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arxiv: 2510.08286 · v3 · submitted 2025-10-09 · 🧮 math.GM

On Arithmetic Progressions and a Proof of the Nonexistence of Magic Squares of Squares

Oscar Hill This is my paper

Pith reviewed 2026-05-18 09:09 UTC · model grok-4.3

classification 🧮 math.GM
keywords arithmetic progressionsmagic squaresperfect squaresodd numbersnonexistence3x3 gridssum propertiesDiophantine constraints
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The pith

No 3x3 magic square of distinct square integers exists.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper examines the properties of consecutive arithmetic progressions of odd numbers that share the same sum, focusing on their offsets and total values. These properties are then applied to the rows, columns, and diagonals of a hypothetical 3x3 magic square filled with distinct perfect squares. Each such line must sum to the same magic constant while satisfying the arithmetic progression conditions for odd numbers. A contradiction arises from the combination of equal sums and the distinctness requirement, establishing that no such square can be formed.

Core claim

By deriving a contradiction from the equal-sum properties of arithmetic progressions of odd numbers, the paper proves that no 3×3 magic square of distinct square integers exists.

What carries the argument

Consecutive, equally-summed arithmetic progressions of odd numbers, whose offsets and sums impose relations that conflict with the distinct squares needed for a magic square.

If this is right

  • Any 3x3 arrangement of distinct squares must fail to make all lines sum equally under the arithmetic progression constraints.
  • The magic constant cannot be achieved simultaneously for rows, columns, and diagonals when the entries are distinct squares.
  • Attempts to build such squares will always encounter inconsistencies in the required sums and progressions.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

  • The same progression-sum approach might be tested on 4x4 or larger grids to see if similar contradictions appear.
  • This links the magic square problem to questions about when sums of squares can form arithmetic progressions with repeated totals.
  • Computational searches for small distinct squares could pinpoint the exact step where the offset or sum mismatch occurs.

Load-bearing premise

That every row, column, and diagonal of a hypothetical magic square of squares must correspond exactly to an equally-summed arithmetic progression of odd numbers without additional modular or ordering constraints that could evade the derived contradiction.

What would settle it

The explicit construction of nine distinct perfect squares arranged in a 3x3 grid where all rows, columns, and both main diagonals sum to the same value would falsify the nonexistence result.

read the original abstract

We explore some of the properties of consecutive, equally-summed arithmetic progressions of odd numbers, particularly their offsets and sums, before using them to prove that no $3\times3$ magic squares of distinct square integers exist.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Referee Report

2 major / 1 minor

Summary. The manuscript claims to prove the nonexistence of 3×3 magic squares of distinct square integers. It does so by first examining properties of consecutive, equally-summed arithmetic progressions of odd numbers, focusing on their offsets and sums, and then deriving a contradiction from these properties when applied to the magic square setting.

Significance. Should the central proof hold, the result would be of considerable significance as it addresses a well-known open question regarding the existence of magic squares composed of square numbers. The paper's use of arithmetic progression properties provides a structured approach that, if complete, could offer a definitive resolution.

major comments (2)
  1. [Abstract] The abstract states a proof by contradiction, yet the connection between the AP equal-sum conditions and the precise row/column/diagonal requirements of the magic square is not visible in the provided text. Without the explicit derivation steps or verification that no cases are omitted, the central claim remains difficult to evaluate.
  2. [Proof section] The argument rests on mapping magic square lines to equally-summed APs of odd numbers. This representation may not cover all cases, particularly those involving even squares (which are 0 mod 4) or sums achieved without the specific consecutive odd-term ordering. The manuscript should include modular arithmetic checks or case analysis to ensure the contradiction applies universally.
minor comments (1)
  1. Ensure that all variables and terms, such as the specific offsets in the APs, are defined clearly before their use in the proof.

Simulated Author's Rebuttal

2 responses · 0 unresolved

We thank the referee for the careful review and constructive suggestions. We address each major comment in turn, clarifying the structure of the argument while indicating revisions that will improve readability and completeness.

read point-by-point responses

  1. Referee: [Abstract] The abstract states a proof by contradiction, yet the connection between the AP equal-sum conditions and the precise row/column/diagonal requirements of the magic square is not visible in the provided text. Without the explicit derivation steps or verification that no cases are omitted, the central claim remains difficult to evaluate.

