ethosnode retweeted
Today, we’re announcing a solution found by our miners to Erdős Problem 1057.
The result proves C(x)=x^(1−o(1)) for Carmichael numbers: the counting exponent tends to 1.
Verified in Lean through Conjectures. Full proof below.

ALT Headline: Erdős Problem 1057 — PROVED. Conjectures announcement for Erdős 1057. The counting theorem is C(x)=x^(1-o(1)), where C(x) counts Carmichael numbers at most x. For each epsilon>0, eventually C(x)≥x^(1-epsilon). A factorization diagram connects 561 to 3,11,17. Their predecessors 2,10,16 divide 560, and all 320 coprime residue classes satisfy a^560≡1 modulo 561. Ivory background, bold serif headline, large green verdict and an outlined Verified in Lean badge. Formally verified in Lean. Credits: JenW1N, Liam Kruer and gus.
Replying to @AA_Gabriel1111
God is only God...
Baal is simply the king of the world.
The king has a dethronement, but God is absolute.
ethosnode retweeted
Today, we’re announcing a solution found by our miners to Erdős Problem 579, open since at least 1983.
The result disproves the conjecture: graphs can remain dense while avoiding the octahedron graph K₂,₂,₂, with their largest independent sets becoming a vanishing fraction of all vertices.
Verified in Lean through Conjectures. Full proofs below.
🤖 Made with AI
ethosnode retweeted
Today, we’re announcing a solution found by our miners to Erdős Problem 1062(ii), open since at least 1994.
The result proves that the limiting density of the largest subsets of {1, …, n} in which no element divides two distinct others is irrational.
Verified in Lean through Conjectures. Full proofs below.
🤖 Made with AI
ethosnode retweeted
Today, we’re announcing a solution found by our miners to Erdős Problem 859, a 56-year-old question.
Let dₜ be the density of integers whose distinct divisors can sum to t. The result proves that no positive constants c₁ and c₂ satisfy dₜ ∼ c₁/(log t)^c₂, disproving Erdős’s proposed asymptotic.
Verified in Lean through Conjectures. Full proof below.
🤖 Made with AI
ethosnode retweeted
Today, we’re announcing a solution found by our miners to Erdős Problem 416(i), a 52-year-old question.
The result proves that doubling the cutoff asymptotically doubles the number of distinct Euler totient values below it: V(2x)/V(x) → 2.
Verified in Lean through Conjectures. Full proofs below.
🤖 Made with AI
ethosnode retweeted
The second is Erdős Problem 653, a 29-year-old question.
The result proves that n points in the plane can have almost n different counts of distinct distances from individual points, as n grows.
Both verified in Lean through Conjectures. Full proofs below.
ethosnode retweeted
Conjectures has done a great job showing how incentive systems don't just parallelize search over solutions — they parallelize search over search systems themselves.
Traditionally, a search pipeline looks something like:
team → search algorithm → compute → solution
With incentives, this expands to:
[verifier + reward] → N competing search systems → solution
The latter is meta-search.
A traditional team can search over parameters inside its system: optimizers, architectures, prompts, agents, heuristics, search procedures, etc.
But many things remain fixed outside the search.
The team itself. Its ingenuity. Its hardware. Its capital. Its energy costs. Its infrastructure. Even the assumptions determining which algorithms it chooses to try.
An incentive system can push that boundary outward.
If anyone can compete and rewards are paid only for verified marginal progress, then the network searches not only over solutions, but over algorithms, teams, compute, infrastructure, and ultimately the organization of the search itself.
The abstraction is powerful because dimensions that are constants inside a single research organization become variables at the network level.
That larger search space has a cost.
But when the bottleneck to finding a solution lies in one of those additional dimensions, expanding the search space can dramatically accelerate convergence.
Congrats to the team show casing this to work so effectively with Conjectures.io









