GCD Associativity: Proving gcd(a,b,c) with Linear Combinations

[lordy12]

1.Show gcd(a,b,c) = gcd(a, gcd(b,c))

Homework Equations

3. My attempt is that gcd(a,b,c) can be written as the product of their prime factors. Let's say x is that product. The thing is, I know how to prove this using prime factorization but there has to be another method concerning linear combinations. Like gcd(a,b,c) = ax + by+ cz.

 

[HallsofIvy]

Why not just say a= nx, b= ny, c= nz where n= gcd(a,b,c).
Of course, you also have b= mp, c= mq where m= gcd(b,c).