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Math-751-A4 · revision r1 · on Math-751

gpt pro 5.4 attempt on Math-751

No claim recorded · LLM-generated · gpt pro 5.4 ·

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. Problem #302
  2. 1) ROUND-2 OBJECTIVE
  3. 2) Round-1 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-2)
  5. 4) ATTACK PLAN (ROUND-2)
  6. 5) WORK (ROUND-2)
  7. 6) ADVERSARIAL VERIFICATION
  8. 7) FINAL (EXACTLY ONE)
  9. 8) COMPLETION ESTIMATE
  10. 9) REFERENCES

Problem #302

1) ROUND-2 OBJECTIVE

Path C (obstruction/correction). Round 1 already showed that the specific conjecture \[ f(N)=(1/2+o(1))N \] is false by proving the explicit lower bound $f(N)\ge (5/8+o(1))N$. The most promising round-2 task is therefore not to re-open the false $1/2$-claim, but to sharpen the corrected picture. In this round I prove a new family of solution-free sets, containing the parity construction as a special case, and I show that within this whole family the density $5/8$ is already optimal.

2) Round-1 FOUNDATION USED

I use the following Round 1 results without re-proving them.
  1. If $1\le a,b,c\le N$ are pairwise distinct and $1/a=1/b+1/c$, then $a<N/2$.
  2. The identity $bc=a(b+c)$ is equivalent to $1/a=1/b+1/c$.
  3. The Round 1 construction \[ [\lceil N/2\rceil,N]\ \cup\ \{n\le N/4:\ n\text{ odd}\} \] is solution-free and has size $(5/8+o(1))N$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a single-residue-class modular obstruction. Instead of taking the low block to be the odd numbers, I take it to be one fixed invertible residue class modulo an arbitrary modulus $m\ge 2$. The parity construction is exactly the case $(m,r)=(2,1)$. This gives a larger theorem and, after optimizing the interval parameter, proves that $5/8$ is the best density obtainable from any construction of this one-residue/one-interval type.

4) ATTACK PLAN (ROUND-2)

The gap left after Round 1 was that the $(5/8)$-construction looked somewhat ad hoc. To improve on that, I will:
  1. prove a general theorem for sets of the form \[ \{n\le \beta N/2:\ n\equiv r\pmod m\}\cup [\beta N,N], \] with $1/2\le \beta\le 1$ and $(r,m)=1$;
  2. optimize the resulting density over $\beta$ and $m$;
  3. check carefully that the modular argument still works when $m$ is composite and one of $b,c$ is not coprime to $m$.

5) WORK (ROUND-2)

Theorem. Fix integers $m\ge 2$ and $r$ with $(r,m)=1$, and fix a real number $\beta$ with $1/2\le \beta\le 1$. Define \[ L_N(m,r,\beta):=\{n\in\{1,\dots,N\}: n\le \lfloor \beta N/2\rfloor,\ n\equiv r\pmod m\}, \] \[ U_N(\beta):=\{\lceil \beta N\rceil,\lceil \beta N\rceil+1,\dots,N\}, \] and \[ A_N(m,r,\beta):=L_N(m,r,\beta)\cup U_N(\beta). \] Then $A_N(m,r,\beta)$ contains no pairwise distinct $a,b,c$ with \[ \frac1a=\frac1b+\frac1c. \] Consequently, \[ f(N)\ge \left(1-\beta+\frac{\beta}{2m}\right)N+O(1). \] Within this family the density is maximized at $\beta=1/2$, where it becomes \[ \frac12+\frac{1}{4m}. \] Hence the best member of this entire family is the parity construction $(m,r,\beta)=(2,1,1/2)$, which recovers density $5/8$.

Proof. Assume for contradiction that distinct $a,b,c\in A_N(m,r,\beta)$ satisfy $1/a=1/b+1/c$. By the Round 1 lemma, the smallest denominator in any such triple satisfies $a<N/2$. Since $\beta\ge 1/2$, every element of $U_N(\beta)$ is at least $N/2$, so necessarily \[ a\in L_N(m,r,\beta). \] In particular, \[ a\le \beta N/2, \qquad a\equiv r\pmod m, \qquad (a,m)=1. \]

I now show that neither $b$ nor $c$ can lie in $L_N(m,r,\beta)$. Suppose first that $b\in L_N(m,r,\beta)$. Then also $b\equiv r\pmod m$ and $(b,m)=1$. There are two cases.

