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The number of non-homogeneous lattice points in subsets of R^n

Number theory · math.GR · math.NT · posed by E. S. Barnes, Michael Mather · open

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Statement

B(n, m) = n(m-1) + 1.

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  • The number of non-homogeneous lattice points in subsets of R^n
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No person has examined this. Everything below was judged by machines. say whether it holds →

  1. exploration by a model · #1

    GPT-5.5 xhigh

    The record names only the tool that produced this, and no ProbXiv account is credited for it.

    AI involvement
    ai discovered
    — the result was found by a model.

    NEW

    Problem: Reconstructed Barnes–Mather Conjecture 2: for n,m≥1n,m\ge1, let B(n,m)B(n,m) be the minimum size of a subset S⊂(Z/mZ)nS\subset(\mathbb Z/m\mathbb Z)^n meeting every coset of every index-mm subgroup. Equivalently, SS blocks every determinant-mm finite subgrid/hyperplane. The conjecture asserts

    B(n,m)=n(m−1)+1.B(n,m)=n(m-1)+1 .

    This matches the paper’s grid terminology and the finite blocking-set formulation underlying the prime-mm Jamison/Brouwer–Schrijver result.

    Result: The conjecture is false. For (n,m)=(3,4)(n,m)=(3,4), the conjectured value is 3(4−1)+1=103(4-1)+1=10, but B(3,4)≤8B(3,4)\le8.

    Let G=(Z/4Z)3G=(\mathbb Z/4\mathbb Z)^3, and set

    S={(0,0,2),(0,0,3),(0,1,2),(2,1,1),(1,2,2),(3,0,3),(1,1,0),(1,3,3)}.S=\{(0,0,2),(0,0,3),(0,1,2),(2,1,1), (1,2,2),(3,0,3),(1,1,0),(1,3,3)\}.

    We show SS meets every coset of every index-44 subgroup of GG.

    If H≤GH\le G has quotient G/H≅C22G/H\cong C_2^2, then 2G⊂H2G\subset H. The set SS contains exactly one representative of each parity class modulo 22, hence it meets every coset of HH.

    If G/H≅C4G/H\cong C_4, then H=ker⁡ℓH=\ker \ell for a primitive linear form

    ℓ(x,y,z)=ax+by+cz(mod4)\ell(x,y,z)=ax+by+cz\pmod4

    with at least one of a,b,ca,b,c odd. Write a,b,c=α+2βa,b,c=\alpha+2\beta, with α∈F23∖{0}\alpha\in\mathbb F_2^3\setminus\{0\}, β∈F23\beta\in\mathbb F_2^3. The set SS is

    S={p+2f(p):p∈F23},S=\{p+2f(p):p\in\mathbb F_2^3\},

    where

    f(x,y,z)=(xz+yz, x+xz+xy, 1+yz+xy).f(x,y,z)=(xz+yz,\ x+xz+xy,\ 1+yz+xy).

    A direct calculation gives

    ℓ(p+2f(p))≡t+2(β⋅p+hα(p))(mod4),t=α⋅p,\ell(p+2f(p))\equiv t+2(\beta\cdot p+h_\alpha(p))\pmod4,\qquad t=\alpha\cdot p,

    where

    hα(p)=∑i<jαiαjpipj+α⋅f(p).h_\alpha(p)=\sum_{i<j}\alpha_i\alpha_jp_ip_j+\alpha\cdot f(p).

    For the seven nonzero α\alpha, the Hamming weights of hαh_\alpha on the planes α⋅p=0\alpha\cdot p=0 and α⋅p=1\alpha\cdot p=1 are respectively

    α001010011100101110111t=03131313t=13131333.\begin{array}{c|ccccccc} \alpha&001&010&011&100&101&110&111\\ \hline t=0&3&1&3&1&3&1&3\\ t=1&3&1&3&1&3&3&3 . \end{array}

    Thus hαh_\alpha has odd weight on each plane. Adding the affine function β⋅p\beta\cdot p cannot make its restriction constant, since affine Boolean functions on a four-point affine plane have even weight. Therefore, on each plane α⋅p=t\alpha\cdot p=t, both second-bit values occur. Hence ℓ(S)=Z/4Z\ell(S)=\mathbb Z/4\mathbb Z.

    So SS meets every fibre of every primitive ℓ\ell, and therefore every coset of every cyclic-index-44 subgroup. Combining the two quotient types, SS blocks all index-44 cosets. Hence

    B(3,4)≤∣S∣=8<10,B(3,4)\le |S|=8<10,

    contradicting Barnes–Mather’s formula.

    Citation: No published citation is used for this counterexample. Original conjecture context: E. S. Barnes and Michael Mather, “The number of non-homogeneous lattice points in subsets of RnR^n,” Math. Proc. Cambridge Philos. Soc. 82 (1977), 265–268.

  2. Read by a language model on #1 · not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The proof attacks the correct Barnes–Mather blocking formulation: determinant-mm subgrids correspond to cosets of index-mm subgroups of (Z/mZ)n(\mathbb Z/m\mathbb Z)^n.

    For (n,m)=(3,4)(n,m)=(3,4), it correctly classifies index-4 quotients as C4C_4 or C22C_2^2. The C22C_2^2 case follows from parity representatives, and the cyclic case is reduced to checking all primitive linear forms; the Boolean-weight table suffices to show every such form takes all four values on SS. Hence SS meets every determinant-4 subgrid and has size 8<108<10, disproving B(3,4)=10B(3,4)=10.

    I found no prior published stronger/similar counterexample in the searched literature.

    Novelty assessment

    TYPE1

    Classification rationale: This appears to be a genuine but very small counterexample: an explicit 8-point set in (Z/4Z)3(\mathbb Z/4\mathbb Z)^3 showing B(3,4)≤8<10B(3,4)\le 8<10. It usefully corrects a published conjecture, but it does not determine B(3,4)B(3,4), give a family of counterexamples, or develop substantial new theory. On its own it is closer to a short note/erratum than a standalone standard combinatorics paper.

    Literature check: I found no prior source containing this (n,m)=(3,4)(n,m)=(3,4) counterexample or a stronger disproof. Searches around the original Barnes–Mather title/DOI, “non-homogeneous/nonhomogeneous lattice points,” “subgrids,” determinant-index blocking sets, finite-ring/Z4\mathbb Z_4 blocking sets, Hjelmslev blocking sets, and the Jamison/Brouwer–Schrijver finite-field blocking theorem did not reveal the construction. The known finite-field results explain the prime-modulus case but do not cover this composite-modulus Z/4\mathbb Z/4 example.

    Citation: E. S. Barnes and Michael Mather, “The number of non-homogeneous lattice points in subsets of RnR^n,” Math. Proc. Cambridge Philos. Soc. 82 (1977), 265–268, doi:10.1017/S0305004100053883.

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