Prove that for any given $c_1,c_2,c_3\in \mathbb{Z}$,the equations set has integral solution. $$ \left\{
\begin{aligned}
c_1 & = a_2b_3-b_2a_3 \\
c_2 & = a_3b_1-b_3a_1 \\
c_3 & = a_1b_2-b_1a_2
\end{aligned}
\right.
$$
$c_1,c_2,c_3\in  \mathbb{Z}$ is given,prove that $\exists a_1,a_2,a_3,b_1,b_2,b_3\in\mathbb{Z}$ meet the equations set.
Apparently,the question equal to how to decompose the integer vector into the cross product (vector product) of two integer vector.And the real question I want to ask just is it.
 A: This is only a proof of a special case of the problem.
The problem becomes simple if we have in addition the assumption that $(c_{1}, c_{2}) \mid c_{3}.$ For, taking $a_{2} := c_{1},$ $b_{3} := 1,$ $a_{3} := 0,$  $a_{1} := -c_{2}$ fulfills the first two equations and from the third we have $c_{1}b_{1} + c_{2}b_{2} + c_{3} = 0.$ Since $(c_{1}, c_{2}) \mid c_{3},$ it follows that there are infinitely many integers $b_{1}, b_{2}$ for which the third holds. 
A: In fact the result holds for any dimension. 
Take a primitive vector $c$ (coordinates coprime).
Then there is $L\in GL(Z)$ such that $Lc$ be the unit vector of the canonical basis. 
You take the cross product of all the lines of $L$ not corresponding to the $0$ of the vector and you get your initial vector $c$.
At the very basis of it is Euclid. You can transform your vector with only integer transformations to get to a unit vector. 
A: First let us assume that $c_1, c_2, c_3$ are coprime, that is $(c_1, c_2, c_3) = 1$.
Suppose first that there exist integer numbers $u_1, u_2, u_3$ and $v_1, v_2, v_3$ such that
$$
\begin{vmatrix}
c_1 & c_2 & c_3 \\ 
u_1 & u_2 & u_3 \\ 
v_1 & v_2 & v_3
\end{vmatrix}
 = 1
$$
In that case we can take $a = c \times u$ and $b = c \times v$. Then $a \times b = (c \times u) \times (c \times v) = (c \cdot (u \times v))c$ (this is one of triple product properties, see http://en.wikipedia.org/wiki/Triple_product for details). Now $c \cdot (u \times v)$ is equal to the determinant above and hence is equal to one. So $a \times b = c$.
Now let us go back and prove the assumption. Let $d = (c_2, c_3)$ and pick $u_2, u_3$ such that $c_2 u_3 - c_3 u_2 = d$. Also pick $u_1 = 0$. Now we have
$$
\begin{vmatrix}
c_1 & c_2 & c_3 \\ 
u_1 & u_2 & u_3 \\ 
v_1 & v_2 & v_3
\end{vmatrix}
=
v_1 d - v_2 (c_1 u_3 - c_3 u_1) + v_3 (c_1 u_2 - c_2 u_1) = v_1 d + c_1(v_3 u_2 - v_2 u_3).
$$
Now $(c_1, d) = 1$ and $(u_2, u_3) = 1$. Let us pick $v_2$ and $v_3$ such that $v_3 u_2 - v_2 u_3 \equiv c_1^{-1} \pmod d$. In that case the right hand side is congruent to $1$ modulo $d$. Finally let us pick $v_1$ such that it is equal to $1$.
A: The given integral vector v has at least one rational vector w such that  w ,v are perpendicular. We take the cross product u of v and w and this is rational. Now we find a suitable common multiple of u and w to eliminate the denominators. Thus we get new integral vectors u',w' perpendicular to v. 
