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?

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 $min$.
