• anonymous
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
• Stacey Warren - Expert brainly.com
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SOLVED
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