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Math-6044 · mathematics · Mathematics

Erdős Problem #990

Status Solved · erdosproblems.comLLM status Solved from source status

Statement

Let f(x)=$\sum_{j=0}^d$ a_j x^j be a complex polynomial of degree d with a_0$\ne$ 0 and a_d$\ne$ 0. Let n be the number of nonzero coefficients a_j (so 2<=n<=d+1). Define

M := $\frac{\sum_{j=0}^d |a_j|}{\sqrt{|a_0 a_d|}}$.

Let z_1,...,z_d be the roots of f (counted with multiplicity), and write z_i = r_i $e^{i \theta_i}$ with $\theta_i$ $\in$ [0,2$\pi$).

Conjectured discrepancy bound: For every interval I$\subset$[0,2$\pi$],

\[ \Big| \#\{i: \theta_i \in I\} - \frac{|I|}{2\pi} d \Big| \ll \sqrt{n \log M}, \]

where the implied constant is absolute (independent of d,n,M) and |I| is the length of I.

Restatement from Math-6044-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 2

Math-6044-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-6044

Erdős Problem 990: a sparse polynomial counterexample

Statement and scope

For a polynomial $f(z)=\sum_{j=0}^{d}a_jz^j$ with $a_0a_d\ne0$, let $\nu(f)$ be its number of nonzero coefficients and \[ M(f)=\frac{\sum_j|a_j|}{\sqrt{|a_0a_d|}}. \] The proposed angular discrepancy bound $O(\sqrt{\nu(f)\log M(f)})$ is false. We reconstruct the complete counterexample in Section 5 of Alexeev--Putterman--Sawhney--Sellke--Valiant, arXiv:2604.06609, which credits an internal OpenAI model. No novelty is claimed.

An interpolation identity

For distinct real numbers $\lambda_0<\cdots<\lambda_{s-1}$, put $P_j=\prod_{\ell\ne j}(\lambda_j-\lambda_\ell)$ and $\Delta_j=|P_j|$. Comparing the coefficient of $X^{s-1}$ in the Lagrange interpolation formula for $X^m$ gives \[ \sum_{j=0}^{s-1}\frac{\lambda_j^m}{P_j} = \begin{cases}0,&0\le m<s-1,\\1,&m=s-1.\end{cases} \tag{1} \] This identity will prescribe the multiplicity exactly.

Exponents and coefficients

Fix $N\ge1$ and let an integer $K\to\infty$. Set $s=N+2$, $d=K^N$, and choose the $s$ exponents \[ m_0=0,\quad m_i=K^{i-1}\ (1\le i\le N),\quad m_{N+1}=d, \qquad \lambda_j=m_j/d. \] Define \[ \tau=-\log(2\Delta_{N+1}/\Delta_0),\qquad f_{N,K}(z)=\sum_{j=0}^{N+1} \frac{(-1)^{N+1-j}\Delta_0}{\Delta_j}e^{-\lambda_j\tau}z^{m_j}. \tag{2} \] Every coefficient is nonzero, the constant coefficient has modulus $1$, and the leading coefficient is exactly $2$. The exponential polynomial $F(u)=f_{N,K}(e^{u/d})$ satisfies, by $1$, $F^{(m)}(\tau)=0$ for $0\le m\le N$ and $F^{(N+1)}(\tau)=\Delta_0\ne0$. Because $z=e^{u/d}$ is locally invertible at $u=\tau$, the positive number $z_0=e^{\tau/d}$ is a root of multiplicity exactly $N+1$.

The coefficient height stays below three

Put $\varepsilon=K^{-1}$ and $x_i=\lambda_i=\varepsilon^{N+1-i}$ for $1\le i\le N$. For fixed $N$, \[ \Delta_0=\prod_{i=1}^{N}x_i=\varepsilon^{N(N+1)/2}, \qquad \Delta_{N+1}=\prod_{i=1}^{N}(1-x_i)=1+O_N(\varepsilon). \] Factoring the differences $x_i-x_j$ according to $j<i$ and $j>i$ gives \[ \Delta_i=x_i^i\prod_{j>i}x_j\,(1+O_N(\varepsilon)),\qquad \frac{\Delta_0}{\Delta_i} =\varepsilon^{i(i-1)/2}(1+O_N(\varepsilon)). \] Also $|\tau|=O_N(|\log\varepsilon|)$ and $x_i\le\varepsilon$, so $e^{-x_i\tau}=1+o_N(1)$ for every middle coefficient. The coefficient for $m_1=1$ consequently has modulus tending to $1$, while each middle coefficient with $i\ge2$ has modulus tending to zero. Including the two endpoints, \[ \sum_j|a_j|\longrightarrow 1+1+2=4,\qquad M(f_{N,K})\longrightarrow 2\sqrt2<3. \] Choose $K=K(N)$ sufficiently large that $M(f_{N,K})<3$; no uniform choice of $K$ in $N$ is needed.

