Exercise 5.5.3 (Equation $3x^2 + 5y^2 + 7y^2 + 9xy + 11 yz + 13 zx = 0$)

Question

Determine whether the equation
$$3x^2 + 5y^2 + 7y^2 + 9xy + 11 yz + 13 zx = 0.$$
has a nontrivial integral solution.

Richard Ganaye · Accepted Answer

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof} 
We obtain a general formula (if $a
e 0, 4ab - d^2 
e 0$) for a ternary quadratic form
$$f(x,y) = ax^2 + by^2 + cz^2 + dxy + eyz +fzx,$$
useful for the three following problems.
\begin{align*}
f(x,y) &= a \left( x^2 + \frac{d}{a} xy + \frac{f}{a} xzight) + by^2 + cz^2 + eyz\
&=a\left[ \left( x + \frac{d}{2a} y + \frac{f}{2a} zight)^2  -\frac{d^2}{4a^2}y^2 - \frac{f^2}{4a^2} z^2 - \frac{df}{2a^2} yz ight] + by^2 + cz^2 + eyz\
&=a  \left( x + \frac{d}{2a} y + \frac{f}{2a} zight)^2 +  \frac{4ab - d^2}{4a} y^2 + \frac{4ac - f^2}{4a} z^2 + \frac{2ae -d f}{2a} yz\
&=a  \left( x + \frac{d}{2a} y + \frac{f}{2a} zight)^2 + \frac{4ab - d^2}{4a} \left( y^2 + \frac{4ae - 2df}{4ab - d^2} yz + \frac{4ac - f^2}{4ab - d^2} z^2 ight)\
&=a  \left( x + \frac{d}{2a} y + \frac{f}{2a} zight)^2 +\frac{4ab - d^2}{4a}  \left\{ \left(y + \frac{2ae -df}{4ab -d^2} z ight)^2 + \left[ \frac{4ac - f^2}{4ab -d^2} - \left(\frac{2ae - df}{4ab - d^2} ight)^2ight] z^2  ight\}\
&=a  \left( x + \frac{d}{2a} y + \frac{f}{2a} zight)^2 +\frac{4ab - d^2}{4a} \left(y + \frac{2ae -df}{4ab -d^2} z ight)^2 + \left(\frac{4ac - f^2}{4a} - \frac{2ae - df)^2}{4a(4ab - d^2)} ight) z^2.
\end{align*}
This gives (if $a
e 0, 4ab - d^2 
e 0$)
\begin{align}
f(x,y) &= a  \left( x + \frac{d}{2a} y + \frac{f}{2a} zight)^2 +\frac{4ab - d^2}{4a} \left(y + \frac{2ae -df}{4ab -d^2} z ight)^2 + \left(\frac{4ac - f^2}{4a} - \frac{2ae - df)^2}{4a(4ab - d^2)} ight) z^2.
\end{align}
If $f(x,y) = 3x^2 + 5y^2 + 7y^2 + 9xy + 11 yz + 13 zx$,  the formula (1) gives (with the help of Sagemath, see Note 2)
\begin{align}
84 f(x,y) &=  7 \, {\left(6 \, x + 9 \, y + 13 \, zight)}^{2} - 3 \, {\left(7 \, y +17 \, zight)}^{2} + 272 \, z^{2}\
&=7 \, {\left(6 \, x + 9 \, y + 13 \, zight)}^{2} - 3 \, {\left(7 \, y +17 \, zight)}^{2} + 17 (4z)^2.
\end{align}
Consider the form 
$$g(X,Y) =7X^2 - 3 Y^2 + 17Z^2.$$
By (3), $f$ has a nontrivial rational solution if and only if $g$ has a nontrivial rational solution. Therefore $f$ has a nontrivial integral solution if and only if $g$ has a nontrivial integral solution.

Here $7, 3$ and $17$ are prime numbers. Note that
$$\legendre{21}{17} = \legendre{51}{7} = \legendre{-119}{3} = 1.$$
\begin{verbatim}
sage: kronecker(21,17), kronecker(51,7), kronecker(-119,3)
(1, 1, 1)
\end{verbatim}
Then by Legendre's Theorem, $g$ has nontrivial integer solutions, so $f$ has nontrivial integer solutions.
\end{proof}

Note 1: Consider some integral solution of $g$, for instance $(X,Y,Z) = (5,9,2)$: $$7\cdot 5^2 - 3\cdot 9^2 + 17\cdot 2^2 = 0.$$

To obtain a corresponding solution $(x,y,z)$ of $f(x,y,z) = 0$, we solve the system
$$
\begin{pmatrix} X \ Y \  Z\end{pmatrix} = 
\begin{pmatrix} 6 & 9 & 13\ 0 & 7 & 17 \ 0 & 0 & 4 \end{pmatrix}
\begin{pmatrix} x \ y \  z\end{pmatrix}
$$
If  $A = \begin{pmatrix} 6 & 9 & 13\ 0 & 7 & 17 \ 0 & 0 & 4 \end{pmatrix}$, then $A^{-1} = \frac{1}{84} \begin{pmatrix} 14 &-18 & 31\  0 & 12& -51\  0&   0&  21\end{pmatrix}$, and
\begin{align*}
&\begin{pmatrix} x \ y \  z\end{pmatrix}
=  \frac{1}{84} \begin{pmatrix} 14 &-18 & 31\  0 & 12& -51\  0&   0&  21\end{pmatrix}
\begin{pmatrix} 5 \ 9 \  2\end{pmatrix} 
= \frac{1}{84} \begin{pmatrix} -30 \ 6  \  42\end{pmatrix} = \frac{1}{14} \begin{pmatrix} -5 \ 1 \  7\end{pmatrix}
\end{align*}
So $(x,y,z) = (-5,1,7)$ is an integral solution of $3x^2 + 5y^2 + 7y^2 + 9xy + 11 yz + 13 zx = 0$:
$$3\cdot 5^2 + 5\cdot1^2 + 7\cdot 7^2 -9\cdot 5\cdot 1 + 11\cdot 1\cdot 7 -13 \cdot 7  \cdot 5 = 0.$$

Note 2: With Sagemath:
\begin{verbatim}
(a,b,c,d,e,f) = (3,5,7,9,11,13)
var('x,y,z')   
g = a * (x + (d/(2*a))*y + (f/(2*a))*z)^2 \
     + ((4*a*b - d^2)/(4*a))* (y + ((2*a*e-d*f)/(4*a*b - d^2))*z)^2  \
     + ((4*a*c -f^2)/(4*a) - (2*a*e-d*f)^2/(16*a^2*b - 4*a*d^2))*z^2
84 * g
\end{verbatim}
$$ 7 \, {\left(6 \, x + 9 \, y + 13 \, zight)}^{2} - 3 \, {\left(7 \, y +
17 \, zight)}^{2} + 272 \, z^{2}$$

Exercise 5.5.3 (Equation $3x^2 + 5y^2 + 7y^2 + 9xy + 11 yz + 13 zx = 0$)

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Exercise 5.5.3 (Equation $3x^2 + 5y^2 + 7y^2 + 9xy + 11 yz + 13 zx = 0$)

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