Saturday, 18 July 2015

elementary number theory - Show that if gcd(abc,d2)=1, then gcd(a,d)=gcd(b,d)=gcd(c,d)=1.

Let a,b,c be integers. Show that if gcd, then \gcd(a,d)=\gcd(b,d)=\gcd(c,d)=1.




Here is my way of approaching this question:



Suppose \gcd(abc,d^2)=1, there exist integers x,y such that abcx+d^2y=1



a(bcx)+d(dy)=1, which implies that \gcd(a,d)=1



b(acx)+d(dy)=1, which implies that \gcd(b,d)=1



c(abx)+d(dy)=1, which implies that \gcd(c,d)=1




So far I don't really know if this is the way to answer this question. Any help would be appreciated.

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