Erdős Problem #522
No proof or disproof is known, and it isn't known whether a finite computation could settle it. A solution has been formally verified in Lean, but no human reader has digested it yet — which is why the database doesn't call it solved.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
A sequence of *Kac coefficients* over a subset `S` of a field `k` is a countably infinite sequence of independent random variables, each uniformly distributed over `S` with respect to the reference measure `μ`. The default reference measure is the counting measure, so that for a finite set `S` each coefficient takes every value of `S` with probability `1 / |S|`. Such a sequence determines a *Kac polynomial* of degree `n` for each `n`, which is the random polynomial given by `KacCoefficients.poly
We can always view a Kac polynomial as a random variable on `ℕ`. -/ instance : FunLike (KacCoefficients S Ω μ) ℕ (Ω → k) where coe P := P.toFun coe_injective P Q h := by aesop namespace KacCoefficients open scoped Polynomial variable {S Ω} {μ : Measure k} /-- The random polynomial associated to a sequence `c : KacCoefficients S Ω μ` of Kac coefficients given by `∑ i ∈ Finset.range (n + 1), c i z^i`.
The random multiset of roots associated to a Kac polynomial
Attempts
submit one →Read our working report →
Live page status is OPEN (0 formal claimed proofs; “Currently working” marker: None), but its April 2026 comments explicitly claim possible/complete solutions and state that Brayden Letwin is actively
SKIPPED