I know i can divide with $6$ so I get $2x-5y+4z=3$, I replaced $2x-5y$ with $u$ so I got $u+4z=3$. I see that $u=3$ and $z=0$ is a solution but I got stuck. How can I continue? Is there a way to solve this using extended euclidean algorithm?
Thank you for your help.
On
Experiments in a spreadsheet show that $x$ $y$ $z$ combinations only add up to a given value if we increment $y$ by $4$ and $z$ by $5$ for a given value of $x$. Further experimentation shows offsets of $1$ and $2$, respectively to make the result zero.
There are infinite solutions starting with $x\in\mathbb{Z}$ and $$y = 4 n + 2 x + 1,\qquad z = 5 n + 2 x + 2\qquad n \in Z$$ Here are samples with $x\in\{-2,-1,0,1,2,3\}$
For $n=-1$ \begin{equation} \cdots\quad (-2,-7,-7)\quad (-1,-5,-5)\quad (0,-3,-3)\quad (1,-1,-1)\quad (2,1,1)\quad (3,3,3)\quad\cdots \end{equation} For $n=0$ \begin{equation} \cdots\quad (-2,-3,-2)\quad (-1,-1,0)\quad (0,1,2)\quad (1,3,4)\quad (2,5,6)\quad (3,7,8)\quad\cdots \end{equation}
For $n=1$ \begin{equation} \cdots\quad (-2,1,3)\quad (-1,3,5)\quad (0,5,7)\quad (1,7,9)\quad (2,9,11)\quad (3,11,13)\quad\cdots \end{equation} for $n=2$ \begin{equation} \cdots\quad (-2,5,8)\quad (-1,7,10)\quad (0,9,12)\quad (1,11,14)\quad (2,13,16)\quad (3,15,18)\quad\cdots \end{equation} For $n=3$ \begin{equation} \cdots\quad (-2,9,13)\quad (-1,11,15)\quad (0,13,17)\quad (1,15,19)\quad (2,17,21)\quad (3,19,23)\quad\cdots \end{equation} and so on.
$(-1,-1,0)$ is a solution.
and then we can add multiples $(2,0,-1)$ to any known solution to find other solutions. Similarly we can add multiples of $(5,2,0)$
$(x,y,z) = (-1+2m + 5n,-1 + 2n, -m)$ should span the space of integer solutions.