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1 · DiscoverYou are hereMeet the questionLearn when it appeared and why it still holds people’s attention.
2 · ChallengeTest one caseMake your first observation in five minutes with a drawing, calculation, or colors.
3 · DevelopBuild on ideasPublish an observation and grow it through comparisons, conjectures, and counterexamples.

Solved1965 → 2026

Erdős #1196 — From an AI-Suggested Method to a Collaborative Proof

How large can the Erdős sum of a primitive set containing only large integers be? In 2026, eight authors checked and developed a method suggested by GPT-5.4 Pro output into a proof of a 1960s conjecture.

Challenge passport

First posed

1965

Time it held mathematicians

61 years to solve

Starting level

Deep dive

Connect a few background ideas before the real wall comes into view.

The starting level measures how easily you can understand and test small cases. It is not the difficulty of a complete proof.

Jump to your first five minutes

Solved by:

Alexeev, Barreto, Li, Lichtman, Price, Shah, Tang, and Tao (method suggested by GPT-5.4 Pro output)

The public problem record and the 2026 collaborative preprint document the solution process.

See the puzzle visually

Primitive set of large integers

A ⊂ [x,∞)

a ∤ b, b ∤ a

Upper bound for the Erdős sum

f(A) = Σ 1/(a log a)

sup f(A) ≤ 1 + O(1/log x)

As x grows

x → ∞ : 1

Asymptotic upper bound

Erdős #1196 does not say that one set’s sum converges to one. It says that the largest possible sum among primitive sets whose members are all at least x approaches one as x→∞.

Problem statement

A set is primitive when no distinct member divides another. Erdős Problem #1196 asks whether the supremum of $f(A)=∑_{a∈A}1/(a,log a)$ over primitive $A⊂[x,∞)$ is $1+o(1)$ as $x→∞$ .

The story of this puzzle

Erdős had shown that $f(A)$ is uniformly bounded for primitive sets. Problem #1196 asks a different asymptotic question: when every member is at least $x$ , does the largest possible sum approach one? It does not say that the sum over the primes is one, nor that merely adding more elements forces convergence to one.

The May 2026 preprint Primitive sets and von Mangoldt chains explicitly credits a method suggested by GPT-5.4 Pro output. A von Mangoldt-weighted Markov chain follows divisibility relations; because a primitive set meets a divisibility chain at most once, mass-counting yields $f(A)≤1+O(1/log x)$ .

The output was a starting point, not a substitute for verification. Eight authors—Boris Alexeev, Kevin Barreto, Yanyang Li, Jared Duker Lichtman, Liam Price, Jibran Iqbal Shah, Quanyu Tang, and Terence Tao—organized and extended the method into a 35-page preprint and further results. The interesting question is how proposing a connection and turning it into checkable mathematics complement each other.

Try it yourself

Mini challenge

Start by checking primitivity. In $A={6,10,15}$ no member divides another, but adding 30 breaks the rule because 6, 10, and 15 divide it. Now draw several divisibility chains that begin at 1 and repeatedly move to a multiple. Why can one chain meet a primitive set at most once? That short observation is the intuitive core of the 2026 probabilistic method.

Beyond MathVoyage

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