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Every natural number is a sum of distinct semiprime unit fractions

This paper proves that every natural number can be expressed as a finite sum of distinct unit fractions with semiprime denominators by adapting the Butler-Erdős-Graham induction to the challenging $\omega=2$ case, while also extending these results to rational numbers and providing the first complete proof for the $\omega=3$ case.

Original authors: Shisheng Li

Published 2026-06-16

📖 6 min read🧠 Deep dive

CC BY 4.0

Original authors: Shisheng Li

Original paper licensed under CC BY 4.0 (http://creativecommons.org/licenses/by/4.0/). ✨ This is an AI-generated explanation of the paper below. It is not written or endorsed by the authors. For technical accuracy, refer to the original paper. Read full disclaimer

The Big Picture: The "Lego" Problem

Imagine you have an infinite supply of Lego bricks. But there's a catch: you can only use bricks that are made of exactly two different colors of plastic fused together. In math terms, these are called semiprimes (numbers like 6, which is $2 \times 3$ , or 15, which is $3 \times 5$ ).

The paper asks a question about Egyptian Fractions. An Egyptian fraction is a way of writing a number as a sum of simple fractions like $1/2, 1/3, 1/4$ , etc., where all the denominators (the bottom numbers) are different.

The Question: Can you build any whole number (1, 2, 3, 100, etc.) by adding up these special fractions, where the bottom numbers are only our "two-color" semiprime bricks?

The Answer: Yes. The author, Shisheng Li, proves that you can build any whole number this way.

The Background: A Puzzle Left Unsolved

This problem wasn't invented by Li. It comes from a famous conjecture by mathematicians Paul Erdős and Ronald Graham. They asked: "If we restrict our bricks to numbers made of exactly $\omega$ (omega) different primes, can we build any number?"

$\omega = 3$ (Three colors): In 2015, other mathematicians (Butler, Erdős, and Graham) proved this works. If your bricks are made of three colors (like $2 \times 3 \times 5 = 30$ ), you can build any number.
$\omega = 2$ (Two colors): They conjectured this also works, but they couldn't prove it. They stopped there.

Why was it harder?
Think of it like a construction crew.

When you have three colors ( $\omega=3$ ), you have a huge, thick pile of bricks. It's easy to find the right combination to fill a gap.
When you drop down to two colors ( $\omega=2$ ), the pile of bricks gets much thinner. The "supply chain" is weak. The methods that worked for the thick pile (three colors) broke down because the thin pile (two colors) didn't have enough variety to fill the gaps easily.

The Solution: A New Way to Bridge the Gap

Li's paper is essentially a manual on how to keep building even when the supply of bricks is very thin.

1. The "Feeding" Problem
The proof uses a method called induction. Imagine you are building a wall, and you want to prove you can build it forever. You show that if you can build a wall of height $N$ , you can definitely build a wall of height $N+1$ .

To do this, you need a "feed" of bricks to fill the new layer.
For $\omega=3$ , the feed is rich and full.
For $\omega=2$ , the feed is very thin. The old method tried to force the thin feed to act like a thick one, and it failed.

2. The "Window" Trick
Li realized that instead of trying to make the thin feed look thick, he could look at the problem differently.

Imagine a window that slides along the wall.
Li proved that for the $\omega=2$ case, this window is always wider than the entire supply of bricks.
Because the window is so wide, it must catch the very first brick (0) or the very last brick (the maximum sum). It can't slip through the cracks.
This means you don't need a complex, thick pile of bricks; you just need to make sure the "start" and "end" of your brick pile cover all the necessary mathematical "remainders" (residues).

3. The Heavy Lifting (Computation and Logic)
To prove this works for every number, Li had to do two things:

The "Checklist" (Computation): He used a computer to check the first 300 steps of the construction manually. It's like checking the first few floors of a skyscraper with a magnifying glass to ensure the foundation is solid. He verified that the "thin feed" works perfectly for these initial steps.
The "Mathematical Safety Net" (Analysis): For the rest of the infinite building (steps 300 to infinity), he used standard mathematical inequalities (Chebyshev bounds) to prove that the bricks are distributed well enough that the "window" will always catch a valid combination.

