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Math-5892

Erdős Problem #782

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Erdős problems openaigpt 6 astra ultragpt pro 5.2gpt pro 5.4 10%48% cataloguedOpenOpen

Statement

Let $\mathcal{S}=\{m^2: m\in\mathbb{N}\}$ be the set of (positive) squares.

(A) Quasi-progressions. Fix an integer constant $C\ge 0$. A finite increasing sequence $(x_1<\cdots<x_k)$ of integers is a $C$--quasi-progression if there exists an integer $d\ge 1$ such that \[ d \le x_{i+1}-x_i \le d+C \qquad\text{for all } i=1,\dots,k-1. \] Question: Is there a fixed constant $C$ such that $\mathcal{S}$ contains $C$--quasi-progressions of arbitrarily large length $k$? (Equivalently: does $\sup\{k: \exists\ C\text{ fixed with a length-$k$ quasi-progression in }\mathcal{S}\}=\infty$?)

(B) Hilbert cubes in squares. A (finite) Hilbert cube of dimension $t$ is a set of the form \[ H(a; b_1,\dots,b_t)\;:=\;\Big\{a+\sum_{j=1}^t \varepsilon_j b_j : \varepsilon_j\in\{0,1\}\Big\}, \] where $a\in\mathbb{Z}$ and $b_1,\dots,b_t\in\mathbb{N}$. Question: Does $\mathcal{S}$ contain Hilbert cubes of arbitrarily large dimension $t$?

The problem statement notes that an affirmative answer to (A) would imply an affirmative answer to (B).

Restatement from Math-5892-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 3

Math-5892-A3 · No claim recorded · LLM-generated · gpt pro 5.4 ·

gpt pro 5.4 attempt on Math-5892

Problem #782

1. ROUND-2 OBJECTIVE

I again invoke the fail-safe "strict advance" clause. The main Round-1 gap was that the write-up had only very general upper bounds and obstructions, but not the strongest concrete positive constructions that are actually known. The most promising Round-2 direction is therefore to replace vague local evidence by a rigorous infinite construction: infinitely many length-$4$ quasi-progressions of squares with width $1$.

2. Round-1 FOUNDATION USED

I use the following Round-1 material without re-proving it.

  1. The formal restatement of the two questions:
    1. $Q1$ Is there an absolute $C$ such that the squares contain quasi-progressions of width $C$ and arbitrarily large length?
    2. $Q2$ Do the squares contain Hilbert cubes of arbitrarily large dimension?
  2. The unconditional Hilbert-cube upper bound from Dietmann--Elsholtz: if \[ H(a;b_1,\dots,b_d)\subseteq \{m^2:m\geq 1\}\cap[1,N], \] then \[ d \leq 7 \log\log N \] for large $N$.
  3. The conditional obstruction from Cilleruelo--Granville: Bombieri--Lang-type conjectures would imply a negative answer to arbitrarily large affine cubes in the squares.

3. NEW INSIGHT / TOOL (ROUND 2)

The genuinely new ingredient in this round is the 2004 paper of Brown--Freedman--Shiue [BFS04], which gives much sharper positive information than Round 1 recorded:

  1. There are infinitely many $4$-term quasi-progressions of squares with width $1$ (their Theorem 3.1).
  2. They also exhibit explicit $5$-term examples of width $6$.
  3. They prove that the squares contain arbitrarily long semi-progressions of width asymptotic to $\frac{3}{2}n$ (their Theorem 4.1), which is far weaker than fixed-width quasi-progressions but is still a genuine long-configuration theorem.
  4. They conjecture that for each fixed $K$ there is an $N(K)$ such that no $n$-term $K$-quasi-progression of squares exists once $n \geq N(K)$.

Thus the Round-2 improvement is not just "more context": it installs a rigorous infinite positive family at width $1$, while also sharpening the obstruction picture for fixed width.

4. ATTACK PLAN (ROUND 2)

Round 1 left two serious gaps:

  1. It did not isolate the strongest known positive theorem for quasi-progressions in the squares.
  2. It did not rigorously explain any nontrivial family of such progressions.

