# General solution for a Diophantine equation with more than two variables

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.

• You can find the solution to this problem in "Diophantine Equations" by L.J. Mordell. Academic Press p. 30-31 (1969) – Piquito Oct 17 '18 at 19:07

$$a = a_0 + k_1 t_1 + k_2 t_2 + …$$
$$b = b_0 - k_0 t_1$$
$$c = c_0 - k_0 t_2$$