gpt pro 5.2 attempt on Math-753

Problem #306

1) FORMAL RESTATEMENT

Let $a/b\in\mathbb{Q}_{>0}$ with $b$ squarefree. The question asks whether there always exist an integer $k\ge 1$ and integers \[ 1<n_1<n_2<\cdots<n_k \] so that \[ \frac{a}{b} = \frac{1}{n_1}+\frac{1}{n_2}+\cdots+\frac{1}{n_k}, \] and each denominator $n_i$ is the product of two distinct primes (equivalently, $n_i$ is a squarefree semiprime).

2) QUICK LITERATURE/CONTEXT CHECK

From the ErdosProblems #306 entry and the Simons Foundation discussion of Butler--Graham $2015$:

• Erdős and Graham conjectured that every integer can be written as a sum of distinct unit fractions whose denominators have exactly two distinct prime factors; this "two primes" version remains open.
• Butler and Graham proved the "three distinct prime factors" version for integers (published 2015).
• An explicit identity (attributed to Graham/Erdős) expresses $1$ as a sum of $48$ reciprocals of semiprimes.

The general question here asks for  every rational with squarefree denominator, which appears to remain open.

3) ATTACK PLAN

1. Necessary conditions. Identify obvious obstructions (e.g. why $b$ being squarefree is necessary).
2. Explicit base identities. Use known identities (e.g. a semiprime decomposition of $1$) to generate many other rationals via subset sums.
3. Search small cases. For small squarefree $b$ and $a$, search for explicit decompositions with bounded denominators.
4. General strategy. If a systematic method existed, it would likely resemble a "covering" argument: build a finite set of allowed denominators whose reciprocal sum is an integer and whose subset sums are rich enough to hit all residue classes needed.

4) WORK

4.1. Squarefree denominator is necessary.

If each $n_i$ is a product of two distinct primes, then each $n_i$ is squarefree. Let $L=\operatorname{lcm}(n_1,\dots,n_k)$. Since the lcm of squarefree integers is squarefree, $L$ is squarefree. Writing the sum over the common denominator $L$ gives $ \sum_i 1/n_i = m/L $ for some integer $m$, hence the reduced denominator divides $L$ and is also squarefree. Therefore any rational representable as such a sum must have squarefree denominator in lowest terms; this justifies the hypothesis on $b$.

4.2. An explicit 48-term semiprime decomposition of $1$.

The Simons Foundation article (Dec 10, 2015) records the identity \[ 1=\sum_{j=1}^{48} \frac{1}{d_j} \] with the following $48$ denominators, each a product of two distinct primes: \begin{align*} &6,21,34,46,58,77,87,115,155,215,287,391,\\ &10,22,35,51,62,82,91,119,187,221,299,689,\\ &14,26,38,55,65,85,93,123,203,247,319,731,\\ &15,33,39,57,69,86,95,133,209,265,323,901. \end{align*} (One can verify by direct exact arithmetic that the sum of reciprocals is $1$.)

4.3. Many other rationals via subset sums (explicit examples).

Because the above list sums to $1$, any subset of the $48$ denominators gives a rational with squarefree denominator. In particular, the following exact identities hold (all denominators are semiprimes and distinct): \begin{align*} \frac{1}{3} &= \frac{1}{6}+\frac{1}{10}+\frac{1}{15},\\ \frac{1}{5} &= \frac{1}{10}+\frac{1}{14}+\frac{1}{35},\\ \frac{1}{2} &= \frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{26}+\frac{1}{35}+\frac{1}{39}+\frac{1}{65}+\frac{1}{91},\\ \frac{2}{5} &= \frac{1}{6}+\frac{1}{10}+\frac{1}{14}+\frac{1}{38}+\frac{1}{57}+\frac{1}{95}+\frac{1}{133}. \end{align*} From these and the $1$-identity one can also get, e.g., $\frac{2}{3}=1-\frac{1}{3}$ by taking the complement of the denominators $\{6,10,15\}$ inside the 48-term representation of $1$; similarly $\frac{3}{5}=1-\frac{2}{5}$. These give (long but explicit) semiprime Egyptian decompositions of $2/3$ and $3/5$.

4.4. Computational small-case search.

A bounded search over semiprime denominators up to $5000$ found short decompositions for several small fractions (including the ones displayed above), suggesting the conjecture is plausible for small squarefree denominators. However, no general constructive proof for arbitrary squarefree $b$ was obtained.

5) VERIFICATION

• The "squarefree denominator" necessity argument is rigorous: the lcm of squarefree integers is squarefree and any reduced denominator divides the lcm.
• Each displayed denominator in the 48-term list factors as $pq$ with $p\ne q$ primes (e.g. $6=2\cdot 3$, $21=3\cdot 7$, $34=2\cdot 17$, etc.).
• The identities for $1/3$, $1/5$, $1/2$, $2/5$ can be verified by clearing denominators (or exact arithmetic); for instance $1/6+1/10+1/15=(5+3+2)/30=10/30=1/3$.

6) FINAL

UNRESOLVED.

1. Progress achieved. I proved that squarefree denominators are necessary, recorded an explicit 48-term decomposition of $1$ into reciprocals of semiprimes, and exhibited explicit decompositions for several other rationals (including $1/2,1/3,1/5,2/5$, and hence also $2/3,3/5$ via complements).
2. Where it got stuck. No method was found to handle an arbitrary target $a/b$ with arbitrary squarefree $b$.
3. What next steps would look like. One would try to develop a general "covering"/iteration scheme (similar in spirit to Butler--Graham for the three-prime case) that generates, for each squarefree $b$, a finite set of allowed denominators whose reciprocal sums can realize every residue class $a\bmod b$ (after scaling to a common squarefree denominator).
4. Falsifiable conjecture. For every squarefree $b$ and every $1\le a<b$, there exists a semiprime Egyptian fraction expansion of $a/b$ with distinct denominators.

7) COMPLETION ESTIMATE

To resolve the problem affirmatively, one needs a uniform construction that takes $(a,b)$ with squarefree $b$ and outputs an explicit semiprime-denominator Egyptian expansion. To resolve it negatively, one needs an explicit squarefree $b$ and $a$ for which one can prove that no such expansion exists (e.g. via congruence obstructions), which would require identifying an invariant that all semiprime-denominator sums satisfy but some $a/b$ violates.