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I know that the question doesn't really match the level of this site, but I will be very grateful if someone showed me the proof:

$\forall x,y,z \in \mathbb{Z}$ if $xz \hspace{4 pt} \vdots \hspace{4 pt} (z - y)$ then $xy \hspace{4 pt} \vdots \hspace{4 pt} (z - y)$,

When $x \hspace{4 pt} \vdots \hspace{4 pt} (z - y)$, it's obvious that for any $y \in \mathbb{Z}$ this holds. But how do I prove for all the other cases?

Thank you!

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    Your question *does* match the level of this site!2012-04-07

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In the notations I know, $x \vdots y$ is equivalent to $y |x$. I'm not sure what you mean, but if this is it, then the proof is obvious:

$y-z$ divides $xz=xy-(y-z)x$

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    I see now: $xz = xy - (y - z)x \hspace{2 pt} \vdots \hspace{2 pt}(y - z) \implies \exists c \in\mathbb{Z} : xy - (y - z)x \hspace{2 pt} = \hspace{2 pt}(y - z)c \implies xy = (y - z)c + (y-z)x = (y-z)(c+x) = -(z - y)(c+x) \implies xy \hspace{2 pt} \vdots \hspace{2 pt} (z-y)$. Thank you! You helped me so much!. Definitely accepted!2012-04-07
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Hint $\rm\ \ x\:(z-y)\: =\: xz - xy\:\ \Rightarrow\:\ z-y\ |\ xz\!\iff\! z-y\ |\ xy$

i.e. $\rm\ mod\ z-y\!:\:\ z\equiv y\ \: \Rightarrow\:\ xz\equiv xy\:\ \Rightarrow\:\ xz\equiv 0\!\iff\! xy\equiv 0$