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BCCG Theorem 1: the counting polynomial of semi-magic squares

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MagicSquares.semi_magic_polynomial

by Yuxuan Xu · Sep 19, 2026 · Mathlib 0df444a (Lean v4.33.1)

combinatoricsehrhartenumerative-combinatoricsmagic-squares

Goal. For every order n≥1n\ge 1n≥1 there is a polynomial p∈Q[X]p\in\mathbb{Q}[X]p∈Q[X] which, simultaneously, has degree exactly (n−1)2(n-1)^{2}(n−1)2, agrees with the counting function HnH_{n}Hn​ at every nonnegative integer, satisfies the reciprocity law p(−n−t)=(−1)n−1p(t)p(-n-t)=(-1)^{n-1}p(t)p(−n−t)=(−1)n−1p(t) for every integer ttt, and vanishes at −1,−2,…,−(n−1)-1,-2,\dots,-(n-1)−1,−2,…,−(n−1). This is Theorem 1 of Beck--Cohen--Cuomo--Gribelyuk, proved by Ehrhart and by Stanley in 1973 and given an elementary proof by Spencer in 1980. It separates the semi-magic squares from the magic, symmetric and pandiagonal classes, whose counting functions are quasi-polynomials rather than polynomials.

Preamble
import Mathlib
import Definitions.Def_MagicSquares
open MagicSquares
Formal statement
namespace MagicSquares

theorem semi_magic_polynomial (n : ℕ) (hn : 1 ≤ n) :
    ∃ p : Polynomial ℚ,
      p.natDegree = (n - 1) ^ 2 ∧
        (∀ t : ℕ, p.eval (t : ℚ) = (semiMagicCount n t : ℚ)) ∧
          (∀ t : ℤ, p.eval (((-(n : ℤ) - t : ℤ) : ℚ))
            = (-1 : ℚ) ^ (n - 1) * p.eval ((t : ℤ) : ℚ)) ∧
            (∀ k : ℤ, 1 ≤ k → k ≤ (n : ℤ) - 1 → p.eval (-(k : ℚ)) = 0) := by
  sorry

end MagicSquares
Source
M. Beck, M. Cohen, J. Cuomo and P. Gribelyuk, The number of "magic" squares, cubes and hypercubes, Amer. Math. Monthly 110 (2003), 707--717 (arXiv:math/0201013); E. Ehrhart (1973); R. P. Stanley, Duke Math. J. 40 (1973) 607--632; J. Spencer, Amer. Math. Monthly 87 (1980) 397--399; M. Beck and D. Paxton, The Ehrhart polynomial of the Birkhoff polytope (arXiv:math.CO/0202267).
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What the Lean code literally says, in plain math · deepseek-v4.1-flash (WorkBuddy blind sub-agent)

Let nnn be a natural number with 1≤n1\le n1≤n. Then there exists a polynomial p∈Q[X]p\in\mathbb{Q}[X]p∈Q[X] satisfying all of the following simultaneously:

  • its natural degree equals (n−1)2(n-1)^2(n−1)2, where n−1n-1n−1 is the natural-number subtraction (so ppp is necessarily non-zero of ordinary degree (n−1)2(n-1)^2(n−1)2);
  • for every natural number ttt, p(t)=semiMagicCount⁡(n,t)p(t)=\operatorname{semiMagicCount}(n,t)p(t)=semiMagicCount(n,t) in Q\mathbb{Q}Q;
  • for every integer ttt, p((−n)−t)=(−1) n−1 p(t)p((-n)-t)=(-1)^{\,n-1}\,p(t)p((−n)−t)=(−1)n−1p(t), where (−n)−t(-n)-t(−n)−t is −n-n−n minus ttt (lifted to Q\mathbb{Q}Q) and the exponent n−1n-1n−1 is natural subtraction;
  • for every integer kkk, if 1≤k1\le k1≤k and k≤n−1k\le n-1k≤n−1 (integer subtraction) then p(−k)=0p(-k)=0p(−k)=0.

Here semiMagicCount⁡(n,t)\operatorname{semiMagicCount}(n,t)semiMagicCount(n,t) counts the n×nn\times nn×n arrays over Fin⁡(t+1)\operatorname{Fin}(t+1)Fin(t+1) (entries integers 0,…,t0,\dots,t0,…,t) whose every row and column sums to ttt. When n=1n=1n=1 the stated degree is 000, the reciprocity factor is (−1)0=1(-1)^0=1(−1)0=1, and the vanishing condition is vacuously true.

Human review
  • Endorsed by Shuze Chen · Sep 19, 2026

  • Endorsed by Yuxuan Xu · Sep 19, 2026

    Confirmed by the mission captain (proposal self-audit).

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