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1. The question and its assumptions

Count ordered pairs from two rational cyclic grids whose sum and difference lie on two specified target grids, using classical Smith normal form. A union in the first input is handled by inclusion-exclusion.

• Every grid order d, b, y, k, a and r appearing in a formula is a positive integer. Orders equal to 1 are permitted.
• G_n is an unshifted finite subgroup of the rational circle Q/Z. Addition and subtraction ignore whole turns, that is, are taken modulo 1.
• The count is of ordered pairs (u,h), with the first and second coordinates restricted to their specified input grids. It is not a count of distinct sums, distinct differences or unordered pairs.
• The basic count uses u in G_d and h in G_b, with both conditions u+h in G_y and u-h in G_k imposed simultaneously.
• The orders y and k are arbitrary positive integers; neither is required to divide L.
• The union variant replaces only the first input grid by G_a union G_r and counts each admissible ordered pair once.
• Zero, negative and nonintegral grid orders, and shifted grids, are outside the stated definitions.

Notation

Q/Z — Rational numbers modulo integers: two rational coordinates represent the same point when their difference is an integer.

G_n — The unshifted n-point rational cyclic grid. A point t in Q/Z belongs to G_n exactly when n*t=0 in Q/Z.

G_n = {j/n + Z : j is an integer, 0 <= j < n}

C(d,b;y,k) — Number of admissible ordered pairs from G_d x G_b; see equation (1).

Q(a,b;y,k;r) — The union count. This function Q is distinct from the symbol Q for rational numbers in Q/Z.

Q(a,b;y,k;r) = #{(u,h) in (G_a union G_r) x G_b : u+h in G_y and u-h in G_k}

gcd — Positive greatest common divisor; signs of entries do not affect the gcd.

lcm — Positive least common multiple.

Z/nZ — The additive group of integer residue classes modulo n.

#S or |S| — The number of elements of a finite set S.

kernel(f) — The input elements mapped to zero by the group homomorphism f.

image(f) — The set of output elements reached by f.

Abar — The square-domain map given by the integer matrix A reduced modulo L, not the rectangular-domain map phi.

2. Equations (1)–(8)

Equation numbers refer to the PDF. In the selectable ASCII text, * means multiplication, ^T denotes a column transpose and |S| denotes finite cardinality.

C(d,b;y,k) = #{(u,h) in G_d x G_b : u+h in G_y and u-h in G_k}

L = lcm(d,b);
alpha = L/d;
beta = L/b;
A = [[y*alpha, y*beta], [k*alpha, -k*beta]]

s1 = gcd(y*alpha, y*beta, k*alpha, k*beta);
s2 = (2*y*k*alpha*beta)/s1;
J_L(s) = L/gcd(L,s)

C(d,b;y,k) = (d*b)/(J_L(s1)*J_L(s2))
             = ((d*b)/(L*L))*gcd(L,s1)*gcd(L,s2)

y*(alpha*i + beta*j) == 0 (mod L);
k*(alpha*i - beta*j) == 0 (mod L)

image(phi) = image(Abar)

phi: Z/dZ x Z/bZ -> (Z/LZ)^2; phi(i,j) = A*(i,j)^T modulo L

Abar: (Z/LZ)^2 -> (Z/LZ)^2; Abar(x,z) = A*(x,z)^T modulo L

|kernel(Abar)| = ((L*L)/(d*b))*|kernel(phi)|

tau: (Z/LZ)^2 -> Z/dZ x Z/bZ; tau(x,z) = (x modulo d, z modulo b)

Abar = phi composed with tau

Every fiber of tau has (L/d)*(L/b) = (L*L)/(d*b) elements.

Q(a,b;y,k;r) = C(a,b;y,k) + C(r,b;y,k) - C(gcd(a,r),b;y,k)

3. Why the count follows

A condensed restatement of the proof. The rectangular input domain and the auxiliary square domain must be distinguished.

1. Translate membership without changing the inputs

   Represent u uniquely by i/d with 0<=i<d and h uniquely by j/b with 0<=j<b, modulo 1. Membership t in G_n is equivalent to n*t being an integer. With L=lcm(d,b), multiplication of the two membership conditions by L gives exactly equation (5). Thus the admissible pairs correspond bijectively to kernel(phi).

   PDF pages 2, 3

2. Verify the rectangular-domain homomorphism

   Multiplying the first column of A by d gives (y*L,k*L)^T, and multiplying the second column by b gives (y*L,-k*L)^T. Both vanish modulo L. Changing i by d or j by b therefore leaves A*(i,j)^T modulo L unchanged. This proves that phi on Z/dZ x Z/bZ is well defined. It is a group homomorphism because multiplication by A respects addition.

   PDF pages 3

3. Compare images before using Smith normal form

   Every rectangular-domain output is a square-domain output because representatives 0<=i<d and 0<=j<b are also available modulo L. Conversely, for any square-domain input (x,z), reduce x modulo d and z modulo b. The column periods just proved ensure that this reduction does not change the output modulo L. The two maps consequently have the same image, as in equation (6).

   PDF pages 3

4. Identify the Smith invariant factors

   The determinant is -2*y*k*alpha*beta, which is nonzero for the positive orders in the statement. For a full-rank 2-by-2 integer matrix, Smith normal form gives unimodular integer matrices U,V with U*A*V=diag(s1,s2), where s1 is the gcd of the entries and s1*s2=abs(det(A)). These facts give equation (3). Since every entry is divisible by s1, the determinant is divisible by s1*s1; hence s2 is an integer and s1 divides s2.

