Can You Get This With One Chance?

Since $x^2 \leq x^2 + y^2 + z^2 = 3$ , hence $|x| \leq \sqrt{3}$ . Since $x$ is an integer, we have $x= -1, 0, 1$ . However, if $x=0$ , then we can't get $y^2 + z^2 = 3$ . Hence, $x$ must be either 1 or -1. We have the same conclusion for $y, z$ , thus $x,y,z$ can be either $1$ or $-1$ .

Now there are $2 \times 2 \times 2 = 8$ solutions.

$(1,1,1)$

$(1,1,-1)$

$(1,-1,1)$

$(1,-1,-1)$

$(-1,1,1)$

$(-1,1,-1)$

$(-1,-1,1)$

$(-1,-1,-1)$

Note that $x, y, z \in \{ -1, 1 \}$ , so there are $2^3 = \boxed {8}$ solutions.

$-2 > x,y,z > 2$ because $2^{2} > 3$ and $(-2)^{2} > 3$

$x, y, z = 1, -1, 0$

Assuming without loss of generalities that $x = 0, 0 + y^{2} + z^{2} = 3$ has no integer solutions.

Assuming WLOG that $x = 0$ and $y = 0, 0 + 0 + z^{2} = 3$ means that $z = \sqrt{3}$ , which is irrational.

$x \neq$ $0$

Thus, $x, y, z, = 1, -1$

Therefore, there are two combinations for each variable $2 \times 2 \times 2 = 8$

there are theree spots x, y and z to fit -1 and 1, which means we have 2 x 2 x 2 = 8 opitons in which -1 and 1 can be arranged into x, y, z to satisfy the given equation