Let a,b,c are in Z with (a,b)=1 and a divides bc. Prove that a divides c. My work: ax+by=1 bc=am (ax+by=1)C cax+cby=c ax(am/b)+by(am/b)=c a^2xm/b+amy=c a(axm/n+my)=c dont know if this is correct?
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Hello, Originally Posted by rmpatel5 Let a,b,c are in Z with (a,b)=1 and a divides bc. Prove that a divides c. My work: ax+by=1 bc=am (ax+by=1)C cax+cby=c ax(am/b)+by(am/b)=c a^2xm/b+amy=c a(axm/b+my)=c dont know if this is correct? Yes it is, but you have to say that $\displaystyle \frac{amx}{b}+my$ is an integer ! Which is true because am/b=c.. actually, you could do this way : (ax+by=1)*c acx+cby=c acx+by(am/b)=c << keep acx a(cx+my)=c cx+my is obviously an integer.
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