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A covering system is a set of congruences $\{a_1 \mod n_1, \dots a_k \mod n_k\}$ covering $\mathbb{Z}$. The Mirsky-Newman theorem says that a covering system cannot cover disjoint sets and have distinct moduli. The proof seems to assume we cover only the natural numbers. Why is this? Is it obvious that a congruence system covers $\mathbb{Z}$ iff it covers $\mathbb{N}$?

TorsionSquid
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2 Answers2

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It's not true that a congruence system that covers the naturals necessarily covers the integers. E.g., $0\bmod2,1\bmod4,3\bmod8,7\bmod{16},\dots$ covers the naturals but not $-1$. But a finite congruence system that covers the naturals must cover the integers. Let $m$ be the least common multiple of the moduli, then the whole shebang is periodic mod $m$, so if it covers everything from $1$ to $m$, it covers everything.

Gerry Myerson
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Consider the finite set of natural numbers under mod $n$ $$\{0,1,2,\ldots,n-1\}$$

I think it is sufficient if we show that above set is congruent to the set $$\{-0,-1,-2,\ldots,-(n-1)\}$$ in some order.

Take a look at thistheorem

AgentS
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