Case 1: $(c,m)=1$. Then $a,b,c$ are all invertible modulo $m$, and from \[ bc=a(b+c) \] we may divide by $abc$ modulo $m$ to get \[ a^{-1}\equiv b^{-1}+c^{-1}\pmod m. \] But $a\equiv b\equiv r\pmod m$, so $a^{-1}\equiv b^{-1}\equiv r^{-1}\pmod m$, hence \[ r^{-1}\equiv r^{-1}+c^{-1}\pmod m, \] which gives $c^{-1}\equiv 0\pmod m$, impossible.

Case 2: $(c,m)>1$. Choose a prime $q\mid (c,m)$. Since $a\equiv b\equiv r\pmod m$ and $(r,m)=1$, we have $q\nmid a$ and $q\nmid b$, whereas $q\mid c$. Reducing \[ bc=a(b+c) \] modulo $q$, the left-hand side is $0$, while the right-hand side is congruent to $ab\not\equiv 0\pmod q$. Contradiction.

Thus $b\notin L_N(m,r,\beta)$. By symmetry $c\notin L_N(m,r,\beta)$ as well. Therefore both $b$ and $c$ must lie in $U_N(\beta)$, so \[ b,c\ge \beta N. \] Assume $b\le c$. Then, because the denominators are distinct, \[ \frac1a=\frac1b+\frac1c<\frac{2}{b}\le \frac{2}{\beta N}, \] so \[ a>\beta N/2. \] But we already know $a\le \beta N/2$, contradiction. This proves that $A_N(m,r,\beta)$ is solution-free.

For the size estimate, we have \[ |U_N(\beta)|=(1-\beta)N+O(1), \qquad |L_N(m,r,\beta)|=\frac{\beta}{2m}N+O(1), \] so \[ |A_N(m,r,\beta)|=\left(1-\beta+\frac{\beta}{2m}\right)N+O(1). \] Since $m\ge 2$, the coefficient of $\beta$ here is \[ -1+\frac{1}{2m}<0, \] so the density is maximized when $\beta$ is as small as allowed, namely $\beta=1/2$. This gives the best density inside the family as \[ \frac12+\frac{1}{4m}. \] Finally, this is maximized over $m\ge 2$ at $m=2$, giving $5/8$. $\square$

6) ADVERSARIAL VERIFICATION

  1. Could the modular step fail for composite $m$? No. If $(c,m)=1$, inverses modulo $m$ exist and the proof is direct. If $(c,m)>1$, one passes to a prime divisor $q\mid (c,m)$ and obtains a contradiction modulo $q$.
  2. Could $b$ and $c$ both be in the upper block without contradiction when $b=\beta N$? No. Distinctness gives \[ \frac1b+\frac1c<\frac2b\le \frac{2}{\beta N}, \] so the resulting inequality for $a$ is strict.
  3. Did I accidentally use a larger family than I actually proved? No. The theorem covers exactly one residue class modulo $m$ in the low interval, not an arbitrary set of residues. I am not claiming optimality beyond this family.
  4. Does the density computation double-count an overlap between low and upper blocks? No. Since $\beta/2<\beta$, the intervals $[1,\beta N/2]$ and $[\beta N,N]$ are disjoint for every $\beta>0$.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED). Round 2 proves a genuine new theorem: the Round 1 parity construction is the optimal member of the entire one-residue/one-interval family \[ \{n\le \beta N/2:\ n\equiv r\pmod m\}\cup [\beta N,N]. \] This does not determine the true asymptotic of $f(N)$, but it does explain rigorously why the Round 1 $(5/8)$-construction is the best possible within a broad modular template.

8) COMPLETION ESTIMATE

COMPLETION: 40%

9) REFERENCES

  • Round 1 output in this conversation (used as fixed foundation).
  • T. F. Bloom, Erdős Problem #302.