If $\gcd(c,m)=1$ and if $ca≡cb \pmod m$ for some $a$ and $b$, one may argue that $a≡b \pmod m$. Note that this claim says that there are situations where we can "cancel" $c\mod m$.
Congruences, elementary number theory
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elementary-number-theory
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1Is there a question here? – 2013-08-27