Why is the answer set limited here?

Question

This question is based on pp $67$ - $68$ of Ash and Gross's "Elliptic Tales".

Here the authors discuss points on a curve in the projective plane.

We have an equation $f(x,y) = x^2+y^2$

We can homogenise this as $F(x,y,z) = x^2 + y^2 + z^2$

Now, the book points out that if we look for solutions to $F(x, y, z) = 0$ from the finite field $F_2$ (which has members 0, 1) then we have the solutions (1:1:0), (1:0:1), (0:1:1) - and I follow that, at least I think I do (e.g., $1^2 + 1^2 + 0^2 = 0 \mod 2$)

The book then sets as an exercise finding the points in $F_3$ (ie the set 0, 1, 2) and lists

(1:1:1), (1:1:2), (1:2:1) and (1:2:2) (ruling out (2:2:2) for reasons of homothety)

My question is why are (2:1:2) and (2:1:1) not also solutions?

Answer 1

simply: $(2,1,2)=2(1,2,1)$ and $(2,1,1)=2(1,2,2)$.