Solution of a diophantine linear equation system

Question

I want to find out for what triples $(a,b,c) \in \mathbb{Z}^3 $ the following system of linear diophantine equations has a solution:

$$\begin{cases} 24x+\phantom{1}6z=a, \\ 24x+16y=b, \\ 32x+\phantom{1}8z=c. \end{cases}$$

I can find solutions by trying numbers, but cannot figure out how to obtain a general solution.

Answer 1

By computing Smith Normal Form, we get$\newcommand\bmat[1]{\begin{bmatrix}#1\end{bmatrix}}$ $$\bmat{2&0&0\\0&8&0\\0&0&0}\bmat{3&0&4\\0&1&0\\1&0&1}=\bmat{0&0&1\\1&-1&-4\\-2&3&8}\bmat{24&24&32\\0&16&0\\6&0&8}.$$ Then the image of the homomorphism \begin{align} &\Bbb Z^3\to\Bbb Z^3& &[x,y,z]\mapsto\bmat{x,y,z}\bmat{24&24&32\\0&16&0\\6&0&8} \end{align} is given by $$\bmat{a,b,c}=\bmat{x,y,z}\bmat{2&0&0\\0&8&0\\0&0&0}\bmat{3&0&4\\0&1&0\\1&0&1}=\bmat{6x,8y,8x}.$$

Answer 2

Clearly $6|a, a=6A$(say) $A=4x+z$

$B=3x+2y$

$C=4x+z$

$z=4x-C$

$3x+2y=(3-2)B\iff3(x-B)=-2(y+B)$

$\dfrac{2(y+B)}3=-(x+B)$ which is an integer

$\implies3|2(y+B)\iff3|(y+B), y+B=3D$(say) $\iff y=3D-B$

$-(x+B)=2D$ where $A,B,C,D$ are arbitrary integers

Answer 3

Put $a=6A,b=8B,c=8C$. Then we have $4x+z=A=C$, $3x+2y=B$. Now take any $x$ st $B-3x$ is even (ie take $x$ odd if $B$ is odd, and even if $B$ is even. Then we have $z=A-4x,y=(B-3x)/2$.

So the requirements on $a,b,c$ are:

we have $a=6A, c=8A$ for some integer $A$; $b$ can be any multiple of 8.