Question

If gcd(a,b)=1 and c|a, then gcd(b,c)=1

Answer 1

since g.c.d.(a,b)=1 then there exist integers x0 and y0 such that
\[ \large 1=ax_0+by_0 \]
also since \(c\mid a\) then there exists k integer such that
\[ \large a=ck \].
Combining both equations we have
\[ \large 1=ax_0+by_0=(ck)x_0+by_0=c(kx_0)+by_0 \]
then g.c.d.(c,b)=1