   PDF pages 3, 4

5. Count the square-domain image

   Unimodular U and V have integer inverses, so their reductions modulo L are bijections. They preserve the image cardinality of the square-domain map. For scalar multiplication by s on Z/LZ, set g=gcd(L,s), L=g*Lprime and s=g*sprime. Since gcd(Lprime,sprime)=1, L divides s*x exactly when Lprime divides x. There are g such residue classes modulo L, so the scalar kernel has g elements and the scalar image has L/g=J_L(s) elements. The two diagonal coordinates act independently, giving |image(Abar)|=J_L(s1)*J_L(s2).

   PDF pages 3, 4

6. Recover the required count and its integrality

   The domain of phi has d*b elements. The finite-group first isomorphism theorem gives |kernel(phi)|*|image(phi)|=d*b. Combining image equality with the diagonal image size yields C(d,b;y,k)=(d*b)/(J_L(s1)*J_L(s2)), and substitution of J_L gives the second expression in equation (4). This group identity establishes exact integrality of the complete count.

   PDF pages 4

7. Check the alternative fiber normalization

   Because d and b divide L, the reduction map tau in equation (7) is well defined and onto. Each residue modulo d has L/d lifts modulo L, and each residue modulo b has L/b lifts, so each fiber has L*L/(d*b) elements. The column-period calculation shows Abar=phi composed with tau. Thus kernel(Abar) is a union of exactly |kernel(phi)| fibers, proving equation (7). Smith normal form gives the square-domain kernel size gcd(L,s1)*gcd(L,s2). Dividing that count by the fiber size gives equation (4), not the unnormalized square-domain kernel count.

   PDF pages 4

8. Justify inclusion-exclusion for a union

   Let g=gcd(a,r). A class t lies in both G_a and G_r exactly when a*t=0 and r*t=0 in Q/Z. Bezout's identity gives integers c,e with c*a+e*r=g, so both equalities imply g*t=0. Conversely, g*t=0 implies both equalities since g divides a and r. Hence G_a intersect G_r=G_g. The intersection of the two admissible ordered-pair sets therefore has size C(g,b;y,k). Inclusion-exclusion gives equation (8). When a=r the count reduces to one grid. When a divides r, G_a is contained in G_r and the formula reduces to C(r,b;y,k); the analogous statement holds when r divides a. G_1={0}, so unit orders are included.

   PDF pages 4, 5

4. The original worked examples

These are the source PDF examples, not additional scientific runs.

(d,b;y,k)=(6,4;3,2)
L=12; alpha=2; beta=3; A=[[6, 9], [4, -6]]
(s1,s2)=(1,72); (J_L(s1),J_L(s2))=(12,1)
C(6,4;3,2)=24/(12*1)=2; equivalently, square kernel 12 divided by fiber size 6 gives 2.
Input size=24; image size=12
Square-kernel size=12; fiber size=6

Admissible pairs: (0, 0), (1/2, 1/2)

(d,b;y,k)=(6,6;3,3)
L=6; alpha=1; beta=1; A=[[3, 3], [3, -3]]
(s1,s2)=(3,6); (J_L(s1),J_L(s2))=(2,1)
C(6,6;3,3)=36/(2*1)=18.
Input size=36; image size=2
Square-kernel size=18; fiber size=1

a=4; r=6; b=4; y=3; k=6; gcd(a,r)=2

Q(4,4;3,6;6)=C(4,4;3,6)+C(6,4;3,6)-C(2,4;3,6)=4+6-2=8.

(0, 0), (1/4, 3/4), (1/2, 1/2), (3/4, 1/4), (1/3, 0), (2/3, 0), (1/6, 1/2), (5/6, 1/2)

5. Reading the formula accurately

6. Scope, assistance and review status

An exposition and application of established Smith normal form and congruence-counting methods, with explicit rational-grid notation, rectangular-domain normalization and inclusion-exclusion examples. No new general congruence-counting theorem or priority for the specialization is claimed.

The source preprint discloses AI assistance with drafting, editing and mathematical inspection. This condensed companion was also prepared with AI assistance.

Independent human peer review and proof-assistant kernel verification are not claimed. This companion is a reading aid, not evidence of either.

This companion does not report new scientific runs. Previously reported finite computational checks and their provenance are described separately in the existing Human Review Brief. Finite checks illustrate agreement on tested inputs; they do not establish the universal result or replace proof review.

A useful mathematical review identifies the source version, page and equation, then supplies the precise inference, missing assumption or counterexample under examination.

Read the independent-review brief · Contact for mathematical feedback

7. Sources and document identity

1. A. T. Butson and B. M. Stewart. Systems of Linear Congruences. Canadian Journal of Mathematics 7 (1955), 358-368. Section 5, printed pages 360-361.

   Classical common-modulus congruence solution counting. The rational-grid notation and fiber normalization are the preprint's explicit specialization, not notation attributed to this source.

2. Richard P. Stanley. Smith Normal Form in Combinatorics. arXiv:1602.00166, version 2 (2016). Theorems 2.3-2.4.

   Smith normal form, quotient interpretation and determinantal-divisor facts.

Cite the source preprint

Walke, Rushikesh Murlidhar. (2026). Counting intersections of rational cyclic grids: an application of Smith normal form (Version 9.1) [Preprint]. Zenodo. DOI: 10.5281/zenodo.23184983.

This DOI identifies the source preprint, not this condensed reading companion. The preprint is AI-assisted and has not been independently peer-reviewed.

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