$\gcd(ca,cb)\mid ca$ and $\gcd(ca,cb)\mid cb \to \gcd(ca,cb)\mid cd$. > Let $(ca)x + (cb)y = cd$ where $d = (a, b).$ > > Then since $\gcd(ca,cb)\mid ca$ and $\gcd(ca,cb)\mid cb \to \gcd(ca,cb)\mid cd$. I don't get how they deduced the conclusion. For one thing, $\gcd(ca,cb)\mid ca$ and $ \gcd(ca,cb)\mid cb \to \gcd(ca,cb)\mid (c(a + b)) $ where $a + b = d$ only if $a + b = ax + by$ meaning $x = y = 1$. Please, elaborate on this.

I can't make head or tails of your objection starting with "For one thing..." It is true that $\gcd(ac,bc)\mid c(a+b)$. Why is that a problem?

You are making this _much_ harder than it is. If $D\mid A$ and $D\mid B$ then $D\mid Ax+By$ for any integers $x,y$.

Now let $A=ac, B=bc,$ and $D=\gcd(A,B)=\gcd(ac,bc)$.