Prove that for any integers $x,y,z$ there exist $a,b,c$ such that $ax+by+cz=0$

Question

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?

Answer 1

$$ax+by+cz=0$$

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 $$A(Bw-Cv)+B(Cu-Aw)+C(Av-Bu)= ABw-ACv+BCu-ABw+ACv-BCu= 0$$

Answer 2

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 $a=b=c=1$.

Answer 3

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.