How to prove $x +y + z =0$ has infinite solutions? I first thought to approach this with the Euclidean algorithm because that's how I had been proving this type of equation before, but I am unsure how to approach it when the equation equals zero. 
I am thinking I could assume one is zero and then it is easy to show the other two are infinite such as:
$y + z =0$
$y = -z$
$-z + z =0$
And then the equation obviously stands for all $y \in \mathbb{Z}$.
Is there a way to do this without assuming one variable is zero? Or is this the correct approach?
 A: For any $y$ and $z$, $(-y-z, y,z)$ is a solution.
A: Let $a,b,c$ be any three real numbers. Then the triplet $x=a-b$, $y=b-c$, $z=c-a$ are
infinitely many solutions of $x+y+z=0$.
A: If you can prove that $x+y+z=0$ has infinite solutions when $x =0$ then that that is enough to prove that $x+y+z = 0$ will have at least as many solutions when when $x$ may or may not be $0$, and you have nothing more to do.
But if $x$ is any integer you want and all, and $z$ is any integer you want at all then if $z = -x -y$ then $x + y + z=0$.  ANd there are any infinite number of choices for $x$ and $y$.
Although I particularly like Dr. Zafar Ahmed's answer too.  $x = a-b$ and $y=b-c$ and $z =c-a$ will be a solution for any $a,b,c$ of which you have infinite choices.
And of course you don't need $x = 0$.  You could have $x =35$.  Then if $z = -y-35$ you have....
... or you could be weird.  Let $a,b,c$ be any three; let $x = a+2b; y = c-3b; z = b-a-c$.... or whatever....
A: You can also say that if you know a solution $(x,y,z)$ then $(kx,ky,kz)$ is also a solution $\forall k\in\mathbb{R}$.
A: Let $x+y+z$ to be in the form $$n+(-\frac{n}{2})+(-\frac{n}{2})=n-n=0;\;for\;n\;being\;any\;real\;number$$We know that for any number of $n$ this holds true and as $x+y+z$ can be written in the above form and that will equal to $0$. Hence there can be infinite number for solutions for the above posed problem.