Angular discrepancy

Take arguments in $[0,2\pi)$ and the interval $I_N=[0,\pi/d)$. All $N+1$ copies of the positive root lie in this interval, while the uniform prediction is $d|I_N|/(2\pi)=1/2$. Thus the discrepancy is at least $N+1/2$. Were the proposed inequality true with an absolute constant $C$, it would imply \[ N+\tfrac12\le C\sqrt{(N+2)\log3} \] for every $N$, a contradiction.

Conclusion and verification scope

The conjecture is disproved by an explicit family, including a proof of exact root multiplicity and bounded height. Taking a positive-length interval subtracts its expected count; the argument does not incorrectly claim discrepancy at least the full multiplicity. Sources: https://www.erdosproblems.com/990 and https://arxiv.org/abs/2604.06609. No local Lean verification is claimed.

COMPLETION ESTIMATE

COMPLETION: 100%

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Math-6044-A2 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-6044

1) FORMAL RESTATEMENT

Let f(x)=$\sum_{j=0}^d$ a_j x^j be a complex polynomial of degree d with a_0$\ne$ 0 and a_d$\ne$ 0. Let n be the number of nonzero coefficients a_j (so 2<=n<=d+1). Define

M := $\frac{\sum_{j=0}^d |a_j|}{\sqrt{|a_0 a_d|}}$.

Let z_1,...,z_d be the roots of f (counted with multiplicity), and write z_i = r_i $e^{i \theta_i}$ with $\theta_i$ $\in$ [0,2$\pi$).

Conjectured discrepancy bound: For every interval I$\subset$[0,2$\pi$],

\[ \Big| \#\{i: \theta_i \in I\} - \frac{|I|}{2\pi} d \Big| \ll \sqrt{n \log M}, \]

where the implied constant is absolute (independent of d,n,M) and |I| is the length of I.

2) QUICK LITERATURE/CONTEXT CHECK

The problem file states that Erdős--Turan proved the same inequality with n replaced by the degree d (i.e. discrepancy $\ll$ $\sqrt{d \log M}$). The question is whether one can replace d by the number n of nonzero coefficients, i.e. obtain a sparsity-sensitive bound. No further resolution is stated in the problem file.

3) ATTACK PLAN

Two complementary routes: (A) Special cases: prove the inequality for very sparse polynomials (binomials, trinomials, etc.) and see how n and M enter. (B) Disproof attempt: try to build a polynomial with small n and moderate M whose roots have angles highly concentrated in a short arc, forcing discrepancy $\gg$ d.

In this writeup I prove two elementary lemmas (one about M itself, one fully resolving the binomial case n=2), and I report small numerical sanity checks for random sparse polynomials.

4) WORK

FAST REALITY CHECK (small computations)

I sampled random sparse polynomials with a_0=a_d=1, and the remaining $n-2$ coefficients chosen as small random nonzero integers at random intermediate exponents. For each polynomial I computed its roots numerically, then computed the exact angular discrepancy max_{interval I} |#{$\theta_i\in$ I} - $|I|/(2\pi)$d|. I also computed the ratio discrepancy / sqrt(n log M).

Exact outputs for degrees 20,30 and n=3,4 (5 trials each; fixed random seed 0):

--- d=20, n=3 (a0=ad=1, other coeffs random integers) ---
trial 0: M=5.000000, disc=1.000000, disc/sqrt(n log M)=0.455
trial 1: M=4.000000, disc=0.745607, disc/sqrt(n log M)=0.366
trial 2: M=3.000000, disc=0.666667, disc/sqrt(n log M)=0.367
trial 3: M=5.000000, disc=2.000000, disc/sqrt(n log M)=0.910
trial 4: M=3.000000, disc=0.642143, disc/sqrt(n log M)=0.354

--- d=20, n=4 (a0=ad=1, other coeffs random integers) ---
trial 0: M=7.000000, disc=1.000000, disc/sqrt(n log M)=0.358
trial 1: M=6.000000, disc=1.000000, disc/sqrt(n log M)=0.374
trial 2: M=6.000000, disc=2.000000, disc/sqrt(n log M)=0.747
trial 3: M=5.000000, disc=0.887882, disc/sqrt(n log M)=0.350
trial 4: M=7.000000, disc=1.602152, disc/sqrt(n log M)=0.574

--- d=30, n=3 (a0=ad=1, other coeffs random integers) ---
trial 0: M=5.000000, disc=0.828288, disc/sqrt(n log M)=0.377
trial 1: M=5.000000, disc=0.942166, disc/sqrt(n log M)=0.429
trial 2: M=5.000000, disc=1.000000, disc/sqrt(n log M)=0.455
trial 3: M=4.000000, disc=1.000000, disc/sqrt(n log M)=0.490
trial 4: M=4.000000, disc=1.000000, disc/sqrt(n log M)=0.490

--- d=30, n=4 (a0=ad=1, other coeffs random integers) ---
trial 0: M=6.000000, disc=1.000000, disc/sqrt(n log M)=0.374
trial 1: M=8.000000, disc=1.333281, disc/sqrt(n log M)=0.462
trial 2: M=8.000000, disc=1.326786, disc/sqrt(n log M)=0.460
trial 3: M=5.000000, disc=1.260708, disc/sqrt(n log M)=0.497
trial 4: M=7.000000, disc=0.958378, disc/sqrt(n log M)=0.344

These experiments are not evidence either way (degrees are tiny and roots are numeric), but at least they do not immediately contradict the conjectured scale $\sqrt{n\log M}$.

