How to prove $\gcd(a,b) \cdot \gcd(c,d)=\gcd(ac,ad,bc,bd)$? I want to prove the identity $\gcd(a,b) \cdot \gcd(c,d)=\gcd(ac,ad,bc,bd)$. I tried this: if $x=\gcd(a,b)$ and $y=\gcd(c,d)$ then I must show $xy=\gcd(ac,ad,bc,bd)$ so I think I have to use the property $\gcd(ar,br)=r\cdot \gcd(a,b)$ but I don't know how to apply it. Can someone help me?

Another way to think about this problem is the following fact: two integers $a$ and $b$ are equal when $r \mid a \implies r \mid b$ and $r \mid b \implies r \mid a$.
(This is similar to how equality works for sets, where $S=T$ when $x \in S \implies x \in T$ and $x \in T \implies x \in S$.)
Here's how you can prove one direction. Suppose $r \mid \gcd(a,b) \cdot \gcd(c,d)$. This means that either $r \mid \gcd(a,b)$ or else $r \mid \gcd(c,d)$.


*

*If $r \mid \gcd(a,b)$, then $r \mid a$ and $r \mid b$. Therefore $r \mid ac, r \mid ad, r \mid bc, r \mid bd$. From this, we can conclude that $r$ is a common divisor of $\{ac, ad, bc, bd\}$, so $r \mid \gcd(ac, ad, bc, bd)$.

*If $r \mid \gcd(c,d)$, then $r \mid c$ and $r \mid d$. So in this case, too, $r \mid ac, r \mid ad, r \mid bc, r \mid bd$. Once again, we can conclude that $r$ is a common divisor of $\{ac, ad, bc, bd\}$, so $r \mid \gcd(ac, ad, bc, bd)$.


Can you prove the other direction: that if $r \mid \gcd(ac, ad, bc, bd)$, then $r \mid \gcd(a,b) \cdot \gcd(c,d)$?
