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?

Question

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?

anonymous · Answer

Nobody besides James ever helps... :/

anonymous · Answer

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

anonymous · Answer

I understand every step besides "$b|(b+sm,m)$." Does anyone know why this is the case?

anonymous · Answer

I have the feeling that he may have typoed and should have written $d|(b+sm,m)$ instead. What do you think?

kinggeorge · Answer

(b, m) is the greatest common divisor of b and m correct?

anonymous · Answer

That's correct.

kinggeorge · Answer

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?

anonymous · Answer

Thank you. :) You write \.(\) instead of \.[\] (without the dot).

kinggeorge · Answer

Thanks.