Lemma 990.1 (basic properties of M)

(i) Scale invariance: For any nonzero complex constant c, if g(x)=c f(x) then M(g)=M(f). (ii) Lower bound: M$\ge$ 2.

Proof. (i) If g_j=c a_j then $\sum$ |g_j|=|c|$\sum$|a_j| and $\sqrt{|g_0 g_d|}$=|c|$\sqrt{|a_0 a_d|}$, so the ratio M is unchanged. (ii) Since a_0 and a_d are among the coefficients, $\sum_{j=0}^d$ |a_j| $\ge$ |a_0|+|a_d|. By AM-GM, |a_0|+|a_d| $\ge$ 2$\sqrt{|a_0 a_d|}$. Dividing by $\sqrt{|a_0 a_d|}$ gives M$\ge$ 2. $\square$

Lemma 990.2 (binomial case n=2)

Assume f(x)=a_0+a_d x^d with a_0 a_d $\ne$ 0 (so n=2). Then for every interval I$\subset$[0,2$\pi$],

\[ \Big| \#\{i: \theta_i \in I\} - \frac{|I|}{2\pi} d \Big| \le 1. \]

In particular, the conjectured inequality holds for n=2 with implied constant 1.

Proof. The roots satisfy x^d = -a_0/a_d. Write -a_0/a_d = R $e^{i\phi}$ with R>0 and $\phi\in$[0,2$\pi$). Then the d roots are z_m = R^{1/d} exp$i(\phi+2\pi m)/d$, m=0,1,...,d-1. Thus the arguments are an arithmetic progression modulo 2$\pi$ with step 2$\pi$/d; equivalently, there is exactly one root in each half-open arc of length 2$\pi$/d.

Fix an interval I of length L=|I|. The expected number under uniform distribution is d L/$2\pi$. Because the points are equally spaced with spacing 2$\pi$/d, the actual count differs from the expected count by at most 1 (a standard lattice-point-in-interval fact on the circle). Concretely: if we identify angles with residues m/d in [0,1), then counting $\theta_m$ in I is counting integers m in an interval of length d L/$2\pi$, which differs from its length by <1. Hence the discrepancy is <=1.

Now compute M. Here $\sum$ |a_j|=|a_0|+|a_d| and n=2. By Lemma 990.1(ii), M>=2, so sqrt{n log M} = sqrt{2 log M} >= sqrt{2 log 2} > 1. Therefore discrepancy <=1 <= sqrt{2 log M} = sqrt{n log M}, which is the claimed inequality with implied constant 1. $\square$

5) VERIFICATION

- Lemma 990.2 uses only the explicit root formula for a binomial; multiplicities do not occur because x^d=constant has d distinct solutions when the constant is nonzero. - The AM-GM step in Lemma 990.1(ii) is exact.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result here: - Full proof of the conjectured inequality for n=2 (binomials), with discrepancy <=1. - Structural facts: M is scale-invariant and M>=2 always. - Small numerical sanity checks for random sparse polynomials show discrepancy often of size O$1$ when coefficients are tiny, consistent with $\sqrt{n\log M}$.

(ii) First gap (crisp statement): Prove that in the Erdős--Turan method controlling angular discrepancy, the quantity d can be replaced (up to absolute constants) by the number n of nonzero coefficients. Equivalently, find a Fourier-analytic bound on the relevant trigonometric sums in terms of n and M only.

(iii) Top 3 next moves (concrete): 1. Work out the Erdős--Turan proof line-by-line and locate the precise point where the degree d appears; attempt to replace that step by a sparsity-sensitive estimate. 2. Study the trinomial family f(x)=a_0+a_m x^m + a_d x^d: attempt to prove the bound with n=3 or produce a family where angles are highly non-uniform while M stays moderate. 3. Search systematically (computationally) for sparse polynomials with small n and small M but large angular discrepancy, to guide either a proof strategy or a disproof.

(iv) Minimal counterexample structure: A family of polynomials with fixed small n (say n=3 or 4), with M growing at most moderately (e.g. log M = O(log d)), but whose root arguments concentrate in an arc of length o$1$, forcing discrepancy $\gg$ d, contradicting O$\sqrt{n\log M}$.

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