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Erdős problem 990

Let f=a0++adxdC[x]f=a_0+\cdots+a_dx^d\in \mathbb{C}[x] be a polynomial. Is it true that, if ff has roots z1,,zdz_1,\ldots,z_d with corresponding arguments θ1,,θd[0,2π]\theta_1,\ldots,\theta_d\in [0,2\pi], then for all intervals I[0,2π]I\subseteq [0,2\pi] (#θiI)I2πd(nlogM)1/2, \left\lvert (\# \theta_i \in I) - \frac{\lvert I\rvert}{2\pi}d\right\rvert \ll \left(n\log M\right)^{1/2}, where nn is the number of non-zero coefficients of ff and M=a0++ad(a0ad)1/2. M=\frac{\lvert a_0\rvert+\cdots +\lvert a_d\rvert}{(\lvert a_0\rvert\lvert a_d\rvert)^{1/2}}.

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4 retained statements2415f78e850a

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FormalConjectures/ErdosProblems/

990.lean

Retained formal statement2 of 4

An internal OpenAI model (see [APSSV26b]) has disproved the conjecture, constructing, for every n1n\geq 1, a polynomial ff with nn non-zero coefficients such that M<3M<3 and with a positive real zero of multiplicity n1n-1.

FormalConjectures/ErdosProblems/990.leanErdos990.erdos_990.variants.counterexample3 linesExact file
∀ (n : ℕ),  1 ≤ nf, f.coeff 0 ≠ 0 ∧ f.support.card = nErdos990.M f < 3 ∧ ∃ x, 0 < xPolynomial.rootMultiplicity (↑x) f = n - 1
SolvedStatement only, no proof

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