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?
 A: Indeed you use the formula $\gcd(ar,br)=r\cdot gcd(a,b)$ twice like this:
$$\begin{align} \gcd(a,b)\cdot\gcd(c,d) &= \gcd(a\cdot\gcd(c,d), b\cdot \gcd(c,d)) \\
&= \gcd(\gcd(ac,ad),\gcd(bc,bd)) \\
&= \gcd(ac,ad,bc,bd) \end{align}$$
A: $$\begin{align*}
gcd(a,b) \cdot gcd(c,d) &= gcd(c\cdot gcd(a,b),d\cdot gcd(a,b)) \\
& = gcd( gcd(ca,cb), gcd(ad,bd)) \\
& =  gcd(ca,cb,ad,bd)\end{align*}$$
A: If you have access to the ideal-theoretic characterization of $\gcd$ in a principal ideal domain:
$$ \langle \gcd(S) \rangle = \langle S \rangle $$
for any set $S$ of elements.
We can compute the product of ideals
$$ \langle a,b\rangle\langle c,d\rangle = \langle ac,ad,bc,bd \rangle $$
And the product of principal ideals is 
$$ \langle x \rangle \langle y \rangle = \langle x y \rangle $$
so we compute
$$ \langle \gcd(a,b) \gcd(c,d) \rangle
=  \langle \gcd(a,b) \rangle \langle \gcd(c,d) \rangle
\\= \langle a,b \rangle\langle c,d \rangle 
\\= \langle ac,ad,bc,bd \rangle
\\= \langle \gcd(ac,ad,bc,bd) \rangle $$
A: 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 $r \mid \gcd(a,b) \cdot \gcd(c,d)$?
A: For any prime $p$ and $n\in\mathbb{N}$, let $e_p(n)$ be the exponent of $p$ in $n$. Then $e_p(gcd(a,b)) = min(e_p(a),e_p(b))$ and similarly for $c,d$. Also $e_p(m\cdot n) = e_p(m)+e_p(n)$. Your result follows from distributivity of $+$ over $min$.
