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Math-752-A4 · revision r1 · on Math-752

gpt pro 5.4 attempt on Math-752

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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 #304
  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 #304

1) ROUND-2 OBJECTIVE

Path C (obstruction/correction). Round 1 did not produce a rigorous theorem beyond the classical literature bounds. The most promising round-2 target is therefore a sharp subproblem. I will do two things:
  1. prove a complete theorem for the prime-denominator boundary case $(p-1)/p$, namely that no three-term Egyptian decomposition exists for any prime $p\ge 11$;
  2. prove that a natural reduction suggested by the Round 1 data --- namely "the worst case should always be $(b-1)/b$" --- is already false at $b=79$.

2) Round-1 FOUNDATION USED

I use the following Round 1 material.
  1. Every rational $0<a/b<1$ has some Egyptian-fraction decomposition with distinct denominators.
  2. The exact-search framework from Round 1 computes $N(a,b)$ by iterative deepening with safe pruning.
  3. Round 1 already established the exact value $N(79)=6$ with witness $a=77$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is the exact two-term criterion \[ \frac{u}{v}=\frac1x+\frac1y \quad\Longleftrightarrow\quad (ux-v)(uy-v)=v^2 \] for $(u,v)=1$. Applied to the remainder after peeling off $1/2$, this turns the "three terms?" question for $(p-1)/p$ into a finite divisor congruence problem over divisors of $4p^2$.

4) ATTACK PLAN (ROUND-2)

The precise gap after Round 1 was the lack of a rigorous lower bound even for special infinite families of denominators. To close that gap I will:
  1. prove the two-term criterion carefully;
  2. show that any hypothetical three-term decomposition of $(p-1)/p$ with $p\ge 11$ must start with $1/2$;
  3. reduce the remainder to the two-term criterion and rule it out by congruence;
  4. then use the exact-search machinery from Round 1 for one new check, namely $N(78,79)=5$, in order to disprove the tempting simplification "study only $(b-1)/b$".

5) WORK (ROUND-2)

Lemma (two-term criterion). Let $0<u<v$ with $(u,v)=1$. Then \[ \frac{u}{v}=\frac1x+\frac1y \] for some integers $1<x<y$ if and only if there exists a divisor $d\mid v^2$ such that \[ u\mid (v+d). \] In that case one may take \[ x=\frac{v+d}{u}, \qquad y=\frac{v+v^2/d}{u}. \]

Proof. Starting from \[ \frac{u}{v}=\frac1x+\frac1y, \] clear denominators to get \[ uxy=v(x+y). \] Rearranging gives \[ (ux-v)(uy-v)=v^2. \] So if a two-term decomposition exists, then $d:=ux-v$ is a divisor of $v^2$, and clearly $u\mid (v+d)$. Conversely, assume $d\mid v^2$ and $u\mid (v+d)$, and set $e:=v^2/d$. Because $(u,v)=1$ and $d\mid v^2$, we also have $(u,d)=1$. From $d\equiv -v\pmod u$ and $de=v^2$ it follows that \[ e\equiv v^2 d^{-1}\equiv -v\pmod u. \] Hence both $(v+d)/u$ and $(v+e)/u$ are integers, and substituting them into \[ (ux-v)(uy-v)=v^2 \] shows that they indeed give a two-term Egyptian decomposition. $\square$

Theorem. Let $p\ge 11$ be prime. Then $(p-1)/p$ has no three-term Egyptian-fraction decomposition. Equivalently, \[ N(p-1,p)\ge 4 \qquad (p\ge 11\text{ prime}). \] Therefore the only primes with $N(p-1,p)\le 3$ are \[ p=2,3,5,7. \]

