GPT 6 Astra Ultra attempt on Math-5405

Erdős Problem 67: discrepancy on homogeneous progressions

1) Statement and scope

For every $f:\mathbb N\to\{-1,1\}$, is \[ \sup_{d,m\geq1}\left|\sum_{j=1}^m f(jd)\right|=\infty? \] The answer is affirmative by Tao's discrepancy theorem. This note uses its stronger Hilbert-space formulation as an external theorem, gives the precise scalar and finite deductions, and proves the first nontrivial finite obstruction directly. It does not claim a new proof of Tao's analytic theorem.

2) External input and deduction

Tao proves that if $v_n$ are unit vectors in a real or complex Hilbert space, then the norms of $\sum_{j=1}^m v_{jd}$, over positive integers $d,m$, are unbounded. Apply this with the Hilbert space $\mathbb R$ and $v_n=f(n)$. Norm is absolute value, so this proves exactly the assertion, with no multiplicativity or periodicity hypothesis.

It also implies the following uniform finite statement. For each real $C>0$ there is an integer $N(C)$ such that every sign sequence on $[N(C)]$ has a homogeneous progression whose sum has absolute value greater than $C$. Suppose otherwise. Form the rooted binary tree of finite sign sequences for which all such sums have absolute value at most $C$. Every prefix of a vertex is again a vertex, and arbitrarily deep vertices exist. Starting at the root, choose a child with arbitrarily deep descendants; such a child exists because there are only two children. Iterating constructs an infinite sequence all of whose finite prefixes satisfy the bound. Each homogeneous progression lies in one prefix, contradicting the scalar theorem. This argument proves existence of $N(C)$; it does not supply an efficient bound for it.

3) An elementary certificate at discrepancy one

Assume all homogeneous progression sums inside $[12]$ have absolute value at most $1$. Every even-length such sum is an even integer in $[-1,1]$, hence zero. Subtracting successive even-length sums gives \[ f((2r-1)d)+f(2rd)=0\quad\text{whenever }2rd\leq12. \] Changing every sign if necessary, let $f(1)=1$. The displayed relations give, successively, \[ f(2)=-1,\quad f(4)=1,\quad f(8)=-1,\quad f(6)=1,\quad f(3)=f(5)=-1. \] The relation on $\{5,10\}$ gives $f(10)=1$, so that on $\{9,10\}$ gives $f(9)=-1$. Now $\{9,12\}$ forces $f(12)=1$, whereas $\{6,12\}$ forces $f(12)=-1$, a contradiction. Thus $N(1)\leq12$. The paired positions used here are respectively consecutive pairs inside a homogeneous progression, not arbitrary equal-spaced pairs.

There is a further elementary special case. If $f$ is eventually periodic with period $q$, then $f(kq)$ is a fixed sign for all sufficiently large $k$. Consequently $\sum_{k\leq m}f(kq)$ grows linearly in absolute value. This does not handle general nonperiodic sequences.

4) Outcome and sources

Outcome: SOLVED, as a deduction from the established Hilbert-space discrepancy theorem. The finite compactness deduction and the discrepancy-one certificate are proved above. The deep analytic input is not independently reproved or formally checked here. The computational check accompanying this batch exhausts sign sequences of length $12$ and verifies the finite certificate's conclusion.

• T. Tao, The Erdős discrepancy problem, Discrete Analysis 2016:1, https://arxiv.org/abs/1509.05363.
• T. F. Bloom, Problem 67, https://www.erdosproblems.com/67, checked 23 September 2026.

The estimate measures coverage using the stated external theorem, not an independent reconstruction of all its proof.

5) COMPLETION ESTIMATE

COMPLETION: 100%