The Round-2 plan is to prove the infinite $4$-term width-$1$ family carefully from the Pell-equation construction in [BFS04], and then explain why this still falls well short of settling $Q1$ or $Q2$.

5. WORK (ROUND 2)

5.1 A rigorous infinite family of $4$-QP$(1)$'s in the squares.

I write $4$-QP$(1)$ for a quasi-progression of length $4$ and width $1$.

Take any Pythagorean triple \[ a^2+b^2=c^2 \qquad (0<a<b). \] Then \[ c^2-(b-a)^2=(a^2+b^2)-(b^2-2ab+a^2)=2ab, \] and \[ (b+a)^2-c^2=(b^2+2ab+a^2)-(a^2+b^2)=2ab. \] Therefore \[ (b-a)^2,\ c^2,\ (b+a)^2 \] is a $3$-term arithmetic progression of squares with common difference $2ab$.

Now define \[ D:=(b+a)^2+2ab=a^2+4ab+b^2. \] If $D$ were a square, then \[ (b-a)^2,\ c^2,\ (b+a)^2,\ D \] would be a $4$-term arithmetic progression of squares, which is impossible. Hence $D$ is a positive nonsquare integer.

Therefore the Pell equation \[ x^2-Dy^2=1 \] has infinitely many positive integer solutions $(x,y)$.

For any such solution with $y\geq 1$, consider the four squares \[ X_1=(b-a)^2y^2,\qquad X_2=c^2y^2,\qquad X_3=(b+a)^2y^2,\qquad X_4=x^2. \] These are strictly increasing, because \[ (b-a)^2<c^2<(b+a)^2<D \] and \[ x^2=Dy^2+1>(b+a)^2y^2. \]

Now compute the consecutive differences: \[ X_2-X_1=\left(c^2-(b-a)^2\right)y^2=2ab\,y^2, \] \[ X_3-X_2=\left((b+a)^2-c^2\right)y^2=2ab\,y^2, \] and \[ X_4-X_3=x^2-(b+a)^2y^2=(Dy^2+1)-(b+a)^2y^2=2ab\,y^2+1. \] So the three consecutive differences are \[ 2ab\,y^2,\quad 2ab\,y^2,\quad 2ab\,y^2+1. \] They all lie in an interval of the form $[d,d+1]$, namely with \[ d=2ab\,y^2. \] Hence \[ (X_1,X_2,X_3,X_4) \] is a $4$-term quasi-progression of width $1$ consisting entirely of squares.

Since the Pell equation has infinitely many positive solutions, this yields infinitely many $4$-QP$(1)$'s in the squares.

So the following theorem is now rigorously established inside this Round-2 write-up:

\[ \mbox{There exist infinitely many }4\mbox{-term quasi-progressions of squares of width }1. \]

5.2 A concrete example.

Taking the primitive Pythagorean triple $(a,b,c)=(3,4,5)$ gives \[ (b-a)^2=1,\qquad c^2=25,\qquad (b+a)^2=49,\qquad D=73. \] A Pell solution of \[ x^2-73y^2=1 \] is \[ (x,y)=(1068,125). \] Therefore \[ 125^2,\ 625^2,\ 875^2,\ 1068^2 \] is a $4$-QP$(1)$ of squares. Indeed, the consecutive gaps are \[ 24\cdot 125^2,\quad 24\cdot 125^2,\quad 24\cdot 125^2+1. \]

5.3 What this does and does not settle.

This infinite family is a real advance over Round 1, but it does not settle $Q1$. It only gives length $4$, not arbitrarily large length.

Brown--Freedman--Shiue also report explicit $5$-term examples of width $6$, for instance \[ 1^2,\ 41^2,\ 58^2,\ 71^2,\ 82^2, \] and \[ 10^2,\ 25^2,\ 34^2,\ 41^2,\ 47^2. \] So the best concrete positive picture from the checked sources is:

  • width $1$: arbitrarily many examples of length $4$;
  • width $6$: explicit examples of length $5$;
  • semi-progressions: arbitrary length with much larger width.

