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A sharp 5/8 bound for an Erd\H{o}s-Sós pairwise-sums problem

This paper resolves Erdős Problem 865 by proving that the minimum size $f_3(N)$ required for a subset of $\{1,\dots,N\}$ to contain three distinct elements whose pairwise sums are also in the set is exactly $5N/8 + O(1)$ , establishing a sharp bound that matches a known construction.

Original authors: Ricky Cipollini

Published 2026-06-30

📖 4 min read🧠 Deep dive

CC BY 4.0

Original authors: Ricky Cipollini

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 "No-Three-Team" Rule

Imagine you are organizing a party with guests numbered from 1 to $N$ . You want to invite as many people as possible, but you have a very strict rule: You cannot have three guests (let's call them Alice, Bob, and Charlie) such that if you pair them up, their "combined numbers" are also guests at the party.

For example, if Alice is #2 and Bob is #3, their sum is #5. If #5 is also at the party, that's a problem. The rule says: You cannot have a trio where every possible pair (Alice+Bob, Alice+Charlie, Bob+Charlie) results in a number that is also a guest at the party.

Mathematicians call this a "pairwise-sum triple." The paper asks a simple question: What is the maximum number of people you can invite to this party before you are forced to accidentally create one of these forbidden trios?

The Answer: The 5/8 Threshold

The paper solves a famous puzzle (Erdős Problem 865) by proving a precise limit.

Think of the total number of guests ( $N$ ) as a giant pizza. The paper proves that if you invite more than 5/8ths of the pizza (plus a tiny, negligible crumb), you cannot avoid having a forbidden trio.

The Lower Bound (The "Bad" Construction): The authors show a specific way to invite exactly 5/8ths of the guests without breaking the rule. They do this by inviting people from two specific slices of the pizza:
1. The slice from 1/8 to 1/4 of the way through.
2. The slice from 1/2 to the very end.
  If you only pick people from these two zones, their "sums" never land back in the guest list. This proves you can get to 5/8.
The Upper Bound (The "Good" Proof): The main work of the paper is proving that you cannot go any higher than 5/8. If you try to invite even one more person beyond that 5/8 mark, the math guarantees that a forbidden trio will appear.

So, the answer is exactly 5/8. It's a sharp, precise line in the sand.

How They Proved It: The "Folding" Trick

To prove that you can't go higher than 5/8, the authors use a clever mental trick called "Folding."

Imagine your list of guests is a long strip of paper.

Pick a Pivot: Choose a specific guest (let's call them the "Pivot") to stand in the middle.
Fold the Paper: Imagine folding the strip of paper so that the numbers below the Pivot line up with the numbers above the Pivot.
- If the Pivot is guest #100, guest #101 folds onto #99, #102 folds onto #98, and so on.
The Collision: When you fold the paper, some numbers might land on top of each other. The authors analyze what happens when these "folded" numbers interact.

They discovered that if you have too many guests, the "folded" numbers create a mathematical collision that forces a forbidden trio to exist. It's like trying to pack too many suitcases into a car; eventually, the geometry of the car forces two suitcases to smash into each other.

The "Lean" Formalization (The Robot Check)

The paper mentions that a part of the proof was checked by a computer program called Lean 4.

Think of the proof as a complex bridge. The authors built it by hand. Then, they handed the blueprints to a super-precise robot (Lean) to check every single bolt and beam. The robot confirmed that the bridge is solid, with no hidden cracks or "sorry, I forgot a step" moments. This gives the mathematical community extra confidence that the 5/8 limit is absolutely correct.

Summary

The Problem: How many numbers can you pick from 1 to $N$ without creating a specific "sum-trio"?
The Result: You can pick up to 5/8 of the numbers. If you pick more, you are mathematically guaranteed to create the trio.
The Method: They used a "folding" technique to show that any attempt to exceed this limit causes a logical contradiction.
The Significance: This solves a decades-old problem (Erdős Problem 865) and confirms that the "5/8" limit is the absolute best possible answer.

Technical Summary: A Sharp 5/8 Bound for an Erdős–Sós Pairwise-Sums Problem

Problem Statement
The paper addresses a specific case of the Erdős–Sós conjecture concerning pairwise sums. Let $f_3(N)$ denote the least integer such that every subset $A \subseteq \{1, \dots, N\}$ with $|A| \geq f_3(N)$ contains distinct elements $a, b, c \in A$ satisfying the condition that their pairwise sums $a+b$ , $a+c$ , and $b+c$ are all contained within $A$ . Such a set is termed "pairwise-sum triple-free" if no such triple exists. The central problem is to determine the asymptotic behavior of $f_3(N)$ .

