The main problem that I want to solve is the Diophantine equation $$c_1x_1 + c_2 x_2 + c_3 x_3 + c_4 x_4 = 0\;\;\;\text{(I)},$$
when $c_1+c_2+c_3+c_4=0$. The problem that I'm facing is that the solution that I'm finding is not compatible with $(t,t,t,t)$, that is always a solution for every $t\in\mathbb{Z}$.
To solve (I), I first consider $x_1 = t$ and try to solve the equation
$$(c_2+c_3+c_4)t = c_2x_2+ c_3 x_3 + c_4 x_4\;\;\;\text{(II)},$$
for every $t\in\mathbb{Z}$. To solve (II), I solve $$dy + c_4x_4 = (c_2+c_3+c_4)t\;\;\;\text{(III)},$$ where $d=\gcd(c_2,c_3)$, and $$dy=c_2x_2 + c_3x_3\;\;\;\text{(IV)},$$
A particular solution of (III) is $$y = \frac{c_2+c_3}{d}t\;\;\;\text{and}\;\;\; x_4 = t.$$
Hence, the general solution of (II) is $$ y=\frac{c_2+c_3}{d}t + \frac{c_4}{e}u\;\;\;\text{and}\;\;\; x_4 = t-\frac{d}{e}u,$$ (where $e=\mathrm{gcd}(d,c_4)=\mathrm{gcd}(c_2,c_3,c_4)$) for every $u\in\mathbb{Z}$.
By finding $z_2,z_3\in\mathbb{Z}$ such that $d=c_2z_2+c_3z_3$, the general solution of equation (I) is
\begin{cases} x_1 & = t,\\ x_2 & = \frac{(z_2+c_3v)[(c_2+c_3)t + dc_4 u]}{d}, \\ x_3 & = \frac{(z_3-c_2v)[(c_2+c_3)t+c_4u]}{d},\text{ and }\\ x_4 & = t-\frac{d}{e}u.\\\end{cases}
As $(t,t,t,t)$ is always a solution, I try to discover $z_2$ and $z_3$ by pluggin $x_i=t$ for any $i\in\{1,2,3,4\}$. The result is
$$z_2=z_3=\frac{d}{c_2+c_3},$$ which is not an integer. What I'm doing wrong here?
added, Wednesday may 27. There is a one line proof that the process below tells us an integral basis for the lattice. use new names, column vector $w$ with element gcd = 1. The row is $w^T.$ We have $h$ being any lattice element, meaning $w^T h = 0.$ And we construct a square integer unimodular matrix $W$ with $w^T W = (1,0,0,0...,0).$ This gives $w^T =(1,0,0,...,0) W^{-1}.$ Put together with $w^T h = 0,$ we arrive at $$ (1,0,0,0...0) W^{-1} h = 0. $$ In turn, this means the first element in $W^{-1}h$ is zero, the others are a bunch of integers. so $$ W^{-1} h = \left( \begin{array}{c} 0 \\ g_2 \\ g_3 \\ \vdots \\ g_n \\ \end{array} \right) $$ Multiply on the left by $W$ for $$ h = W \left( \begin{array}{c} 0 \\ g_2 \\ g_3 \\ \vdots \\ g_n \\ \end{array} \right) $$ If you write this out, you see it means that $h$ itself is an integer linear combination of the columns of $W$ except the first (left hand) column. We get $h = g_2 c_2 + g_3 c_3 + \cdots + g_n c_n,$ where $c_j$ is column $j$ in the square matrix $W.$ The second through final columns of $W$ really do span the lattice of integer vectors orthogonal to $w^T.$ Furthermore, $\det W = 1$ says that they are independent; the second through $n$th columns of $W$ are a basis.
ORIGINAL: The problem you are doing is usually described as finding a basis for the integral lattice of integer vectors orthogonal to your given vector. I did one, I think the way I do this one is better than my usual way. I begin with row vector $v=(105,70, 42, 30)$ and find a square integer matrix, call it $R,$ of determinant $1$ such that $ vR = (1,0,0,0) .$ It follows that the second, third, and fourth column of $R$ are a basis for the lattice. Finally, I use a command for the LLL basis reduction.
The transpose of that gives the basis as rows, namely $$(-2, 3, 0, 0)$$ $$(-2, 0, 5, 0)$$ $$(-2, 0, 0, 7)$$
The Gram matrix for this basis is
so that the associated quadratic form is $$ 13 x^2 + 29 y^2 + 53 z^2 + 8yz + 8zx +8xy. $$
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