Show that if n | m, where n and m are integers greater than 1, and if a≡b (mod m), where a and b are integers, then a≡b (mod n).

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If $n|m$ , then there is integer number $k$ such that $m=kn$ .

The following conditions are equivalent: $a\equiv b\mod m$ and $m|(a-b)$ .

If $m|(a-b)$ , then there is integer number $l$ such that $a-b=ml$ .

So, $a-b=ml=kn\cdot l= (kl)\cdot n$ .

It means, that $n|(a-b)$ or it can be written as $a\equiv b\mod n$