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Math-753-A4 · revision r1 · on Math-753

gpt pro 5.4 attempt on Math-753

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Manuscript

Contents
  1. Problem #306
  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 #306

1) ROUND-2 OBJECTIVE

Path C (obstruction/correction). The full statement remains open, so the realistic round-2 target is a corrected proven statement covering genuinely new infinite subclasses. In this round I prove two such subclasses:
  1. a graph-theoretic construction for many fractions whose denominator is a product of distinct primes;
  2. a short explicit three-term criterion for certain unit fractions $1/p$ with $p$ prime.
These are strictly stronger than the isolated examples from Round 1.

2) Round-1 FOUNDATION USED

I use the following Round 1 facts.
  1. If a rational number is representable as a sum of reciprocals of distinct semiprimes, then its reduced denominator is squarefree.
  2. Round 1 recorded explicit semiprime decompositions for $1/3,1/5,1/2,2/5$, and a 48-term semiprime decomposition of $1$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new insight is that the allowed denominators $p_i p_j$ naturally correspond to edges of a graph on the prime factors of the target denominator. This gives an immediate exact formula for a large family of fractions. A second new observation is that for prime $p$, a three-term semiprime decomposition of $1/p$ appears whenever \[ (q-1)(r-1)=p+1 \] for suitable distinct primes $q,r$.

4) ATTACK PLAN (ROUND-2)

The specific gap after Round 1 was the lack of any systematic constructive theorem beyond a few hand-picked examples. To close that gap I will:
  1. prove a general graph formula for denominators $P=\prod p_i$;
  2. extract from it a concrete infinite family for three-prime denominators;
  3. prove a separate three-term criterion for $1/p$ and show that it subsumes several Round 1 examples.

5) WORK (ROUND-2)

Theorem A (graph construction). Let $p_1,\dots,p_t$ be distinct primes, and let \[ P:=p_1p_2\cdots p_t. \] For any simple graph $G$ on the vertex set $\{1,\dots,t\}$, define \[ a_G:=\sum_{\{i,j\}\in E(G)} \prod_{k\ne i,j} p_k. \] Then \[ \frac{a_G}{P}=\sum_{\{i,j\}\in E(G)} \frac{1}{p_i p_j}. \] In particular, every such $a_G/P$ has a semiprime Egyptian-fraction decomposition with distinct denominators.

Proof. For each edge $\{i,j\}$, \[ \frac{1}{p_i p_j}=\frac{1}{P}\prod_{k\ne i,j} p_k. \] Summing over all edges of $G$ gives exactly the claimed identity. Since different edges give different unordered pairs $\{i,j\}$, the denominators $p_i p_j$ are distinct semiprimes. $\square$

Corollary A1 (three-prime denominator family). Let $p,q,r$ be distinct primes. Then each of the seven fractions \[ \frac{p}{pqr},\ \frac{q}{pqr},\ \frac{r}{pqr},\ \frac{p+q}{pqr},\ \frac{p+r}{pqr},\ \frac{q+r}{pqr},\ \frac{p+q+r}{pqr} \] has a semiprime Egyptian-fraction decomposition. Indeed, \[ \frac{p}{pqr}=\frac1{qr}, \qquad \frac{q}{pqr}=\frac1{pr}, \qquad \frac{r}{pqr}=\frac1{pq}, \] \[ \frac{p+q}{pqr}=\frac1{pr}+\frac1{qr}, \qquad \frac{p+r}{pqr}=\frac1{pq}+\frac1{qr}, \qquad \frac{q+r}{pqr}=\frac1{pq}+\frac1{pr}, \] \[ \frac{p+q+r}{pqr}=\frac1{pq}+\frac1{pr}+\frac1{qr}. \] This gives an explicit infinite family of squarefree denominators with three prime factors for which several nontrivial numerators are provably representable.

Theorem B (a three-term criterion for $1/p$). Let $p,q,r$ be distinct primes. If \[ (q-1)(r-1)=p+1, \] then \[ \frac1p=\frac1{pq}+\frac1{pr}+\frac1{qr}. \] Equivalently, whenever \[ qr-q-r=p, \] we obtain a three-term semiprime Egyptian-fraction decomposition of $1/p$.

Proof. The condition $(q-1)(r-1)=p+1$ expands to \[ qr-q-r=p. \] Therefore \[ \frac1{pq}+\frac1{pr}+\frac1{qr} =\frac{p+q+r}{pqr} =\frac{qr}{pqr} =\frac1p. \] The denominators are distinct semiprimes because $p,q,r$ are distinct primes. $\square$

Examples from Theorem B.

  1. If $p+2$ is prime, then taking $q=2$ and $r=p+2$ gives \[ \frac1p=\frac1{2p}+\frac1{p(p+2)}+\frac1{2(p+2)}. \] Thus \[ \frac1{17}=\frac1{34}+\frac1{38}+\frac1{323}. \]
  2. If $(p+3)/2$ is prime, then taking $q=3$ and $r=(p+3)/2$ gives \[ \frac1p=\frac1{3p}+\frac1{p(p+3)/2}+\frac1{3(p+3)/2}. \] Thus \[ \frac1{11}=\frac1{21}+\frac1{33}+\frac1{77}, \qquad \frac1{19}=\frac1{33}+\frac1{57}+\frac1{209}. \]
  3. Taking $(p,q,r)=(23,5,7)$ gives \[ \frac1{23}=\frac1{35}+\frac1{115}+\frac1{161}. \]
These formulas subsume several of the isolated small examples from Round 1 and show that there is a genuine infinite-pattern mechanism behind them.

6) ADVERSARIAL VERIFICATION

  1. Are the denominators really distinct in Theorem A? Yes. Different edges correspond to different unordered pairs $\{i,j\}$, and for distinct primes the products $p_i p_j$ are distinct.
  2. Is Theorem A compatible with the original squarefree-denominator hypothesis? Yes. The target denominator before reduction is $P=\prod p_i$, which is squarefree; after reduction, the denominator still divides $P$, so it remains squarefree.
  3. Could Theorem B fail because one denominator repeats another? No. The theorem explicitly assumes $p,q,r$ are distinct primes, so $pq,pr,qr$ are pairwise distinct semiprimes.
  4. Did I accidentally claim necessity in Theorem B? No. The theorem is only a sufficient criterion. It does not classify all three-term semiprime decompositions of $1/p$.
  5. Does any of this solve the full problem? No. The full question asks for every rational with squarefree denominator. The new theorems produce large explicit subclasses, but not all of them.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED). Round 2 proves two new constructive theorems. First, the graph construction gives a systematic infinite family of representable rationals with squarefree denominator $P=\prod p_i$. Second, Theorem B gives a clean explicit three-term criterion for many prime-denominator unit fractions $1/p$. This is a genuine step beyond the Round 1 list of isolated examples, but it still falls short of the full conjecture for arbitrary $a/b$.

8) COMPLETION ESTIMATE

COMPLETION: 34%

9) REFERENCES

  • Round 1 output in this conversation (used as fixed foundation).
  • T. F. Bloom, Erdős Problem #306.
  • S. Butler, P. Erdős and R. Graham, Egyptian fractions with each denominator having three distinct prime divisors.
  • T. Watanabe, New examples of the representation of $1$ by the sum of reciprocals of semiprime numbers, arXiv:2009.03275.