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Fix $m\in\mathbf{Z}$ and consider for $x,y\in\mathbf{Z}$ the relation $\sim$, where $x\sim y$ whenever there is a $z\in\mathbf{Z}$ such that $z^2=(x\cdot y)^m$.

Consider for a fixed $n\in\mathbf{N}$ the relation $\Delta$ on $\mathbf{N}$ where $x\Delta y$ whenever there is a $z\in\mathbf{N}$ such that $z^2=(x\cdot y)^n$. I use the convention $0\notin \mathbf{N}$.

How can I show that $\sim$ is not transitive and $\Delta$ is? And how can I determine the cardinality of the equivalence classes under $\Delta$?

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    If m<0 only +1 and -1 are related with itself and there are not any more relation. If m=0 there are relation between all integers. In both case the relation is transitive, isn't it?2012-06-02

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