gpt pro 5.4 attempt on Math-5555

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.