Thread: gcds

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?

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 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.