## 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 $$d|b+sm$$ 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?

1. anonymous

Nobody besides James ever helps... :/

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. KingGeorge

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

6. anonymous

That's correct.

7. KingGeorge

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?

8. anonymous

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

9. KingGeorge

Thanks.