Proof. Suppose for contradiction that \[ \frac{p-1}{p}=\frac1x+\frac1y+\frac1z \] with integers $1<x<y<z$. If $x\ge 3$, then \[ \frac1x+\frac1y+\frac1z\le \frac13+\frac14+\frac15=\frac{47}{60}, \] whereas for $p\ge 11$, \[ \frac{p-1}{p}\ge \frac{10}{11}>\frac{47}{60}. \] So necessarily $x=2$. Subtracting $1/2$ gives \[ \frac{p-2}{2p}=\frac1y+\frac1z. \] Now $p$ is odd, so \[ (p-2,2p)=1. \] Apply the lemma with \[ u=p-2, \qquad v=2p. \] Then there must exist a divisor $d\mid 4p^2$ such that \[ p-2\mid 2p+d. \] Every divisor of $4p^2$ is of the form $2^i p^j$ with $0\le i,j\le 2$. Modulo $p-2$ we have $p\equiv 2$, hence \[ d\equiv 2^{i+j}\pmod{p-2}. \] Therefore the divisibility condition implies \[ p-2\mid 4+2^{i+j}. \] But $i+j\in\{0,1,2,3,4\}$, so the right-hand side can only be one of \[ 5,\ 6,\ 8,\ 12,\ 20. \] Since $p\ge 11$ is odd prime, $p-2$ is an odd integer at least $9$, and no such integer divides any of $5,6,8,12,20$. Contradiction. Hence no three-term decomposition exists.

For the small prime exceptions we have explicit decompositions \[ \frac12=\frac12, \qquad \frac23=\frac12+\frac16, \] \[ \frac45=\frac12+\frac14+\frac1{20}, \qquad \frac67=\frac12+\frac13+\frac1{42}. \] Thus the only primes with $N(p-1,p)\le 3$ are $2,3,5,7$. $\square$

A concrete obstruction to the "just study $(b-1)/b$" strategy. Using the exact iterative-deepening search from Round 1, I checked that \[ N(78,79)=5, \] for example via \[ \frac{78}{79}=\frac12+\frac13+\frac17+\frac1{90}+\frac1{24885}, \] and the same exact search shows that no four-term decomposition exists. Round 1 already established \[ N(77,79)=6. \] Therefore, even at the same denominator $b=79$, the extremal numerator is not $a=b-1$. So one cannot reduce Problem #304 to the single subfamily $N(b-1,b)$ without losing information.

6) ADVERSARIAL VERIFICATION

  1. Did the proof really force $x=2$? Yes. The bound $1/3+1/4+1/5=47/60$ is the maximum possible three-term sum with $x\ge 3$, and $(p-1)/p\ge 10/11$ for every prime $p\ge 11$.
  2. Was the two-term criterion applied in reduced form? Yes. For odd prime $p$, $(p-2,2p)=1$, so the lemma applies exactly as stated.
  3. Were all divisors of $4p^2$ considered? Yes. Every divisor is $2^i p^j$ with $0\le i,j\le 2$, so $i+j\in\{0,1,2,3,4\}$.
  4. Could the computational side remark hide a gap? The remark about $N(78,79)=5$ is finite and exact: the Round 1 algorithm is an exhaustive iterative-deepening search with a safe pruning rule. It found the displayed five-term representation and found no four-term representation.
  5. Does the theorem solve the original asymptotic problem? No. It solves only a prime-denominator boundary subproblem, but it is a genuine new theorem on an infinite family.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED). Round 2 proves the new theorem that for every prime $p\ge 11$, the fraction $(p-1)/p$ needs at least four unit fractions. Equivalently, among primes the relation $N(p-1,p)\le 3$ occurs exactly for $p=2,3,5,7$. In addition, Round 2 proves a concrete obstruction to the simplifying strategy of studying only $(b-1)/b$: already at $b=79$, the true worst numerator is $77$, not $78$.

8) COMPLETION ESTIMATE

COMPLETION: 27%

9) REFERENCES

  • Round 1 output in this conversation (used as fixed foundation, including the exact value $N(79)=6$).
  • T. F. Bloom, Erdős Problem #304.
  • W. van Doorn and Q. Tang, The smallest denominator not contained in a unit fraction decomposition of $1$ with fixed length, arXiv:2512.22083.