I am interested in finding general form of the solution of linear Diophantine equation
$$ ax + by +cz = d. $$
We will use notation $$ f : \mathbb{Z^3} \to \mathbb{Z}, $$ $$ f(x,y.z) = ax + by + bz; $$ for breifity
It is easy to see that if the equation has a solution $(x_0,y_0,z_0) \in \mathbb{Z}^3$, then any other solution $(x_0',y_0',z_0')$ can be expressed as $$ (x'_0,y'_0,z'_0) = (x_0,y_0,z_0) + (x'_0 - x_0,y'_0 - y_0,z'_0 - z_0). $$
It is clear that the last term belongs to the $\ker f$ . So, if the basis $(e_1,e_2)$ of $\ker f$ was known it would be possible to write for $\alpha, \beta \in \mathbb{Z}$ $$ (x'_0,y'_0,z'_0) = (x_0,y_0,z_0) + \alpha e_1 + \beta e_2. $$
Despite that it is easy to get two linearly independent elements $v_1, v_2 \in \ker f$, as we are working with the ring of scalars which is not a field, it is not true that $(v_1,v_2)$ always form a basis.
What are correct expressions for basis elements $e_1, e_2$?
I can suggest $$ v_1 = \left(0, -\frac{c}{\mathrm{gcd}(c,b)}, \frac{b}{\mathrm{gcd}(c,b)} \right), $$ $$ v_2 = \left( \frac{b}{\mathrm{gcd}(a,b)}, -\frac{a}{\mathrm{gcd}(a.b)} , 0\right); $$ for the start.
Found something that shows things in general: see the comments at https://mathoverflow.net/questions/103152/determinant-of-integer-lattice-basis-of-l-x-1-ldots-x-n-a-1x-1-cdotsa Note that the Gram determinant I talk about is exactly the square of the volume of the fundamental parallelotope ( parallelogram or parallelepiped or...)
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Your pair do not necessarily give a basis for the integral lattice; having finished my part, I will now check the Gram matrix for your basis. If we give new integers $p,q,r,$ I guess also $g, \alpha, \beta,$ with $$ \gcd(a,b,c) = 1, $$ $$ pa + q b = g = \gcd(a,b), $$ $$ a = g \alpha, \; \; \; b = g \beta, $$ then (several) steps lead to basis $$ v_1 = (-\beta, \alpha, 0), $$ $$ v_2 =( -pc, -qc ,g ). $$
Let me give the Gram matrix and determinant, need some time
$$ G = \left( \begin{array}{cc} \alpha^2 + \beta^2 & (p \beta - q \alpha) c \\ (p \beta - q \alpha) c & g^2 + (p^2 + q^2) c^2 \end{array} \right) $$ with $$ \det G = a^2 + b^2 + c^2 $$
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Alright, finished yours: if $g = \gcd(a,b)$ and $h = \gcd(b,c)$ then your determinant is $$ \left( \frac{b^2}{g^2 h^2} \right) \left( a^2 + b^2 + c^2 \right).$$ This is too large, especially in the extreme case when $g=h=1,$ which would happen for the example $a = 11, b = 13, c = 17.$
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Alright, deriving carefully. $$ ax+by +cz = 0. $$ $$ \gcd(a,b,c) = 1, $$ $$ pa + q b = g = \gcd(a,b), $$ $$ a = g \alpha, \; \; \; b = g \beta, $$ $$ p \alpha + q \beta = 1. $$ $$ ax+by = -cz $$ $$ (\alpha x + \beta y) g = - cz. $$ $$ \gcd(g,c) = 1,$$ so that $$ g | z, $$ let $$ z = g w. $$ $$ (\alpha x + \beta y) = - cw. $$ We have $$ p \alpha + q \beta = 1, $$ so $$ (\alpha x + \beta y) = - c(p \alpha + q \beta)w. $$ Move around, $$ (x+pcw) \alpha = - (y+qcw) \beta. $$ Again, $$ \gcd(\alpha, \beta) = 1, $$ so $$ \alpha | (y+qcw). $$ Let $$ y+qcw = \alpha t, $$ $$ x+pcw = - \beta t $$ We have constructed integers $t,w$ with $$ x = \beta t - p c w, $$ $$ y = \alpha t - q c w, $$ $$ z = g w. $$
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