Prior to this work, a standard construction $A = [\frac{N}{8}, \frac{N}{4}] \cup [\frac{N}{2}, N]$ demonstrated that $f_3(N) \geq \frac{5}{8}N + O(1)$ . The goal of the paper is to prove the matching upper bound, thereby resolving Erdős Problem 865 and establishing the exact asymptotic constant.

Methodology
The proof is self-contained and relies on a combination of combinatorial induction, additive number theory, and a "folding" technique. The methodology proceeds through the following key steps:

The Folded Additive Lemma (Lemma 2.1):
The authors first establish a lemma on a "folded" interval. For a set $B \subseteq \{1, \dots, m-1\}$ where no distinct pair sums to $m$ or to another element in $B$ (modulo $m$ ), the lemma bounds the size of $B$ relative to the set of "collisions" $C(B)$ . Specifically, it proves $|B| - |C(B)| \leq \frac{m}{4} + 2$ . The proof utilizes reflection symmetry ( $B \to -B$ ) and an inductive argument on $|B|$ , analyzing cases where the sum of the two smallest elements either lies in $C(B)$ or does not.
The Folding Lemma (Lemma 3.1):
The authors introduce a "folding" operation around a pivot element $h \in A$ . By partitioning the set $A$ into elements smaller than $h$ ( $X$ ) and elements of the form $h+r$ where $r < h$ ( $Y$ ), they relate the size of $A$ to the structure of the intersection $B_h = X \cap Y$ . The lemma demonstrates that if $A$ is triple-free, the sum of the sizes $|X| + |Y|$ is bounded by $\frac{5}{4}h - |E \setminus C(B_h)| + 1$ , where $E$ represents the "gaps" in the interval $[1, h-1]$ . This step effectively reduces the problem of bounding the global set size to analyzing local additive properties modulo $h$ .
Inductive Upper Bound (Section 4):
The main theorem is proven by strong induction on $H$ (where $N=2H$ ). The proof splits into two cases based on the gap between the largest element in the lower half of the interval and the smallest element in the upper half:
- Case 1 ( $s \leq 4e$ ): The set is folded around the pivot $h = q$ (the smallest element in the upper half). The authors show that the gap between the lower and upper parts of $A$ forces a sufficient number of "missing" elements ( $|E \setminus C(B_h)|$ ) to satisfy the bound.
- Case 2 ( $s > 4e$ ): The set is folded around $h = p$ (the largest element in the lower half). Here, the large gap implies that the lower portion of the set is sparse. The authors apply the induction hypothesis to the lower segment and combine this with the folding lemma to derive the bound.

Key Contributions and Results

Sharp Upper Bound: The paper proves that for all $N$ , every subset $A \subseteq [1, N]$ with $|A| \geq \frac{5}{8}N + C$ (for some absolute constant $C$ ) contains a pairwise-sum triple.
Resolution of Erdős Problem 865: By combining the new upper bound with the existing lower-bound construction, the paper establishes that $f_3(N) = \frac{5}{8}N + O(1)$ . This confirms that the constant $5/8$ is best possible.
Formalization: The authors note that an earlier conditional version of the reduction from a coarser bound to the sharp bound has been formalized in Lean 4/Mathlib. This formalization uses no "sorries" (placeholders) and no added axioms, verifying the logical soundness of the reduction steps.
Refined Conjecture: The paper mentions a sharper conjectural formula suggested by Stijn Cambie regarding the exact maximum size for specific $N$ modulo 8, though the primary result remains the asymptotic $5/8$ bound.

Significance
The paper claims to resolve Erdős Problem 865 by providing a self-contained proof of the sharp $5/8$ bound. Its significance lies in closing the gap between the known construction and the theoretical upper limit for this specific pairwise-sum problem. The work is presented as a complete mathematical proof, with the added rigor of a formalized reduction in Lean, ensuring the validity of the combinatorial arguments without reliance on unverified assumptions. The authors explicitly state that the proof is self-contained and that all implicit constants are absolute.