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Lipschitz graphs without three-term progressions

Find a function N → Z with bounded steps whose graph has no three-term progression, or prove none exists (the additive square problem).

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  1. OpenedOct 21, 2026
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  4. Verified0 routes passing
  5. Adopted1 merged

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Is there a function from the natural numbers to the integers that moves by a bounded amount at each step, yet whose graph never contains three equally spaced points on a line? Equivalently: can an infinite sequence drawn from a finite set of integers avoid two adjacent blocks of the same length with the same sum? The analogous question for three adjacent blocks was answered in 2014; for two blocks it is open.

Problem

Does there exist a function f:N→Zf : \mathbb{N} \to \mathbb{Z} and a constant KK with ∣f(n+1)−f(n)∣≤K|f(n+1) - f(n)| \le K for all nn, such that the graph Γ={(n,f(n)):n∈N}⊂Z2\Gamma = \{(n, f(n)) : n \in \mathbb{N}\} \subset \mathbb{Z}^2 contains no three-term arithmetic progression? That is, there should be no n≥1n \ge 1 and d≥1d \ge 1 with

f(n)+f(n+2d)=2f(n+d).f(n) + f(n + 2d) = 2 f(n + d).

Word formulation. Put xn=f(n+1)−f(n)x_n = f(n+1) - f(n), so xn∈{−K,…,K}x_n \in \{-K, \dots, K\}. Then Γ\Gamma has a three-term progression exactly when the word x1x2x3⋯x_1 x_2 x_3 \cdots contains an additive square: two adjacent blocks xm⋯xm+d−1x_{m} \cdots x_{m+d-1} and xm+d⋯xm+2d−1x_{m+d} \cdots x_{m+2d-1} of the same length and the same sum. So the question is whether additive squares can be avoided by an infinite word over some finite set of integers. By a compactness argument, it is the same to ask for arbitrarily long finite words over one fixed finite alphabet.

Background

Green first heard the question from Jacob Fox in 2005 and again from Natasha Morrison, and it was considered in print by Brown, Jungić and Poelstra (2014). In combinatorics on words, Pirillo and Varricchio (1994) asked whether additive squares or additive cubes can be avoided, motivated by questions about semigroups, and Halbeisen and Hungerbühler (2000) asked again about squares. Every abelian square (two adjacent blocks that are rearrangements of each other) is an additive square, so the problem strengthens the classical theory of avoiding abelian powers. Green notes that tempting constructions starting from Behrend-type sets seem to fail.

What is known

  • Additive cubes are avoidable: Cassaigne, Currie, Schaeffer and Shallit (2014) constructed an infinite word over {0,1,3,4}\{0, 1, 3, 4\} with no three adjacent blocks of the same length and sum. Taking partial sums gives a Lipschitz function whose graph has no four-term progression.
  • An alphabet avoiding additive squares needs at least four letters, since abelian squares cannot be avoided over three letters (they can over four: Keränen, 1992).
  • Over four letters {a,b,c,d}\{a, b, c, d\} with a+d=b+ca + d = b + c, every word of length more than 60 contains an additive square (Freedman, as reported by Cassaigne et al.).
  • Zhang (2026, unrefereed preprint) classified the four-letter alphabets over which every additive-square-free word is shorter than 139; the cutoff is sharp, since the longest such word over {0,2,3,7}\{0, 2, 3, 7\} has length exactly 139.
  • The general problem is still described as open in 2025 (Vukusic), who solved a measure-theoretic variant in which it fails.

What would count as progress

  • An infinite word over an explicit finite set of integers with no additive square, for instance a fixed point of a morphism, with a proof that can be checked by computer (as was done for additive cubes, or with automatic-sequence tools). This answers the question positively.
  • Long finite additive-square-free words over small alphabets. Any proposed word of length nn can be checked in O(n2)O(n^2) time using prefix sums.
  • Computer-assisted proofs that every sufficiently long word over a given alphabet contains an additive square, ruling out that alphabet.
  • A formal proof of the Lean statement green_15 in GreensOpenProblems/15.lean (google-deepmind/formal-conjectures), or of its negation.

References

  1. G. Pirillo and S. Varricchio, On uniformly repetitive semigroups, Semigroup Forum 49 (1994), 125–129.
  2. J. Cassaigne, J. D. Currie, L. Schaeffer and J. Shallit, Avoiding three consecutive blocks of the same size and same sum, J. ACM 61 (2014), Art. 10; arXiv:1106.5204.
  3. T. Brown, V. Jungić and A. Poelstra, On double 3-term arithmetic progressions, Integers 14 (2014), A43.
  4. I. Vukusic, A Lebesgue variant of the additive square problem, arXiv:2506.21200 (2025).
  5. E. Zhang, The additive-square-free spectrum below 139, arXiv:2610.02215 (2026).

Source

Imported from Ben Green's 100 open problems 15, curated by Ben Green. Open according to the source as of 2026-10-20. The statement above is written in our own words; see the source for the original wording, history and discussion.

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