Consider the Diophantine equation $$k_0a+k_1b+k_2c+k_3d+\cdots=1$$ where $a,b,c,d,\cdots$ are variables and suppose that a solution obtained through the Euclidean Algorithm is $a_0,b_0,c_0,d_0,\cdots$.
What is the general solution for $a,b,c,d,\cdots$?
If we take a step back and consider such an equation with two variables, then for $$k_0a+k_1b=1$$ we get the general solution $a=a_0+k_1t$ and $b=b_0-k_0t$ for some integer $t$.
How should one deal with this for an equation with more than two variables?
P.S. Finding one solution isn't a problem as we could pair up $(a,b)$, $(c,d)$ and then use the Extended Euclidean Algorithm.