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The statement is:

Let $A=\{(a_s(n_s)\}^k_{s=1}$ be an exact cover of $\mathbb{Z}$ with $1. Then we must have $n_{k-1}=n_k$.

The only proof I could find was here. But I have some difficulties understanding it. It is very short and it goes like this:

Without loss of generality we assume that $0\leq a_s for all $s\in[1,k]$. For $|z|<1$ we have $\sum_{s=1}^k\frac{z^{a_s}}{1-z^{n_s}}=\sum_{s=1}^k\sum_{q=0}^\infty z^{a_s+qn_s}=\sum_{n=0}^\infty z^n=\frac{1}{1-z}.$ If $n_{k-1} then $\infty=\lim_{\substack{z\to e^{2\pi i/n_k}\\ |z|<1}}\frac{z^{a_k}}{1-z^{n_k}}=\lim_{\substack{z\to e^{2\pi i/n_k}\\ |z|<1}}\left(\frac{1}{1-z}-\sum_{s=1}^{k-1}\frac{z^{a_s}}{1-z^{n_s}}\right)<\infty,$ a contradiction!

What I don't get is the following: $\sum_{s=1}^k\sum_{q=0}^\infty z^{a_s+qn_s}=\sum_{n=0}^\infty z^n.$ It just doesn't make sense to me. So what am I misunderstanding?

1 Answers 1

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That equation states that $A$ is an exact cover. On the left, you have all the terms $z^m$ where $m$ satisfies $m\equiv a_s\pmod{n_s}$, with coefficient the number of such congruences $m$ satisfies. On the right, you have $z^n$ for every $n$, with coefficient $1$. So, the equality is saying that every non-negative integer satisfies exactly one congruence.

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    I see now, thank you very much.2012-11-16