Someone wrote this proof: If \(b\equiv c\mod m\), then \((b,m)=(c,m)\). Proof: If \(b\equiv c\mod m\) then \(c=b+sm\) for some \(s\). Now if \(d=(b,m)\) then \(d|b+sm\) so \(b|(b+sm,m)\). Likewise if \(d'=(b+sm,m)\) then \(d'|(b+sm)-sm=b\) so \(d'|(b,m)=d\). Thus \(d=d'\). \(\blacksquare\) But I have no freaking clue how that proves the implication. o_o Could someone help me see how?

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Someone wrote this proof: If \(b\equiv c\mod m\), then \((b,m)=(c,m)\). Proof: If \(b\equiv c\mod m\) then \(c=b+sm\) for some \(s\). Now if \(d=(b,m)\) then \(d|b+sm\) so \(b|(b+sm,m)\). Likewise if \(d'=(b+sm,m)\) then \(d'|(b+sm)-sm=b\) so \(d'|(b,m)=d\). Thus \(d=d'\). \(\blacksquare\) But I have no freaking clue how that proves the implication. o_o Could someone help me see how?

Mathematics
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Nobody besides James ever helps... :/
Because not all the people in the world are smart like iq of 680 ok we try are best to answer questions vitally for free
I understand every step besides "\(b|(b+sm,m)\)." Does anyone know why this is the case?

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I have the feeling that he may have typoed and should have written \(d|(b+sm,m)\) instead. What do you think?
(b, m) is the greatest common divisor of b and m correct?
That's correct.
I think you're right. It only makes sense if it's d instead of b there. By the way, how do you do inline LaTeX here?
Thank you. :) You write \.(\) instead of \.[\] (without the dot).
Thanks.

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