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Let $C \subset \mathbb{P}^2$ be a planar conic curve, defined by a ternary quadratic form $Q(x_1, x_2, x_3)$ say. Suppose that $C(\mathbb{Q}) \ne \emptyset$, or equivalently, that $C$ is everywhere locally soluble (i.e., $C(\mathbb{Q}_p) \ne \emptyset$ for every prime $p$ and $C(\mathbb{R}) \ne \emptyset$). Further, it is easy to show that there exist primitive binary quadratic forms $f_1, f_2, f_3 \in \mathbb{Z}[u,v]$ that parametrize the rational points $C(\mathbb{Q})$. In particular, $(x_1, x_2, x_3) = (f_1(u,v), f_2(u,v), f_3(u,v)), u,v \in \mathbb{Z}$.

For example, if $Q(x_1, x_2, x_3) = x_1^2 + x_2^2 - x_3^2$ then we can parametrize the primitive solutions by $x_1 = 2uv, x_2 = u^2 - v^2, x_3 = u^2 + v^2$.

For a given ternary quadratic form $Q(x_1, x_2, x_3) \in \mathbb{Z}[x_1, x_2, x_3]$ which is everywhere locally soluble, call a triple of binary quadratic forms $(f_1, f_2, f_3)$ a parametrizing triple if $Q(f_1, f_2, f_3) \equiv 0$. Consider the binary sextic form $F_Q = f_1 f_2 f_3$.

In general, what can we say about $F_Q$, given $F$? In the example above, we have $(f_1, f_2, f_3) = (2uv, u^2 - v^2, u^2 + v^2)$ and $F_Q = 2uv(u^2 - v^2)(u^2 + v^2)$. This $F_Q$ is very special: indeed, it is a sextic Klein form, with an exceptionally large $\text{PGL}_2$ automorphism group. Does this hold in general when $Q$ is a diagonal form?

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I don't know if the following will fully answer your question, since I'm not sure if you attach a huge importance to have integral coefficients/primitive solutions. Apologies if it doesn't.

Let $Q$ be any nondegenerate (I assume this is the case of interest) ternary quadratic form such that $Q(\mathbb{Q})\neq 0$. Then the theory of quadratic forms says that $Q$ splits off a hyperbolic plane. Hence $Q\simeq \langle a,1,-1\rangle$ for some nonzero $a$.

Now it is known that $\langle 1,-1\rangle\simeq \langle a,-a\rangle$ for any nonzero $a$, so finally $Q$ is isomorphic to $\langle a ,a ,-a\rangle$.

In other words, there exists a basis of $K^3$ such that $Q(x_1,x_2,x_3)=a(x_1^2+x_2^2-x_3^2)$, where the $x_i'$s are the coordinates of a vector $x\in K^3$ in this basis.

Now $Q(x_1,x_2,x_3)=0\iff x_1^2+x_2^2-x_3^2=0$.

All in all, the case you already covered is the only one. Hope this helps.

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  • $\begingroup$ Yes that is indeed helpful! I managed to prove by hand what I wanted (that the product of the parametrizing binary quadratic forms is sextic Klein form) by hand, but it is nice to see an elegant argument. $\endgroup$ – Stanley Yao Xiao Oct 19 at 11:07

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