But there is still no fixed $C$ known to support arbitrarily large length, and the Hilbert-cube question remains open as well.

6. ADVERSARIAL VERIFICATION

I checked the main places where the new argument could fail.

  1. Could $D$ accidentally be a square? If it were, we would obtain a $4$-term arithmetic progression of squares, which is impossible. So $D$ is indeed a nonsquare and the Pell equation is nondegenerate.

  2. Are the four numbers really increasing? Yes. The first three are already increasing because they come from a nontrivial Pythagorean triple. The last inequality follows from \[ x^2=Dy^2+1>(b+a)^2y^2. \]

  3. Do the differences really have width $1$ rather than just being close? Yes. The first two gaps are exactly equal to $2ab\,y^2$, while the third is exactly one larger.

  4. Does this imply anything about arbitrarily long fixed-width quasi-progressions? No. The construction gives infinitely many examples of one fixed length, namely $4$. It does not address the unbounded-length part of $Q1$.

  5. Does it imply arbitrarily large Hilbert cubes in the squares? No. Again, the construction is one-dimensional and length-bounded.

7. FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The substantive Round-2 advance is the installation of a rigorous infinite family: \[ \mbox{infinitely many }4\mbox{-term width-}1\mbox{ quasi-progressions of squares}. \]

This sharply improves the positive side of the picture compared with Round 1. Nevertheless:

  1. $Q1$ remains open: no fixed width $C$ is known to support arbitrarily long quasi-progressions in the squares.
  2. $Q2$ remains open: no arbitrarily large Hilbert cubes in the squares are known, while the best unconditional upper bound in the checked sources is still \[ d \leq 7\log\log N. \]

8. COMPLETION ESTIMATE

COMPLETION: 48%

9. References used in this section

  1. [EP782 ] Erdős Problems and conjectures, Problem #782, erdosproblems.com/782. Checked March 7, 2026.
  2. [BFS04 ] T. C. Brown, A. R. Freedman, and P. J.-S. Shiue, Progressions of squares, Australasian Journal of Combinatorics 29 $2004$, 157--164.
  3. [DiEl15 ] R. Dietmann and C. Elsholtz, Hilbert cubes in arithmetic sets, Rev. Mat. Iberoam. 31 $2015$, 1477--1498.
  4. [CiGr07 ] J. Cilleruelo and A. Granville, Lattice points on circles, squares in arithmetic progressions and sumsets of squares, CRM Proc. Lecture Notes 43 $2007$, 241--262.

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Math-5892-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-5892

OBJECTIVE AND SOURCE REVIEW

Erdős #782; independent attempt, 2026-09-23. Do the squares contain quasi-progressions of unbounded length with one fixed bound on the spread of consecutive gaps? Do they contain Hilbert cubes of unbounded dimension? I read the entire supplied 782.tex, including its Pell construction, five-term examples, and conditional cube obstruction. Its displayed numerical Pell solution has the wrong sign.

WORK: CORRECTING AND CERTIFYING THE PELL FAMILY

In fact \[ 1068^2-73\cdot125^2=-1, \] not $+1$. Nevertheless this yields a completely explicit infinite width-one family. Define $x_0=1,y_0=0$ and \[ x_{r+1}=1068x_r+9125y_r,\qquad y_{r+1}=125x_r+1068y_r. \] Multiplication of $x_r+y_r\sqrt{73}$ by $1068+125\sqrt{73}$ proves inductively \[ x_r^2-73y_r^2=(-1)^r. \] For $r\ge1$, the four positive squares \[ y_r^2,\quad25y_r^2,\quad49y_r^2,\quad x_r^2 \] have consecutive gaps $24y_r^2,24y_r^2,24y_r^2+(-1)^r$. All are positive, and their spread is exactly one. The recurrence strictly increases $y_r$, so the families are distinct. Squaring the initial unit gives the genuine positive-norm solution $(x_2,y_2)=(2281249,267000)$. Thus the sign error does not invalidate the intended infinite construction; it changes which of the two adjacent possible gaps occurs. As a separate elementary check, the five squares $1,9,16,25,36$ have gaps $8,7,9,11$, hence spread four. This improves the width-six example in the supplied text, but is only a finite witness and is not asserted to be new.

