Problem Statement
Let $f(z)=\sum_{0\leq k\leq n} \epsilon_k z^k$ be a random polynomial, where $\epsilon_k\in \{-1,1\}$ independently uniformly at random for $0\leq k\leq n$.
Is it true that, if $R_n$ is the number of roots of $f(z)$ in $\{ z\in \mathbb{C} : \lvert z\rvert \leq 1\}$, then\[\frac{R_n}{n/2}\to 1\]almost surely?
Is it true that, if $R_n$ is the number of roots of $f(z)$ in $\{ z\in \mathbb{C} : \lvert z\rvert \leq 1\}$, then\[\frac{R_n}{n/2}\to 1\]almost surely?
Categories:
Analysis Polynomials Probability
Progress
Random polynomials with independently identically distributed coefficients are sometimes called Kac polynomials - this problem considers the case of Rademacher coefficients, i.e. independent uniform $\pm 1$ values. Erdős and Offord [EO56] showed that the number of real roots of a random degree $n$ polynomial with $\pm 1$ coefficients is $(\frac{2}{\pi}+o(1))\log n$.There is some ambiguity whether Erdős intended the coefficients to be in $\{-1,1\}$ or $\{0,1\}$ - see the comments section.
A weaker version of this was solved by Yakir [Ya21], who proved that\[\frac{R_n}{n/2}\to 1\]in probability. (This weaker claim was also asked by Erdős, and also appears in a book of Hayman [Ha67].) More precisely,\[\lim_{n\to \infty} \mathbb{P}(\lvert R_n-n/2\rvert \geq n^{9/10}) =0.\]See also [521].
Source: erdosproblems.com/522 | Last verified: January 15, 2026