Prove that for any integers $x,y,z$ there exist $a,b,c$ such that $ax+by+cz=0$ It is rather obvious that for any 3 coprime integers $x,y,z$ there exist 3 non-zero integers $a,b,c$ such that:
$$ax+by+cz=0$$
Any simple argument to prove it?
 A: Assume $z\neq 0$. One of them needs to be for $x,y,z$ to be coprime.
Find $u,v\neq 0$ so that $ux+vy\neq 0$. Then let $a=zu,b=zv,c=-(ux+vy)$.
There is always such $u,v$ unless one of $x,y$ is zero. Assume $x=0$ and $y\neq 0$ then choose $a=1,b=z,c=-y$.
If $z=1$ and $x=y=0$, then you can't find a solution with non-zero $c$.
You don't really need coprime, of course. Just that at least $2$ of $x,y,z$ are non-zero. Also, if all three are zero, you can trivially solve with $a=b=c=1$.
A: I am going to change the language of the problem for clarity: Let a, b and c be the constants that satisfy the property of being coprime. You asked what the solutions are to satisfy:
$$ ax + by = cz $$
For variables x, y, and z in $ \mathbb Z $
Note that in your form, one merely needs make z the opposite sign to yield a solution, then flip the variable names as I did.
We say that $ ax + by $ is a linear combination of cz. If the gcd(x,y) | cz, we have an easy case to do all the solutions with the extended euclidean algorithm. Note that you specified x and y coprime, so the gcd(x, y) = 1. See extended euclidean algorithm.
Otherwise, there are infinitely many solutions for large enough cz. This is evident in the fact that if x and y are coprime, then there exists a linear combination for any value larger than xy - x - y, a fact that can be found here. In fact this is almost a duplicate problem.
If x and y are not coprime, divide both sides by their greatest common divisor and adjust z to be appropriate for the problem. Tada! Now they are coprime.
A: $$ax+by+cz=0$$ 
Let $(a,b,c)=m$
$$a=mA$$
$$b=mB$$
$$c=mC$$
where $(A,B,C)=1$
then
$$Ax+By+Cz=0$$
Then for all integers $u,v,w$, we have:
$$x=Bw-Cv$$
$$y=Cu-Aw$$
$$z=Av-Bu$$
We confirm that
$$A(Bw-Cv)+B(Cu-Aw)+C(Av-Bu)= ABw-ACv+BCu-ABw+ACv-BCu= 0$$