The Results: What Else Did They Find?

The paper doesn't just stop at whole numbers. It extends the logic to fractions (rational numbers).

The Threshold: They proved that for fractions with "square-free" denominators (denominators that don't have repeated prime factors, like 6 or 10, but not 12), you can build them if the fraction is large enough (specifically, larger than about 1/5).
The "Deep Core" Mystery: There is a tiny gap left open. For very small fractions (like 1/1000), the proof doesn't quite reach down yet. The author reduces this remaining mystery to a single, specific guess: "If you keep adding more bricks, the gaps between them eventually disappear." If this guess is true, the proof is complete for all numbers.
The "Three-Color" Bonus: As a side effect of solving the "two-color" problem, Li also provided the first complete proof for the "three-color" problem (sphenic numbers) that the original authors had only guessed at.

Summary Analogy

Imagine you are trying to fill a swimming pool with water using a hose.

The Old Method: Tried to use a hose that only worked if the water pressure was super high (lots of bricks). It worked for big pools (3 colors) but failed for small, tight spaces (2 colors).
Li's Method: Realized that even with a weak hose (2 colors), if you aim it at the right angle and check the first few feet of the pool carefully, the water will naturally fill the whole pool because the hose is actually wider than the gaps in the pool floor.

The Bottom Line: Shisheng Li has solved a 10-year-old mathematical puzzle, proving that you can build any whole number using only fractions with "two-prime" denominators, using a clever mix of computer checking and mathematical logic to bridge the gap where previous methods failed.

Technical Summary: "Every Natural Number is a Sum of Distinct Semiprime Unit Fractions"

Problem Statement
The paper addresses a specific case of the Erdős–Graham problem regarding Egyptian fractions with restricted denominators. The general problem asks whether, for a squarefree denominator $b$ , every rational $a/b$ can be represented as a sum of distinct unit fractions $1/n_i$ where each $n_i$ is a product of exactly $\omega$ distinct primes.

Context: For $\omega \ge 3$ , the integer case ( $b=1$ ) was proven by Butler, Erdős, and Graham (BEG15) in 2015. They conjectured the result holds for $\omega = 2$ (semiprime denominators) but could not prove it, noting that the combinatorial "feed" driving their induction degenerates as $\omega$ decreases.
Goal: This paper resolves the integer case ( $b=1$ ) for $\omega = 2$ , proving that every natural number $m \ge 1$ is a finite sum of distinct unit fractions with semiprime denominators. It further extends this to rational numbers $a/b$ (with squarefree $b$ ) above an explicit threshold and proves the full rational case for $\omega = 3$ .

Methodology
The proof adapts and modifies the induction framework established by BEG15. The core mechanism involves analyzing the set of subset sums of specific "atom" sets derived from primorials.

