Elementary GCD number theory proof

Solution 1:

Simplifying the gcd notation to $(a,b)$, etc., note first that for any four numbers,

$$(a,b,c,d)=((a,b,c),d)=((a,b),c),d)=((a,b),(c,d))$$

with any permutations of the variables. This is because for each prime the gcd picks out the least power that divides the numbers under consideration.

Note next that $m(a,b)=(ma,mb)$ in general. Together, these equalities imply

$$\begin{align} (a,c)(a,d)&=(a(a,c),d(a,c))\quad\text{letting }m=(a,c)\\ &=(a(a,c),(ad,cd))\quad\text{letting }m=d\\ &=(a(a,c),(ad,ab))\quad\text{using }ab=cd\\ &=(a(a,c),a(d,b))\\ &=a((a,c),(d,b))\\ &=a(a,c,d,b)\\ &=a(a,b,c,d) \end{align}$$