[Math] Method of solving extended Euclidean algorithm for three numbers

Question

I already got idea of solving gcd with three numbers. But I am wondering how to solve the extended Euclidean algorithm with three, such as:

47x + 64y + 70z = 1

Could anyone give me a hint? Thanks a lot.

Answer 1

Notice that $gcd(x,y,z)=gcd(x,gcd(y,z))$. First we find $a$, $b$ such that $gcd(x,gcd(y,z))=ax+bgcd(y,z)$, then $c$, $d$ such that $gcd(y,z)=cy+dz$. Finally we obtain $gcd(x,y,z)=ax+bcy+bdz$.