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If $\alpha \in \mathbb{Z}(\omega)$, show that $\alpha$ is congruent to either $0, 1$ or $-1$ modulo $1-\omega$.

Exercise 1 page 134 in the book 'A Classical Introduction to Modern Number Theory' of K. Ireland and M. Rosen.

Thanks a lot.

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    -1 Question ill-posed, without due definitions and without answering back requests for clarification.2012-06-23

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Hint $\rm\: \omega\equiv 1\:\Rightarrow\: 0 = \omega^2\!+\!\omega\!+\!1\equiv 3\ $ so $\rm\ \omega\equiv 1,\ 3\equiv 0\:\Rightarrow\: m\!+\!n\omega \equiv (m\!+\!n)\:\! mod\ 3\equiv 0,\pm1$