Reduction to Subset Sums:
The problem is reduced to showing that for a natural number $m$ , there exists an $N$ such that $P_N m$ (where $P_N$ is the $N$ -th primorial) lies in the set of subset sums $L_2(N)$ of the set $D_2(N) = \{ P_N / (p_i p_j) : 1 \le i < j \le N \}$ .
The Induction Skeleton:
The proof relies on establishing that $L_2(N)$ contains a "central block" of integers $[\lceil \sigma_2(N)/6 \rceil, \lfloor 5\sigma_2(N)/6 \rfloor]$ , where $\sigma_2(N)$ is the sum of elements in $D_2(N)$ .
- Base Case: Verified computationally for $N_0 = 10$ using a packed bitset to check subset-sum reachability over 45 atoms.
- Induction Step ( $N \to N+1$ ): The step uses the splitting recursion $L_2(N+1) = L_1(N) + p_{N+1} L_2(N)$ , where $L_1(N)$ is the set of subset sums of single-prime reciprocals (the "feed").
Overcoming the $\omega=2$ Obstruction:
In the $\omega=3$ case, the feed $L_1(N)$ is a rich set. In the $\omega=2$ case, the feed collapses to the subset sums of single primes, which is structurally "thin." BEG15's method failed here because the "runs-connect" argument required a residue-complete inner set that did not exist in the degenerate case.
- New Insight: The author observes that in the $\omega=2$ recursion, the covering window (the interval of target values covered by a single feed element) is always at least as long as the entire feed range. Consequently, the window must contain one of the feed's endpoints ($0$ or $\sigma_1(N)$ ).
- Strategy: Instead of trying to restore internal structure to the thin feed, the proof uses the feed directly as the residue-complete object. The induction succeeds if the feed attains every residue class modulo the new prime $q = p_{N+1}$ "soon enough" (within a specific strip length).
Analytic and Computational Verification:
The proof reduces the difficulty to a single inequality: $Y_0(N) \le \beta(N)$ , where $Y_0(N)$ is the onset radius for residue completeness of the feed, and $\beta(N)$ is the available strip length.
- Finite Range ( $10 \le N \le 299$ ): Verified via exact integer arithmetic using a Dijkstra-style shortest-path algorithm over the residue group $\mathbb{Z}_q$ .
- Analytic Tail ( $N \ge 300$ ): Proven using elementary Chebyshev bounds on prime gaps and the monotonicity of the sum of prime reciprocals.
- Residue Completeness: For small $N$ where Olson's addition theorem does not strictly apply, residue completeness is verified computationally.

Key Results

Theorem 1.2 (Integer Case): Every natural number $m \ge 1$ can be written as a finite sum of distinct unit fractions with semiprime denominators. The proof is non-constructive and effective but yields enormous bounds on the size of the denominators (roughly $N \sim \exp(\exp(O(\sqrt{m})))$ ).
Rational Extension (Theorem 7.1 & 7.4): For any squarefree denominator $b$ , every rational $a/b \ge \min\{B_{N_b}/6, 1/5\}$ is representable as a sum of distinct semiprime unit fractions. Here $B_{N_b}$ depends on the largest prime factor of $b$ .
Rational $\omega=3$ Theorem (Theorem 8.1): Every positive rational $a/b$ with squarefree $b$ is a sum of distinct unit fractions with sphenic denominators (products of three distinct primes). This is achieved by a "lift" argument: multiplying the target by a large prime $r$ to move it into the representable range of the $\omega=2$ theorem, then dividing by $r$ .
General $\omega \ge 3$ (Corollary 8.6): By descent, the result extends to all $\omega \ge 3$ .

Significance and Open Problems

Resolution of a Conjecture: The paper provides the first complete proof of the $\omega=2$ integer case, a problem left as a conjecture by BEG15. It demonstrates that the "thin feed" regime can be handled by adapting the induction to rely on endpoint coverage rather than internal structure.
Unconditional Results: All results are unconditional, relying on elementary number theory and finite computations verified by independent scripts.
The Remaining Open Core: The paper identifies the only remaining open case as the "deep core" of the $\omega=2$ $ω = 2$ problem: rationals $a/b$ $a / b$ below the threshold $\min\{B_{N_b}/6, 1/5\}$ $min {B_{N_{b}} /6, 1/5}$ .
- The author reduces this to a single explicit conjecture: The Gap-Free Floor Conjecture (GFF), which posits that the gap-free floor of the semiprime subset-sum set tends to zero as $N \to \infty$ . If true, this would resolve the problem for all rationals.
Scope: The paper explicitly states it does not address denominators with square factors (which are impossible to represent under these constraints) or representations using fixed finite sets of primes. The focus remains strictly on the universal additive basis of semiprime reciprocals.

The work is described as an "engine" that drives the proof as far as current methods allow, isolating the specific structural difficulty (the gap-free floor) that prevents a complete solution for small rationals in the $\omega=2$ case.