# 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.

• This can be rewritten into vector cross product notation: $$[a_1, a_2, a_3]\times [b_1, b_2, b_3] = [c_1, c_2, c_3]$$ – Arthur May 22 '15 at 8:12
• Hint: the left hand side is a vector product. – Martigan May 22 '15 at 8:12
• @Arthur yes,the real question I want to ask is how to decompose a integer vector into cross product (vector product) of two integer vector. – lanse2pty May 22 '15 at 8:21
• @Martigan I know it ,but I don't know how to prove it . – lanse2pty May 22 '15 at 8:22
• No other conditions? say some hypothesis on the coprimeness of $c_{1}, c_{2}, c_{3}$? – Megadeth May 22 '15 at 8:54

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.

• So thank you. Can you explain the mean of $(c_1,c_2)|c_3$, I really forgot it .Sorry. – lanse2pty May 22 '15 at 9:14
• No problem. In number theory it is usual to denote the greatest common divisor of two integers $x, y$ by $(x, y)$. – Megadeth May 22 '15 at 9:18
• Thank you. Why $(c_1,c_2)|c_3$ imply $c_1b_1+c_2b_2+c_3=0$ has integer solution?Besides, as the other answer (Adelafif's answer),it is easy to see for having integer solution ,$c_1,c_2,c_3$ not must be $(c_1,c_2)|c_3$. But really thank you . – lanse2pty May 22 '15 at 10:52
• @lanse2pty The number $(c_1, c_2)$ is the smallest non-zero number that can be written as a linear combination of $c_1$ and $c_2$. All other numbers that can be written as a linear combination of $c_1$ and $c_2$ is a multiple of that. And any multiple of the gcd can be written as a linear combination. Since if $mc_1 + nc_2 = (c_1, c_2)$, then the multiple $r(c_1, c_2) = r(mc_1 + nc_2) = (rm)c_1 + (rn)c_2$. – Arthur May 22 '15 at 11:33
• Seemingly, I learned it ,but have forgot,thank you. – lanse2pty May 22 '15 at 11:41

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.

• The $c$ must be coordinates coprime? – lanse2pty May 22 '15 at 11:47

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 $1$.

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.

• The issue is that you have a vector which is proportionnal, not equal... The OP wants to prove that there exists integer vectors such as the direct cross product gives with no factor the initial integer vector. – Martigan May 22 '15 at 9:52
• Although your answer is a little irrelevant ,but I think ,maybe there is a way to prove or deny the question in your answer.Thank you . – lanse2pty May 22 '15 at 10:56