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Math-5555-A3 · revision r1 · on Math-5555

gpt pro 5.2 attempt on Math-5555

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #287
  2. 1) FORMAL RESTATEMENT
  3. 2) QUICK LITERATURE/CONTEXT CHECK
  4. 3) ATTACK PLAN
  5. 4) WORK
  6. 4.1. Proof of the weaker bound $\max(n_{i+1}-n_i)\ge 2$ (no consecutive denominators)
  7. 4.2. Exhaustive search for a counterexample with gaps $\le 2$ and bounded $n_k$
  8. 4.3. Why this does not yet prove the conjecture
  9. 5) VERIFICATION
  10. 6) FINAL
  11. 7) COMPLETION ESTIMATE

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)

Theorem: There do not exist integers $n\ge 2$ and $k\ge 1$ such that \[1=\sum_{j=0}^{k-1}\frac1{n+j}. \] In particular, any Egyptian-fraction representation of $1$ with distinct denominators must have a gap at least $2$.
Proof: Assume for contradiction that $1=\sum_{j=0}^{k-1} \frac1{n+j}$ with $n\ge 2$. Let \[L=\mathrm{lcm}(n,n+1,\dots,n+k-1). \] Because $n\ge 2$, among $n,n+1,\dots,n+k-1$ there is at least one even number, hence $2\mid L$. Let $t=\max_{0\le j\le k-1} v_2(n+j)$ be the maximal $2$-adic valuation among the denominators; then $v_2(L)=t\ge 1$.

Claim: exactly one of the integers $n,n+1,\dots,n+k-1$ has $2$-adic valuation equal to $t$. Indeed, if there were two distinct terms with valuation $t$, their difference would be divisible by $2^{t+1}$ (difference of two odd multiples of $2^t$), so the interval length $k-1$ would be at least $2^{t+1}$. But any block of $2^{t+1}$ consecutive integers contains a multiple of $2^{t+1}$, contradicting maximality of $t$. Hence the maximiser is unique.

Now multiply the assumed identity by $L$: \[L=\sum_{j=0}^{k-1} \frac{L}{n+j}. \] Let $d_*$ be the unique denominator with $v_2(d_*)=t$. Then $L/d_*$ is odd (all factors of $2$ cancel), while for every other denominator $d$ we have $v_2(d)\le t-1$ and hence $L/d$ is even. Therefore the right-hand side is (odd) $+$ (sum of evens), hence odd. But the left-hand side $L$ is even. Contradiction. ∎

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%