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Suppose $n|m$, then $n\cdot d=m,~d\in\Bbb Z$.

If $a\equiv b\mod m$, then $a\equiv b\mod{n\cdot d}$.
Additionally, $d\equiv d\mod{n\cdot d}$.

So we know $ad\equiv bd\mod{n\cdot d}$.
$\therefore a\equiv b\mod n$

Would this be a valid proof?

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  • $\begingroup$ What is the statement that you're proving? $\endgroup$ – Yanko Oct 4 at 21:37
  • $\begingroup$ I'm trying to prove that if n|m where n, m are integers greater than 1, and if a === b (mod m) then a === b (mod n) $\endgroup$ – Paul Silvestri Oct 4 at 21:45
  • $\begingroup$ Ok then your argument is valid. But perhaps you need to prove the last line (like, why can you divide everything by $d$, did you prove this in class?) $\endgroup$ – Yanko Oct 4 at 21:47
  • $\begingroup$ I don't think so. So I would need to prove that? $\endgroup$ – Paul Silvestri Oct 4 at 21:48
  • $\begingroup$ I think you do, if there's a $k$ such that $ad= (nd)\cdot k + bd$ what can you do next? $\endgroup$ – Yanko Oct 4 at 21:49

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