample with gaps $\le 2$ and bounded $n_k$

A brute-force backtracking search was implemented for all strictly increasing sequences $2\le n_1<\cdots<n_k\le N$ with successive gaps in $\{1,2\}$, checking whether $\sum 1/n_i=1$ exactly.

Result: no counterexample exists with $n_k\le 47$. (The search is exhaustive for this range.)

4.3. Why this does not yet prove the conjecture

The problem remains open because the conjecture could still fail only at much larger denominators, and there is currently no general argument ruling out all "gap $\le 2$" patterns. The $2$-adic lcm argument in §4.1 relies heavily on having all denominators consecutive; when isolated denominators are removed, the uniqueness-of-maximum-$v_2$ property can fail in a global set.

5) VERIFICATION

• The proof in §4.1 is self-contained and purely elementary.
• The computation in §4.2 is exhaustive for $n_k\le 47$ by construction (enumerating all admissible gap-$\le 2$ patterns in $[2,N]$).

6) FINAL

UNRESOLVED.

1. Furthest point reached: gave a complete proof of the known weaker statement "no consecutive denominators" (hence gap $\ge 2$), and performed an exhaustive search showing no gap-$\le 2$ counterexample exists with maximum denominator $\le 47$.
2. Key gap: extending the $2$-adic/lcm obstruction to the setting where isolated denominators may be missing from the interval (gaps allowed to be $2$) remains unresolved.
3. Most promising next step: find a prime-power valuation (not necessarily $2$) that is forced to appear in exactly one denominator when gaps are $\le 2$, or relate the conjecture to distribution of Sophie Germain primes (as suggested by the conditional approach).
4. Explicit missing lemma that would close it: a theorem ensuring that in any gap-$\le 2$ denominator set supporting $\sum 1/n_i=1$, there exists a prime $p$ whose exact $p$-adic valuation in the lcm is attained uniquely by one denominator (giving an immediate contradiction mod $p$).

7) COMPLETION ESTIMATE

COMPLETION: 35%