GPT 6 Astra Ultra attempt on Math-5487
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Erdős Problem #192
1. FORMAL RESTATEMENT
Let $a_0,a_1,\ldots\in\mathbb R^d$ satisfy $a_{n+1}-a_n\in\{e_1,\ldots,e_d\}$. For which positive integers $d$ must its vertex set contain a nonconstant three-term arithmetic progression?2. LITERATURE AND SCOPE
The classification is $d\le3$. We prove the exact correspondence with abelian squares and the small-alphabet obstruction in full. For $d\ge4$ we explicitly use Keränen's 1992 theorem that an infinite abelian-square-free word exists on four letters. Its morphism verification is an external theorem, not a computation claimed here.3. WALK--WORD EQUIVALENCE
Encode step $e_j$ by letter $j$. The displacement along a finite block is its vector of letter counts. If $i<j<k$, then \[ a_i+a_k=2a_j \quad\Longleftrightarrow\quad (\text{counts in steps }i,\ldots,j-1) =(\text{counts in steps }j,\ldots,k-1). \] Equality of counts also gives $j-i=k-j$. Hence these consecutive blocks are permutations of each other: an abelian square. Conversely every such square gives the displayed progression. Any three-term progression in the vertex set is caught by this argument, since the sum of the coordinates of $a_n-a_0$ is $n$: its midpoint has the middle index.4. THE SHARP TERNARY BOUND
Every ternary word of length eight contains an abelian square. For completeness the entire finite proof is given below. Rename letters in order of first occurrence as $0,1,2$. Starting with the empty word, append an allowed letter and discard a word whenever its last $2t$ letters split into two blocks with equal count vectors. This tests precisely the new factors created by appending a letter. The surviving canonical words are \[ \begin{array}{c|l} 1&0\\ 2&01\\ 3&010,\ 012\\ 4&0102,\ 0120,\ 0121\\ 5&01020,\ 01021,\ 01201,\ 01202,\ 01210\\ 6&010201,\ 010210,\ 010212,\ 012010,\ 012101\\ 7&0102010,\ 0102101,\ 0121012\\ 8&\varnothing. \end{array} \] Each row follows by at most three extensions of each word in the preceding row and comparison of count vectors for $1\le t\le\lfloor n/2\rfloor$. This is a finite exhaustion with its full state list, not an assumption based on random sampling. A separate exact script checks all $3^8$ words as well.In particular nine consecutive vertices of a three-dimensional positive-unit walk already contain a progression. Dimensions one and two embed into dimension three.
The threshold is eight letters, not seven: $0102010$ avoids abelian squares. The current problem-page sentence asserting a seven-letter threshold is off by one; this does not affect its dimension classification.
5. FOUR AND MORE DIMENSIONS
By Keränen's theorem choose an infinite word $w_0w_1\cdots$ on four letters with no abelian square. Define $a_0=0$ and $a_{n+1}=a_n+e_{w_n}$. The equivalence proved above excludes every three-term progression. Appending zero coordinates gives a counterexample for every $d>4$.6. FINAL AND DEPENDENCY
SOLVED: the forced dimensions are exactly $d\le3$, relative to the explicitly cited Keränen theorem. The finite obstruction and both geometric reductions are proved here; the infinite four-letter word theorem is not reproved. This is an attributed reconstruction, with no novelty claim. The completion estimate refers to the classification with this standard external input.7. SOURCES
Current statement,https://www.erdosproblems.com/192. V. Keränen, Abelian squares are avoidable on 4 letters, ICALP 1992, LNCS 623, 41--52, https://doi.org/10.1007/3-540-55719-9_62; the publisher's abstract explicitly states the infinite-word theorem and its 85-uniform morphism construction.