GPT 6 Astra Ultra attempt on Math-5482
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Erdős Problem 185: avoiding collinear triples in a ternary cube
1) Statement and scope
Let $f_3(n)$ be the largest size of a subset of $\{0,1,2\}^n$ with no three distinct collinear points. Is $f_3(n)=o(3^n)$? We prove the answer is yes using density Hales--Jewett as an explicit external theorem. We also reconstruct the elementary spherical-layer lower bound $f_3(n)\gg3^n/\sqrt n$.2) Deduction of the upper bound
The density Hales--Jewett theorem for three symbols states that for every $\epsilon>0$ there is an $N_0(\epsilon)$ such that, for all $n\geq N_0(\epsilon)$, every subset of $\{0,1,2\}^n$ of size at least $\epsilon3^n$ contains a combinatorial line. Its variable coordinates form a nonempty set $S$. If $b$ has the fixed coordinates of the line and zero in $S$, the three points are \[ b,\quad b+\mathbf1_S,\quad b+2\mathbf1_S. \] They are distinct and collinear in $\mathbb R^n$. Therefore a set with no collinear triple has size less than $\epsilon3^n$ whenever $n\geq N_0(\epsilon)$. Since this holds for every positive $\epsilon$, it is exactly $f_3(n)/3^n\to0$.3) An explicit lower bound from a sphere
For an integer $r$, consider \[ S_r=\left\{x\in\{0,1,2\}^n: \sum_{i=1}^n(x_i-1)^2=r\right\}. \] A Euclidean sphere has no three distinct collinear points: restricting its defining squared norm to a line with nonzero direction gives a quadratic equation with positive leading coefficient, hence at most two roots. Thus every $S_r$ is an admissible set.Choose $x$ uniformly from the cube and let $X=\sum_i(x_i-1)^2$. Its summands are independent, equal to one with probability $2/3$, so $\mathbb EX=2n/3$ and $\operatorname{Var}X=2n/9$. Chebyshev's inequality gives \[ \mathbb P(|X-2n/3|<\sqrt n)\geq7/9. \] The indicated interval contains at most $2\sqrt n+2$ integers. At least one layer therefore has size at least \[ \frac{7\cdot3^n}{9(2\sqrt n+2)} \geq\frac7{36}\frac{3^n}{\sqrt n}\qquad(n\geq1). \] The lower and upper bounds are compatible, since the density of this construction tends to zero. They do not determine an asymptotically exact formula for $f_3(n)$.
4) Outcome and sources
Outcome: SOLVED using density Hales--Jewett. The geometric deduction and the spherical construction are proved in full. The general density theorem is not reproved here; the separate attempt for Problem 171 only proves its two-symbol special case. No novelty or local formal-verification claim is made.- D. H. J. Polymath, A new proof of the density Hales--Jewett theorem,
https://annals.math.princeton.edu/2012/175-3/p06,https://arxiv.org/abs/0910.3926; the original theorem is due to Furstenberg and Katznelson. - T. F. Bloom, Problem 185,
https://www.erdosproblems.com/185, checked 23 September 2026.