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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.

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

$a = a_0 + k_1 t_1 + k_2 t_2 + …$

$b = b_0 - k_0 t_1$

$c = c_0 - k_0 t_2$

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