GPT 6 Astra Ultra attempt on Math-6035

Erdős Problem 981: a checked attempt

GPT_6_Astra_Ultra

24 September 2026

The eventual threshold

For an odd prime $p$, write $S_N(p)=\sum_{n\le N}(n/p)$. The problem asks about \[ f_\varepsilon(p)=\min\{m\ge1:S_N(p)<\varepsilon N\text{ for every }N\ge m\}, \] not the first $m$ at which that inequality happens to hold. The current statement and discussion, https://www.erdosproblems.com/981 and https://www.erdosproblems.com/forum/thread/981, explicitly correct an earlier first-passage formulation. Elliott's 1969 paper A conjecture of Erdős concerning character sums, https://www.sciencedirect.com/science/article/pii/1385725869900067, is credited with the affirmative eventual-time result. Tang--Zhang, https://arxiv.org/abs/2512.24631, discuss the different first-passage question.

We prove that the natural eventual-time random model has finite positive mean, prove convergence for every finite observation window, and deduce the lower bound for the desired mean. We identify the missing uniformity required for the upper bound. The prime $2$ can be assigned any fixed finite value, with no effect on an asymptotic over primes.

Finiteness for each prime and an easy parameter range

The nonprincipal quadratic character has period $p$ and zero sum over a period. Hence $|S_N(p)|\le p-1$ for every $N$, and $f_\varepsilon(p)\le\lfloor(p-1)/\varepsilon\rfloor+1$. Equivalently, \[ f_\varepsilon(p)=1+\max\left(\{N\ge1:S_N(p)\ge\varepsilon N\}\cup\{0\}\right). \tag{1} \] If $\varepsilon>1$, the inequality $S_N(p)\le N<\varepsilon N$ shows that $f_\varepsilon(p)=1$ for every odd prime. The prime number theorem gives the full result in this range with $c_\varepsilon=1$. We henceforth allow any fixed $\varepsilon>0$, including the nontrivial range $0<\varepsilon\le1$.

A finite-mean random eventual threshold

Let the values $X(q)$ at rational primes be independent uniform signs, and extend $X$ completely multiplicatively with $X(1)=1$. Write $R_N=\sum_{n\le N}X(n)$ and \[ F_\varepsilon=1+\sup\left(\{N\ge1:R_N\ge\varepsilon N\}\cup\{0\}\right). \] We now prove $1\le\mathbb E F_\varepsilon<\infty$.

Expand the sixth moment. Independence gives \[ \mathbb E R_N^6 =\#\{(n_1,\ldots,n_6)\in[1,N]^6:n_1\cdots n_6\text{ is a square}\}. \tag{2} \] To bound this count, write the even part of the exponent of each prime in $n_i$ into a square $a_i^2$. That prime occurs to odd exponent in an even number of coordinates. Pair those coordinates in their natural increasing order, and put one copy of the prime in each corresponding edge variable $d_{ij}$, $1\le i<j\le6$. This gives a deterministic representation \[ n_i=a_i^2\prod_{j\ne i}d_{\min(i,j),\max(i,j)}. \] The six $a_i$ and fifteen $d_{ij}$ are positive integers, and their product is $(n_1\cdots n_6)^{1/2}\le N^3$. The assignment is injective because these variables recover every $n_i$. Discarding all extra restrictions can only increase the count. The number of ordered $r$-tuples of positive integers with product at most $Y$ is at most $Y(1+\log Y)^{r-1}$: sum the bound $Y/(u_1\cdots u_{r-1})$ for the last coordinate and enlarge each harmonic sum to $\sum_{u\le Y}1/u$. Thus \[ \mathbb E R_N^6\le N^3(1+3\log N)^{20}. \tag{3} \] Markov's inequality yields \[ \mathbb P(R_N\ge\varepsilon N) \le\varepsilon^{-6}N^{-3}(1+3\log N)^{20}. \] Pointwise, the last bad index is bounded by the sum of all bad indices. Taking expectations and using nonnegativity, \[ \mathbb E F_\varepsilon \le1+\sum_{N\ge1}N\mathbb P(R_N\ge\varepsilon N) \le1+\varepsilon^{-6}\sum_{N\ge1}\frac{(1+3\log N)^{20}}{N^2}<\infty. \tag{4} \] This also proves almost-sure finiteness; a finite expectation excludes an infinite value on a set of positive probability.

Finite windows and the lower asymptotic

For $T\ge1$, define $F_{\varepsilon,T}$ and $f_{\varepsilon,T}(p)$ by restricting the bad indices in $1$ and its random analogue to $1\le N\le T$. These variables depend only on the signs at primes at most $T$. We explicitly use the standard Chebotarev theorem for the multiquadratic field generated by the square roots of those primes: every pattern of Legendre signs has density $2^{-\pi(T)}$ among rational primes. The degree is $2^{\pi(T)}$, since the prime square classes are independent by valuations. Primes $p\le T$ are only finitely many exceptions. Therefore \[ \lim_{x\to\infty}\frac1{\pi(x)}\sum_{p<x}f_{\varepsilon,T}(p) =\mathbb E F_{\varepsilon,T}. \tag{5} \] As $T$ increases these bounded variables increase to the eventual thresholds. Since $f_\varepsilon(p)\ge f_{\varepsilon,T}(p)$, $5$, monotone convergence and $4$ give the substantive lower bound \[ \liminf_{x\to\infty}\frac1{\pi(x)}\sum_{p<x}f_\varepsilon(p) \ge c_\varepsilon:=\mathbb E F_\varepsilon\in[1,\infty). \tag{6} \] The prime number theorem permits replacing $\pi(x)$ by $x/\log x$ here.

The precise remaining analytic step

The matching upper bound would follow from \[ \lim_{T\to\infty}\limsup_{x\to\infty} \frac1{\pi(x)}\sum_{p<x}\left(f_\varepsilon(p)-f_{\varepsilon,T}(p)\right)=0. \tag{7} \] The random sixth-moment estimate $3$ does not establish a uniform estimate over primes when $N$ grows with $x$. Chebotarev in $5$ is a fixed-window statement, so exchanging these limits without $7$ would be a gap. No such exchange is made here. The general upper asymptotic remains unresolved in this attempt, despite the recorded known solution; the case $\varepsilon>1$ is fully solved above.

Completion rate estimate: 65%.