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For all non-zero integers $x,y,z$ clearly there exist infinitely many non-zero integers $a,b,c$ such that $$ax+by+cz=0$$ How can I prove this simple statement?

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    For any given non-zero $x,y,z\in\mathbb{Z}$, let $a,b,c\in\mathbb{Z}$ satisfy $a\neq0,,b\neq0,,c\neq0$ and $ax+by+cz=0$. Then notice that for any non-zero $t\in\mathbb{Z}::(at)x+(bt)y+(ct)y=0$ is also true, so $at,,bt,,ct$ also are solutions (since $t\neq0$, divide both sides by $t$ and arrive to the initial assumption). Now you just need to prove there is a solution. – dbanet Jan 31 '16 at 20:03
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    Just a comment that it feels odd to use $x, y,$ and $z$ as known, integer constants with $a,b,$ an $c$ as your variables. – pjs36 Jan 31 '16 at 20:06

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We assume that at least one of $x,y,z$ is non-zero, without loss of generality $z\neq 0$.

First note there is at least one non-zero solution:

Let $a=-z,b=-z,c=x+y$. we get $-zx+-zy+(x+y)z$.

Now notice if $(a,b,c)$ works then $(ka,kb,kc)$ works.

Asinomás
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