Does $ac\equiv ab \pmod{ad}$ always imply that $c\equiv b \pmod d$ ? If this is not always the case, then when is it ?
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$$ac\equiv ab \mod ad\iff\exists k\in\Bbb Z:ac-ab=adk\iff a(c-b)=a(dk)$$ then, if $a\neq 0$, cancellation of $a$ is valid, since $\Bbb Z$ is an integral domain. Thus, $c-b=dk\iff c\equiv b\mod d$.
Dave
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write your equation in the form $$ac=ab+kad$$ for some integer $k$ then we have $$a(c-b-kd)=0$$ if $a=0$ then all is clear, and if not then we get $$c=b+kd$$ and this is $$c\equiv b \mod d$$
Dr. Sonnhard Graubner
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