# Rational parametrization of points on sphere of irrational radius

I am trying to figure out if a closed form parametrization can exist for finding all of the rational points on a sphere of radius $\sqrt{2}$. ie, x=f(u,v), y=g(u,v), and z=h(u,v) where f,g,and h are all rational functions, and the sum of their squares equals 2.

I have an vague understanding of the notion that for finding rational points on a surface or curve, one starts by identifying a single rational point on said function, then drawing lines of rational slope through that point to find other rational points, but I'm not entirely sure how it works. exactly. One trivial rational point on a sphere of radius $\sqrt{2}$, for example, is (x,y,z)=(1,1,0).... can somebody please show me how to find all the other rational points on the sphere starting from this?

The formula for the sphere is $x^2+y^2+z^2=2$. And we have a known rational point (x,y,z) = (1,1,0) on the sphere. We find lines of rational slope that pass through this point, as I had described above. We can define $x=u z+1$, and $y=v z+1$. Substituting that in $x^2+y^2+z^2=2$, gives ${z}^{2}{u}^{2}+2 zu+2+{z}^{2}{v}^{2}+2zv+{z}^{2}=2$.
Subtracting the $2$ from the both sides gives us ${z}^{2}{u}^{2}+2 zu+{z}^{2}{v}^{2}+2zv+{z}^{2} = 0$. This factors easily to $z \left( z{u}^{2}+2 u+z{v}^{2}+2 v+z \right) = 0$. We already had a solution at $z=0$, so the remaining solution is $z=-2 {\frac {u+v}{{v}^{2}+{u}^{2}+1}}$. From this, we can now trivially solve for x and y, by using $x=u z+1$, and $y=v z+1$ as mentioned above, providing a complete rational parametrization for the sphere.