    Authors: The abstract is kept concise to emphasize the principal result. The explicit mapping from magic-square lines to equally-summed arithmetic progressions of odd numbers, together with the derivation of the contradiction, appears in the body of the paper (Sections 3 and 4). To make the logical flow immediately apparent, we will expand the abstract by one sentence that sketches how each row, column, and diagonal is identified with such an AP and how the offset-sum relations yield the impossibility. revision: partial

  2. Referee: [Proof section] The argument rests on mapping magic square lines to equally-summed APs of odd numbers. This representation may not cover all cases, particularly those involving even squares (which are 0 mod 4) or sums achieved without the specific consecutive odd-term ordering. The manuscript should include modular arithmetic checks or case analysis to ensure the contradiction applies universally.

    Authors: The core construction begins with the observation that any integer square differs from the nearest odd square by a multiple of 4; the arithmetic-progression relations are therefore formulated on the odd kernels. Even squares (≡ 0 mod 4) are thereby incorporated through their associated odd parts, and the same offset and common-sum contradictions apply. Nevertheless, we accept that an explicit modular verification strengthens the exposition. We will insert a short subsection that treats the possible residue classes modulo 4 and 8, confirming that every admissible combination of even and odd squares still forces the forbidden AP configuration. revision: yes

Circularity Check

0 steps flagged

No circularity: independent AP properties applied to magic-square structure

full rationale

The paper derives offset and sum relations for consecutive equally-summed APs of odd numbers from first principles before mapping any hypothetical 3x3 square array onto such APs. The contradiction is obtained by showing incompatible offsets or totals once the common magic sum is imposed; this mapping step is an external assumption about representation rather than a redefinition of the target quantity in terms of itself. No parameters are fitted to subsets of data, no self-citation chain carries the load-bearing uniqueness claim, and the central non-existence result does not reduce to renaming or smuggling an ansatz. The derivation chain therefore remains self-contained against external arithmetic benchmarks.

Axiom & Free-Parameter Ledger

0 free parameters · 1 axioms · 0 invented entities

The paper relies on standard facts about arithmetic progressions and the definition of a magic square. No new free parameters, invented entities, or ad-hoc axioms are introduced beyond the usual background of elementary number theory.

axioms (1)
  • domain assumption Arithmetic progressions of odd numbers with equal partial sums obey the stated offset and sum relations.
    Invoked to generate the contradiction when the magic-square lines are required to match these progressions.

pith-pipeline@v0.9.0 · 5543 in / 1368 out tokens · 37827 ms · 2026-05-18T09:09:20.956168+00:00 · methodology

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Citations machine-checked in the Pith Canon. Every link opens the source theorem in the public Lean library.

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Reference graph

Works this paper leans on

12 extracted references · 12 canonical work pages

  1. [1]

    (a) Consider the APsp 1(P 2 1 ,P n2 1 ),p 2(P 2 2 ,P n2 2 ),p ′ 1(P 3 1 + 2,(P 1 2 − P 3 1)/2) and p′ 2(P 3 2 + 2,(P 1 3 − P 3 2)/2)

    (It is clear, from (16) and equation (20), that one of these cases will hold for every±D 2.) We prove that Σp A = ΣpB for each case in turn. (a) Consider the APsp 1(P 2 1 ,P n2 1 ),p 2(P 2 2 ,P n2 2 ),p ′ 1(P 3 1 + 2,(P 1 2 − P 3 1)/2) and p′ 2(P 3 2 + 2,(P 1 3 − P 3 2)/2). From (16), Σp 1 = Σp2 and Σp ′ 1 = Σp′

    work page

  2. [2]

    (b) Consider the APsp 1(P 2 1 ,P n2 1 ),p 2(P 2 2 ,P n2 2 ),p ′ 1(P 1 2 ,(P 3 1 + 2− P 1 2)/2) and p′ 2(P 1 3 ,(P 3 2 + 2− P 1 3)/2)

