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Let's take a $3$ dimensional sphere of radius $r$, such as $x^2+y^2+z^2 = r^2$.

For $r^2=1$, all the rational points of the sphere (i.e. points for which all coordinates are rational numbers) are characterised by

\begin{equation} x=\frac{2p}{p^2+q^2+1};y=\frac{2q}{p^2+q^2+1};z=\frac{p^2+q^2-1}{p^2+q^2+1} \end{equation}

with $p$ and $q$ rationals.

For $r^2 = 2$, using this method, one can show that

$$x=uz+1; y=vz+1; z=-2 {\frac {u+v}{{v}^{2}+{u}^{2}+1}}$$

with $u,v$ integers.

How would one do to find the rational characterisation when $r^2 = 3$?

Pyrofoux
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    Exactly the same way. Given any degree-2 hypersurface with a rational point (in this case, $(1,1,1)$), there's a bijection between rational points on the hypersurface and rational lines through the chosen rational point. This is likely a duplicate, but I can't quite find the appropriate post to mark it as a duplicate of at the moment. – KReiser Mar 23 '23 at 14:33
  • The general rule is to start with one rational point, and then join rays from that point of rational "slope" and find the other point on the sphere. Starting with $v_0=(-1,-1,-1)$ and an integer vector $a=(p,q,r)$ with $\gcd(p,q,r)=1,$ you want to find another point on the sphere of the form $v_0+at,$ where $t$ is rational. This amounts to solving: $$(-1+pt)^2+(-1+qt)^2+(-1+rt)^2=3$$ or $$2t(p+q+r)=t^2(p^2+q^2+r^2).$$ $t=0$ gives the obvious solution, and there is another solution if $$t=\frac{2(p+q+r)}{p^2+q^2+r^2}.$$ – Thomas Andrews Mar 23 '23 at 14:35
  • This gives $$x=\frac{p^2-q^2-r^2+2pq+2pr}{p^2+q^2+r^2}\y=\frac{q^2-p^2-r^2+2qp+2qr}{p^2+q^2+r^2}\z=\frac{r^2-p^2-q^2+2rp+2rq}{p^2+q^2+r^2}$$ – Thomas Andrews Mar 23 '23 at 14:41
  • @MarkBennet Thank you for the catch, I fixed the title. – Pyrofoux Mar 24 '23 at 09:06
  • @TedShifrin Thanks for the catch, I mixed up notations. The one with $r^2=2$ assumes $u,v$ integers, but the one for $r^2 = 1$ assumes they're rationals. – Pyrofoux Mar 24 '23 at 09:06
  • @ThomasAndrews Would you like to add your comment as an answer, so I could accept it? – Pyrofoux Mar 24 '23 at 16:09

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well. Finding the rational points by stereographic projection will tend to produce fractions that reduce to smaller denominators. Which may not matter to you, of course.

For $a^2 + b^2 + c^2 = 3 d^2 $ in integers, we find all primitive solutions with $\gcd(w,x,y,z) = 1,$ also $w+x+y+z$ odd, finally

$$ a = w^2 + x^2 - y^2 - z^2 + 0 \, w x - 2 w y + 2 x y + 2 w z + 2 x z + 0 \, y z $$ $$ b = w^2 - x^2 + y^2 - z^2 + 2 w x - 0 \, w y + 2 x y - 2 w z + 0 \, x z + 2 y z$$ $$ c = w^2 - x^2 - y^2 + z^2 - 2 w x + 2 w y + 0 \, x y + 0 \, w z + 2 x z + 2 y z$$ $$ d = w^2 + x^2 + y^2 + z^2 $$

The zero coefficients are on purpose, to keep the spacing.

Will Jagy
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