If $\gcd(n,a)=1$, and $\gcd(n,b)=1$, prove $\gcd(n,ab)=1$
2026-04-04 05:42:34.1775281354
If $\gcd(n,a)=1$, and $\gcd(n,b)=1$ prove $\gcd(n,ab)=1$
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Hint: $ua+vn =1, u'b+v'n=1$. $(ua+vn)(u'b+v'n)=1 = uu'(ab)+(uav'+u'vb+vnv')n$