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Source. The unnumbered Theorem on p. 77 (physical p. 1 of the scan); the paper proves it on p. 80 (physical p. 4) by combining Lemmas 1 and 2. The proof below is written here and runs through the paper's two lemmas.

T. Cochrane and G. Myerson, Covering congruences in higher dimensions, Rocky Mountain J. Math. 26 (1996), no. 1, 77–81, doi:10.1216/rmjm/1181072104; the edition read and its page mapping are named on the source card.

Statement

There is a finite family of triples

$$(a_1,b_1,m_1),\ldots,(a_r,b_r,m_r),\qquad 1<m_1<\cdots<m_r,\qquad \gcd(a_j,b_j,m_j)=1, \tag{1}$$

such that every $(x,y)\in\mathbb Z^2$ satisfies at least one homogeneous congruence

$$a_jx-b_jy\equiv0\pmod {m_j}. \tag{2}$$

Proof

Use the twenty-class composite cover in Lemma 2. Every one of its moduli has no prime divisor other than $2$, $3$, or $5$. Apply Lemma 1. It gives the three vertical triples

$$(0,1,2),\qquad(0,1,3),\qquad(0,1,5), \tag{3}$$

together with $(1,a,m)$ for each of the twenty pairs $(a,m)$ in Lemma 2. Thus this construction uses twenty-three congruences.

Lemma 1 proves that the resulting family covers every ordered pair. The three moduli in (3) are distinct primes, while all twenty remaining moduli are distinct composite integers. Hence all twenty-three moduli are different and greater than one. The vertical triples and lifted triples satisfy

$$\gcd(0,1,p)=1, \qquad \gcd(1,a,m)=1.$$

Reorder the triples by their moduli to obtain (1). This proves the theorem.

Scope

The theorem concerns homogeneous congruences in two variables. It does not produce a one-dimensional homogeneous cover: the integer $1$ would fail every congruence $x\equiv0\pmod m$ with $m>1$. The subgroup and matrix consequences and the extension to more variables are proved separately.

Bears on. Higher-dimensional analogs of covering systems. This construction does not settle the one-dimensional odd-covering question in Problem 7.