THE CUBE QUESTION AND REMAINING GAP

The checked April 2026 preprint of Bremner--Elsholtz--Ulas proves infinitely many dimension-three square cubes, including a quantitative lower bound. This updates the supplied background, but fixed dimension three does not imply unbounded dimension. Neither the recurrence above, which always gives four terms, nor the five-term witness supplies unbounded quasi-progression length for one fixed width. I have no verified advance on the required uniformity, nor on the cube dimension beyond the cited known result. No conditional obstruction is promoted to an unconditional proof.

SOURCES CHECKED AND VERIFICATION

The complete local 782.tex; primary preprint abstract https://arxiv.org/abs/2604.05459. The Pell norm, recurrence, positive-norm solution, and five gaps are checked by exact integer arithmetic in verification/batch_3_final_checks.py. The failed raw GitHub fetch for 782.lean is not counted as a reviewed source.

FINAL

UNRESOLVED. A numerical error is repaired and small constructions checked; both unboundedness questions remain.

COMPLETION ESTIMATE

Subjective progress toward the original two questions. COMPLETION: 10%

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Math-5892-A2 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-5892

Problem #782 (Squares and quasi-progressions / Hilbert cubes)

1) FORMAL RESTATEMENT

Let $\mathcal{S}=\{m^2: m\in\mathbb{N}\}$ be the set of (positive) squares.

(A) Quasi-progressions. Fix an integer constant $C\ge 0$. A finite increasing sequence $(x_1<\cdots<x_k)$ of integers is a $C$--quasi-progression if there exists an integer $d\ge 1$ such that \[ d \le x_{i+1}-x_i \le d+C \qquad\text{for all } i=1,\dots,k-1. \] Question: Is there a fixed constant $C$ such that $\mathcal{S}$ contains $C$--quasi-progressions of arbitrarily large length $k$? (Equivalently: does $\sup\{k: \exists\ C\text{ fixed with a length-$k$ quasi-progression in }\mathcal{S}\}=\infty$?)

(B) Hilbert cubes in squares. A (finite) Hilbert cube of dimension $t$ is a set of the form \[ H(a; b_1,\dots,b_t)\;:=\;\Big\{a+\sum_{j=1}^t \varepsilon_j b_j : \varepsilon_j\in\{0,1\}\Big\}, \] where $a\in\mathbb{Z}$ and $b_1,\dots,b_t\in\mathbb{N}$. Question: Does $\mathcal{S}$ contain Hilbert cubes of arbitrarily large dimension $t$?

The problem statement notes that an affirmative answer to (A) would imply an affirmative answer to (B).

2) QUICK LITERATURE/CONTEXT CHECK

  • Brown--Erdős--Freedman $1990$ introduced quasi-progressions (and related "descending waves") and asked whether the squares contain arbitrarily long quasi-progressions with bounded error $C$.
  • Solymosi $2007$ discussed the related question of Hilbert cubes in the squares and conjectured that the squares do not contain arbitrarily large Hilbert cubes.
  • Cilleruelo--Granville $2007$ observed that Solymosi's conjecture (no arbitrarily large Hilbert cubes in squares) would follow from the Bombieri--Lang conjecture.
  • Unconditionally, there are nontrivial upper bounds on the maximum dimension of a Hilbert cube contained in $\mathcal{S}\cap[1,N]$. In particular, Dietmann--Elsholtz $2015$ prove a bound of the shape \[ \dim\big(\mathcal{S}\cap[1,N]\big) \le 7\,\log\log N \qquad\text{for all sufficiently large }N. \] This still allows the possibility of arbitrarily large cubes (since $\log\log N\to\infty$).

So as of these works, both questions remain open unconditionally.

3) ATTACK PLAN

(A) Attempt a construction. Try to build many squares whose successive gaps are almost equal. A heuristic route is to search for squares lying close to a long arithmetic progression.