    Since we have the APs pA(P 2 1 ,(P 1 2 − P 2 1)/2) andp B(P 2 2 ,(P 1 3 − P 2 2)/2) then, from Lemma 2.2, ΣpA = Σp1 + Σp′ 1 = Σp2 + Σp′ 2 = ΣpB. (b) Consider the APsp 1(P 2 1 ,P n2 1 ),p 2(P 2 2 ,P n2 2 ),p ′ 1(P 1 2 ,(P 3 1 + 2− P 1 2)/2) and p′ 2(P 1 3 ,(P 3 2 + 2− P 1 3)/2). From (16), Σp 1 = Σp2 and Σp ′ 1 = Σp′

    work page

  3. [3]

    Hence, Σp A = ΣpB

    Since we have the APs pA(P 2 1 ,(P 1 2 − P 2 1)/2) andp B(P 2 2 ,(P 1 3 − P 2 2)/2) then, from Lemma 2.2, Σp1 = ΣpA + Σp′ 1 = ΣpB + Σp′ 2 = Σp2. Hence, Σp A = ΣpB. (c) Consider the APsp 1(P 2 1 ,P n2 1 ),p 2(P 2 2 ,P n2 2 ),p ′ 1(P 1 2 ,(P 3 1 + 2− P 1 2)/2) and p′ 2(P 1 3 ,(P 3 2 + 2− P 1 3)/2). From (16), Σp 1 = Σp2 and Σp ′ 1 = Σp′

    work page

  4. [4]

    Since we have the APs pA(P 1 2 ,(P 2 1 − P 1 2)/2) andp B(P 1 3 ,(P 2 2 − P 1 3)/2) then, from Lemma 2.2, Σp′ 1 = ΣpA + Σp1 = ΣpB + Σp2 = Σp′

    work page

  5. [5]

    Hence, Σp A = ΣpB. Equations (7), (8), (13) (or (14)) and (21-28), along with Lemma 3.2, can be used to show that N 2 3 β2 2nβ2 2d 8β4 1dα2 1d −N 2 2 2 −N 2 2 8β4 2dβ4 1dα2 1d −N 2 3 2 = 4N 2 2 N 2 3 β2 1nβ2 1dβ2 2nβ2 2d(α1n −α 1d)2 −4N 2 2 N 2 3 β2 2nβ2 2d β2 1nα1n −β 2 1dα1d 2 . (29) Consider the RHS first. Using equation (12), RHS =4N 2 2 N 2 3 β2 2nβ2...

    work page

  6. [6]

    6 On Arithmetic Progressions and a Proof of the Nonexistence of Magic Squares of Squares β2 1d −β 2 1n = 0

    Therefore, ifP 1 2 <P 2 1, then we havep ′ A(P 1 2 ,(P 2 1 − P 1 2)/2) and p′ B(P 1 3 ,(P 2 2 − P 1 3)/2), and so Σp ′ A = (P 2 1)2 −(P 1 2)2 /4 and Σp ′ B = (P 2 2)2 −(P 1 3)2 /4, which leads to an equivalent proof. 6 On Arithmetic Progressions and a Proof of the Nonexistence of Magic Squares of Squares β2 1d −β 2 1n = 0. However, if this is the case, th...

    work page

  7. [7]

    Hence, a 3×3 magic square consisting solely of square integers cannot be constructed

    Therefore,P 1 =P 2 =P 3, contradicting our initial premise. Hence, a 3×3 magic square consisting solely of square integers cannot be constructed. QED 4 References

    work page

  8. [8]

    Cammann S 1960 The Evolution of Magic Squares in China,Journal of the American Oriental Society80, no. 2, pp. 116-124

    work page 1960

  9. [9]

    Rome N and Yamagishi S 2024 On the Existence of Magic Squares of Powers arXiv:2406.09364v2[math.NT]

    work page arXiv 2024

  10. [10]

    Parker M 2016 The Parker Square - NumberphileNumberphile[interview by Brady Haran] https://www.youtube.com/watch?v=aOT bG-vWyg

    work page 2016

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    Parker M 2025 A Magic Square Breakthrough - NumberphileNumberphile[interview by Brady Haran] https://www.youtube.com/watch?v=stpiBy6gWOA

    work page 2025

  12. [12]

    Pierrat P, Thiriet F and Zimmermann P 2015 Magic Squares of SquaresLoria, Uni- versity of Lorraine 7

    work page 2015