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Erdős Problem #990

DISPROVED LEANdatabase updated 2026-04-19

The statement as posed is false. A solution has also been checked by machine, in Lean.

What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).

The number of roots of `f`, counted with multiplicity, whose argument lies in `I`. The argument is normalised to `[0, 2π)`.

For \(f = a_0 + \cdots + a_dx^d\), the quantity \(M=\frac{\lvert a_0\rvert+\cdots +\lvert a_d\rvert}{(\lvert a_0\rvert\lvert a_d\rvert)^{1/2}}\).

Let \(f=a_0+\cdots+a_dx^d\in \mathbb{C}[x]\) be a polynomial. Is it true that, if \(f\) has roots \(z_1,\ldots,z_d\) with corresponding arguments \(\theta_1,\ldots,\theta_d\in [0,2\pi]\), then for all intervals \(I\subseteq [0,2\pi]\) \[ \left\lvert (\# \theta_i \in I) - \frac{\lvert I\rvert}{2\pi}d\right\rvert \ll \left(n\log M\right)^{1/2}, \] where \(n\) is the number of non-zero coefficients of \(f\) and \[ M=\frac{\lvert a_0\rvert+\cdots +\lvert a_d\rvert}{(\lvert a_0\rvert\lvert a_d\rvert)^{1/2}}. \] An internal OpenAI model (see [APSSV26b]) has disproved the conjecture, constructing, for every \(n\geq 1\), a polynomial \(f\) with \(n\) non-zero coefficients such that \(M<3\) and with a positive real zero of multiplicity \(n-1\), whence letting \(I=[0,c/d]\) for a suitably small \(c>0\), \[ \left\lvert (\# \theta_i \in I) - \frac{\lvert I\rvert}{2\pi}d\right\rvert \geq n-1. \]
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research solved · 3 further variants / known partial results in the Lean file. The authoritative wording is on erdosproblems.com/990.
Prizenone
Topicsanalysis
Statement in LeanFormal Conjectures
Solution in Leanyes
Original pageerdosproblems.com/990
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