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Source. Lemma 1 on p. 78, with its proof on pp. 78–79 (physical pp. 2–3 of the scan). The paper's definition of a cover (p. 77) requires distinct moduli greater than one, which is why the statement below says distinct. The proof below is written here, including the distinct-modulus and primitivity checks the paper leaves implicit.

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

Let

C={(ai,mi):1≤i≤r}\mathcal C=\{(a_i,m_i):1\le i\le r\}

be a finite covering system of the integers whose moduli are distinct and composite. Put M=∏imiM=\prod_i m_i, and let p1,…,ptp_1,\ldots,p_t be the distinct prime divisors of MM. Then the triples

(0,1,p1),…,(0,1,pt),(1,a1,m1),…,(1,ar,mr)(1)(0,1,p_1),\ldots,(0,1,p_t), (1,a_1,m_1),\ldots,(1,a_r,m_r) \tag{1}

form a homogeneous cover of Z2\mathbb Z^2: every (x,y)∈Z2(x,y)\in\mathbb Z^2 satisfies one of

−y≡0(modpj),x−aiy≡0(modmi).(2)-y\equiv0\pmod {p_j}, \qquad x-a_i y\equiv0\pmod {m_i}. \tag{2}

All moduli in (1) are distinct and greater than one, and every coefficient triple is primitive in the sense that its three entries have greatest common divisor one.

Proof

Fix (x,y)(x,y). If gcd⁡(y,M)>1\gcd(y,M)>1, choose a prime pjp_j dividing that greatest common divisor. Then y≡0(modpj)y\equiv0\pmod {p_j}, so the corresponding vertical congruence in (2) holds.

Suppose instead that gcd⁡(y,M)=1\gcd(y,M)=1. Choose z∈Zz\in\mathbb Z with

yz≡1(modM).yz\equiv1\pmod M.

The one-dimensional family C\mathcal C covers the integer xzxz, so for some ii,

xz≡ai(modmi).xz\equiv a_i\pmod {m_i}.

Multiplication by yy is valid modulo mim_i, and yz≡1(modmi)yz\equiv1\pmod {m_i}. Therefore x≡aiy(modmi)x\equiv a_i y\pmod {m_i}, which is the corresponding lifted congruence in (2). The two cases cover every ordered pair.

The pjp_j are distinct primes, while the mim_i are distinct composite integers, so no modulus in the first part of (1) equals one in the second. Finally,

gcd⁡(0,1,pj)=gcd⁡(1,ai,mi)=1.\gcd(0,1,p_j)=\gcd(1,a_i,m_i)=1.

Thus (1) satisfies every part of the homogeneous-cover definition.

Bears on. The construction of a finite homogeneous cover in the main theorem.