Exercise 5.5.10* (Representation of a rational by a ternary quadratic form)

Question

Let $Q(x,y,z) = ax^2 + by^2 + cy^2$ where $a,b$, and $c$ are nonzero integers. Suppose that the Diophantine equation $Q(x,y,z) = 0$ has a nontrivial integral solution. Show that for any rational number $g$, there exist rational numbers $x,y,z$ such that $Q(x,y,z) = g$.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
Let $(x_0,y_0,z_0) 
e (0,0,0)$ be a solution of $Q(x,y,z) = 0$, so that
\begin{align}
ax_0^2 + b y_0^2 + c z_0^2 = 0.
\end{align}
Without loss of generality we may assume $x_0 
e 0$ (If $x_0 = 0$ but $y_0 
e 0$ for instance, we exchange the roles of $x$ and $y$).

Let $g$ be any rational number. Since
$$(g+a)^2 - (g-a)^2 = 4ag,$$
we obtain a nontrivial solution $(x_1,y_1,z_1) \in \Q^3$ of
\begin{align}
a x_1^2 - gy_1^2 -az_1^2 = 0,
\end{align}
where
\begin{align}
x_1 = g + a, \qquad y_1 = 2a, \qquad z_1 = g-a.
\end{align}
(Then $a x_1^2 - gy_1^2 -az_1^2 = a (g + a)^2 - 4ga^2 - a (g-a)^2 = 0$.)

By equation (1), multiplying by $z_1^2$,
\begin{align*}
0 &= ax_0^2z_1^2 + by_0^2 z_1^2 + cz_0^2 z_1^2\
&=x_0^2 (ax_1^2 - gy_1)^2 +  by_0^2 z_1^2 + cz_0^2 z_1^2& (	ext{by } (2))\
&=a(x_0 x_1)^2 + b(y_0 z_1)^2 + c(z_0 z_1)^2 - g (x_0y_1)^2.
\end{align*}
Here $x_0 y _1 = 2a x_0 
e 0$ (because $a
e 0$ by hypothesis, and $x_0
e 0$ by asumption), thus
\begin{align*}
g &= a \left(\frac{x_0 x_1}{x_0y_1}ight)^2 + b \left( \frac{y_0z_1}{x_0y_1} ight)^2 + c \left(\frac{z_0z_1}{x_0y_1} ight)^2\
&= a \left(\frac{x_0 (g+a)}{2ax_0}ight)^2 + b \left( \frac{y_0(g-a)}{2ax_0} ight)^2 + c \left(\frac{z_0(g-a)}{2ax_0} ight)^2&(	ext{by }(3)).
\end{align*}
This shows that
$$g = Q\left(\frac{x_0 (g+a)}{2ax_0}, \frac{y_0(g-a)}{2ax_0}, \frac{z_0(g-a)}{2ax_0} ight)$$
is represented by the form $Q$.
\end{proof}

Example (and verification): By Problem (5), $2x^2 +  y^2 - z^2$ has a non trivial solution $(x_0, y_0, z_0) = (2,1,3)$. Take $g = 3/5$. Then
\begin{align*}
\frac{3}{5} &=  Q\left(\frac{2\left(\frac{3}{5} + 2ight)}{8}, \frac{\frac{3}{5} - 2}{8}, \frac{3 \left(\frac{3}{5} - 2)ight)}{8} ight), \
\frac{3}{5}  &= 2  \left(\frac{26}{40}ight)^2 + \left(\frac{7}{40}ight)^2 -  \left(\frac{21}{40}ight)^2\qquad (	ext{True}).
\end{align*}
Note: the hypothesis $a 
e 0, b 
e 0 , c 
e 0$ is essential: for instance the ternary form $Q(x,y,z) = x^2$ has a nontrivial solution $(0,0,1)$, but for $g = 2$, $x^2 = 2$ has no rational solution.

Exercise 5.5.10* (Representation of a rational by a ternary quadratic form)

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Exercise 5.5.10* (Representation of a rational by a ternary quadratic form)

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