On representing the general solution with the special solutions for the diophantine equation


here $a_1 ,a_2, \dotsb,a_n,c\in\Bbb Z,(a_1 ,a_2, \dotsb,a_n)=1$.

Can we find the general solution? We all know, $n=2$, the general solution is $x=x_0-a_2t,y=y_0+a_1t$.

How about $n\geq 3$? It doesn't exist such a formula, does it?


Actually we can find the general solution but i believe it is not so usefull in practice.
I will rewrite J.Hunter's solution which i found in his book Number Theory:

If $d=(a_1,a_2,...,a_n)$ and $d\mid c$ then the general solution is given by
$$\left(\begin{matrix}x_1\\ x_2\\ .\\ .\\ .\\ x_n\\\end{matrix}\right)=\left(\begin{matrix} a_{11} & a_{12} & . . . & a_{1n} \\ a_{21} & a_{22} & . . . & a_{2n} \\ \cdots& \cdots &\cdots &\cdots\\ \cdots &\cdots&\cdots&\cdots\\ \cdots &\cdots&\cdots&\cdots\\ a_{n1} & a_{n2} & . . . & a_{nn} \end{matrix}\right)\cdot \left(\begin{matrix} 1 \\ t_1 \\ . \\ .\\ .\\ t_{n-1} \end{matrix}\right)$$
With all $a_{ij},t_1,...,t_{n-1}$ being integers.
I think that this the best you will find.


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