Prove that if $\operatorname{gcd}(a, b)=1$ then $\operatorname{gcd}(a c, b)=\operatorname{gcd}(c, b).$

Question

• Prove that if $\operatorname{gcd}(a, b)=1$ then $\operatorname{gcd}(a c, b)=\operatorname{gcd}(c, b).$

My Attempt. Assume $\operatorname{gcd}(a, b)=1,$ then $1|a$ and $1|b.$ Let $\operatorname{gcd}(c, b)=d$, that is $d|c$ and $d|b$ whenever an integer $d'|c$ and $d'|b$, we have $d'|d.$ We need to show $\operatorname{gcd}(ac, b)=d.$ Since $1|a$ and $d|c$, we get $d|ac.$ Now,....

How can I complete the proof? Can you help? Thanks...

Answer 1

Let $d= \mathrm{gcd}(c,b)$. Then $d \mid c$, $d \mid b$ and clearly $d \mid ac$, so $d \mid \mathrm{gcd}(ac,b)$.

Since $1= \mathrm{gcd}(a,b)$, then $1=ax+by$ for some $x,y$. Similarly $d=cz+bw$ for some $z,w$.

$$d=1 \cdot d=(ax+by)(cz+bw)=acxz+bcyz+abxw+b^2yw=acX+bY$$ where $X=xz$ and $Y=axw+byw+cyz$.

It follows then that $\mathrm{gcd}(ac,b) \mid d$ and thus $\mathrm{gcd}(ac,b)=\mathrm{gcd}(c,b)$.