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anonymous
 4 years ago
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 \(db+sm\) 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?
anonymous
 4 years ago
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 \(db+sm\) 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nobody besides James ever helps... :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Because not all the people in the world are smart like iq of 680 ok we try are best to answer questions vitally for free

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I understand every step besides "\(b(b+sm,m)\)." Does anyone know why this is the case?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have the feeling that he may have typoed and should have written \(d(b+sm,m)\) instead. What do you think?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.1(b, m) is the greatest common divisor of b and m correct?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.1I 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you. :) You write \.(\) instead of \.[\] (without the dot).
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