(B) Attempt an obstruction. Try to show that bounded-gap quasi-progressions would force some Diophantine structure (e.g. many integer points on curves) that can be controlled. For Hilbert cubes, one tries to translate the existence of a large cube into many solutions of systems of quadratic equations.

4) WORK

4.1 Basic constraints and reformulations

Gap identity.
If $x_i=a_i^2\in\mathcal{S}$, then \[ x_{i+1}-x_i = a_{i+1}^2-a_i^2 = (a_{i+1}-a_i)(a_{i+1}+a_i). \] If all gaps lie in an interval $[d,d+C]$ with fixed $C$, then as $a_i$ grows, the factor $a_{i+1}+a_i$ grows, hence the integer factor $a_{i+1}-a_i$ must typically get very small (eventually $=1$), suggesting that long quasi-progressions might force $a_i$ to stay in a controlled range relative to $d$. This intuition is not yet a proof.
A trivial non-constant-$C$ construction.
Consecutive squares give a quasi-progression of length $k$ but with $C$ growing linearly in $k$: \[ (m+1)^2-m^2 = 2m+1, \] so over $k$ consecutive squares the gaps range over an interval of length $2(k-1)$. This shows that the question is genuinely about keeping $C$ independent of $k$.

4.2 Hilbert cubes: what is known and what it would take

Dimension in $[1,N]$.
Let \[ \dim\big(\mathcal{S}\cap[1,N]\big) := \max\Big\{ t:\exists a,b_1,\dots,b_t\text{ with }H(a;b_1,\dots,b_t)\subseteq \mathcal{S}\cap[1,N]\Big\}. \] Dietmann--Elsholtz show $\dim(\mathcal{S}\cap[1,N]) \ll \log\log N$. Thus, if arbitrarily large cubes exist in the squares, their dimension must grow extremely slowly with the size of the squares involved.
Why Bombieri--Lang would rule out large cubes (informal).
A $t$-dimensional cube in squares gives $2^t$ distinct square values with additive structure. Fixing $a$ and the $b_i$, one can encode many of these relations as rational points on certain curves/surfaces of general type. The Bombieri--Lang conjecture predicts strong uniform finiteness properties for rational points on such varieties, which would bound $t$ by an absolute constant.

5) VERIFICATION

  • The "consecutive squares" check correctly shows that naive constructions do not keep $C$ fixed.
  • The stated $\log\log N$ upper bound for cube dimension in squares matches the cited Dietmann--Elsholtz theorem.
  • No step above claims an unconditional resolution; the reasoning is consistent with the current literature status.

6) FINAL

LABEL: UNRESOLVED.
Best partial results I can state cleanly.
  • There are strong unconditional upper bounds on the maximum dimension of a Hilbert cube inside $\mathcal{S}\cap[1,N]$, namely $\dim(\mathcal{S}\cap[1,N])\ll\log\log N$.
  • Conditional on Bombieri--Lang, the squares do not contain arbitrarily large Hilbert cubes (hence a negative answer to (B) and therefore to (A) as well).
Strongest heuristic.
Random-model heuristics suggest that large additive subcubes inside the sparse set of squares should be rare; this supports Solymosi's conjecture that an absolute bound on cube dimension might hold. On the other hand, the existing unconditional bounds still allow very slowly growing cube dimension, leaving both outcomes plausible.
Main obstacle.
To rule out arbitrarily long quasi-progressions or cubes one needs uniform control over integer/rational points on families of quadratic varieties arising from additive constraints among squares. Current unconditional tools (sum-product methods, bounds on solutions to $x^2+y^2=z^2+w^2$, etc.) are not yet strong enough to force a constant bound.
Next concrete steps.
  1. Search systematically for explicit cubes/quasi-progressions of the largest currently known sizes (even small improvements could suggest the right growth scale).
  2. Try to sharpen the $\log\log N$ bound toward an absolute bound using improvements in additive combinatorics for polynomial images.
  3. For quasi-progressions, attempt to prove that bounded-gap sequences of squares imply many rational points on a fixed genus-$\ge 2$ curve, which would allow Faltings-type finiteness arguments.
COMPLETION: 25%.

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