HV56: Condensed readable mathematical companion to preprint v9.1
Condensed companion, not a full transcript or revised preprint
Companion date: 2026-10-09
Source: Counting intersections of rational cyclic grids: an application of Smith normal form
Author: Rushikesh Murlidhar Walke
Source version: 9.1; prepared 2026-10-06; 7 pages
https://rmwalke.com/hv56/HV56_Preprint_V9_1.pdf
SHA-256: 5fa16b34427e28adb0e74c91554571d709fcd7483bdf615af3fcfdc883e59f68
AI-assisted condensed companion; not an independent mathematical review or review certificate
Equation numbers and page numbers below refer to the v9.1 PDF. The accompanying explanations are condensed restatements, not quotations of the entire paper. Read the PDF together with this companion.

PURPOSE
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.

ASSUMPTIONS
- 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.

EQUATIONS (numbers and pages refer to the source PDF)

(1) Ordered-pair count — PDF page 2
C(d,b;y,k) = #{(u,h) in G_d x G_b : u+h in G_y and u-h in G_k}
All coordinates are classes modulo 1 and all four orders are positive integers.

(2) Common denominator and matrix — PDF page 3
L = lcm(d,b); alpha = L/d; beta = L/b; A = [[y*alpha, y*beta], [k*alpha, -k*beta]]
L is a common denominator; the original input still has d*b elements, not L*L.

(3) Smith invariant factors and scalar image size — PDF page 3
s1 = gcd(y*alpha, y*beta, k*alpha, k*beta); s2 = (2*y*k*alpha*beta)/s1; J_L(s) = L/gcd(L,s)
s1 and s2 are positive Smith invariant factors of A, with s1 dividing s2. Each J_L(s) is an integer.

(4) The count — PDF page 3
C(d,b;y,k) = (d*b)/(J_L(s1)*J_L(s2)) = ((d*b)/(L*L))*gcd(L,s1)*gcd(L,s2)
The complete expression is an exact integer count. The isolated factor (d*b)/(L*L) need not be an integer. No floor or rounding operation is used.

(5) Membership as congruences — PDF page 3
y*(alpha*i + beta*j) == 0 (mod L); k*(alpha*i - beta*j) == 0 (mod L)
u=i/d+Z and h=j/b+Z, where i is modulo d and j is modulo b. In denominator L the points are alpha*i/L and beta*j/L. The symbol == here denotes congruence.

(6) Equality of the two images — PDF page 3
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
The domains differ. Column-period reduction proves equality of their images; this does not assert that a unimodular change of coordinates preserves the rectangular input domain.

(7) Square-kernel normalization — PDF page 4
|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.
tau is well defined and surjective because d and b divide L. The kernel of Abar is the inverse image under tau of the kernel of phi.

(8) Union count — PDF page 5
Q(a,b;y,k;r) = C(a,b;y,k) + C(r,b;y,k) - C(gcd(a,r),b;y,k)
Each occurrence of C uses its own L, alpha, beta, s1 and s2. The grid identity G_a intersect G_r = G_gcd(a,r) determines the subtracted intersection.

PROOF STEPS (condensed restatement)

1. Translate membership without changing the inputs
PDF pages: 2, 3
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).

2. Verify the rectangular-domain homomorphism
PDF pages: 3
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.

3. Compare images before using Smith normal form
PDF pages: 3
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).

4. Identify the Smith invariant factors
PDF pages: 3, 4
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.

5. Count the square-domain image
PDF pages: 3, 4
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).

6. Recover the required count and its integrality
PDF pages: 4
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.

7. Check the alternative fiber normalization
PDF pages: 4
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.

8. Justify inclusion-exclusion for a union
PDF pages: 4, 5
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.

ORIGINAL WORKED EXAMPLES (not new scientific runs)

Unequal denominators — PDF page 5
(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
Pairs: (0, 0), (1/2, 1/2)
These are all admissible ordered pairs. Replacing the rectangular input by the entire square-domain kernel without the fiber normalization would give the wrong count.

Parity on an even grid — PDF page 5
(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
Writing u=i/6 and h=j/6, both conditions require i+j and i-j to be even. Equivalently i and j have the same parity. There are three even and three odd residues modulo 6, giving 3*3+3*3=18. Division by 2 cannot be performed freely in a congruence with an even modulus.

Union example — PDF page 6
a=4; r=6; b=4; y=3; k=6; gcd(a,r)=2
d | L | alpha | beta | s1 | s2 | J1 | J2 | count
4 | 4 | 1 | 1 | 3 | 12 | 4 | 1 | 4
6 | 12 | 2 | 3 | 3 | 72 | 4 | 1 | 6
2 | 4 | 2 | 1 | 3 | 24 | 4 | 1 | 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.
Pairs: (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)
Shared pairs: (0, 0), (1/2, 1/2)
The two shared pairs are included in each of the first two counts and are subtracted once. Every pair in the final list is counted once. Rational coordinates in this companion are exact strings, not decimal approximations.

READING CLARIFICATIONS

Integrality concerns the complete count
The complete expression in equation (4) is an integer. In the second expression, the isolated factor (d*b)/(L*L) need not be an integer: for d=6,b=4 it is 24/144=1/6. It is the product with both gcd factors that gives the integer count. This is an explanatory clarification accompanying the unchanged PDF.

The origin pair is always present
For the stated unshifted grids, (0,0) belongs to every input grid pair, and both its sum and difference are 0 in every target grid. Therefore C(d,b;y,k)>=1. There is no separate zero-count condition in the theorem. Since C is a kernel size, it divides d*b and satisfies 1<=C<=d*b. These are consequences of the definitions and proof, not reports of new computational tests.

The common denominator is not the input size
The input has d*b ordered pairs. The auxiliary square domain has L*L elements. Its kernel count must be divided by the fiber size L*L/(d*b). Equal images do not imply equal kernel sizes for domains of different cardinalities.

Recompute the invariants for each union term
Equation (8) applies the basic formula separately to input orders a, r and gcd(a,r), with the same b,y,k. Reusing a common set of L,alpha,beta,s1,s2 across the three terms would generally change the count.

Use the actual equations and state reading limitations
The original PDF's extracted text can omit mathematical symbols. Use the selectable equations here together with the numbered equations and page images in the PDF. If a displayed formula cannot be read, record that limitation instead of reconstructing an unobserved formula.

SCOPE 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.
https://rmwalke.com/hv56/Human_Review_Brief_EN.txt
A useful mathematical review identifies the source version, page and equation, then supplies the precise inference, missing assumption or counterexample under examination.

CLASSICAL REFERENCES
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
https://doi.org/10.4153/CJM-1955-039-2
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.

Richard P. Stanley. Smith Normal Form in Combinatorics. arXiv:1602.00166, version 2 (2016).
Theorems 2.3-2.4
https://arxiv.org/abs/1602.00166
Smith normal form, quotient interpretation and determinantal-divisor facts.