Exercise 5.5.4 (Equation $5x^2 + 7y^2 + 9y^2 + 11xy + 13 yz + 15 zx = 0$)

Question

Determine whether the equation
$$5x^2 + 7y^2 + 9y^2 + 11xy + 13 yz + 15 zx = 0.$$
has a nontrivial integral solution.

Richard Ganaye · Accepted Answer

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

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

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

\begin{proof} 
Put
$$f(x,y) = x^2 + 7y^2 + 9y^2 + 11xy + 13 yz + 15 zx .$$
Using formula (1) in Problem 3, we obtain
\begin{align*}
380 f(x,y) &= 19 \, {\left(10 \, x + 11 \, y + 15 \, zight)}^{2} + {\left(19 \, y -35 \, zight)}^{2} - 2080 \, z^{2}\
&= 19 \, {\left(10 \, x + 11 \, y + 15 \, zight)}^{2} + {\left(19 \, y -35 \, zight)}^{2} -130 \, (4z)^{2}\
\end{align*}
Consider the form
\begin{align*}
g(X, Y) &= 19 X^2 + Y^2 -130 Z^2\
\end{align*}
Then $f$ has nontrivial integral solutions if and only if $g$ has nontrivial integral solutions.

If $a = 19, b = 1, c = -130$, then $-ab = -19$ is not a square modulo $130$, otherwise $-19$ would be a square modulo the prime $13$, but $\legendre{-19}{13} = -1$.
\begin{verbatim}
sage:  Mod(-19, 130).is_square()
False
\end{verbatim}
By Legendre's Theorem, the equation $5x^2 + 7y^2 + 9y^2 + 11xy + 13 yz + 15 zx = 0$ has no nontrivial integral solution.

\end{proof}

Exercise 5.5.4 (Equation $5x^2 + 7y^2 + 9y^2 + 11xy + 13 yz + 15 zx = 0$)

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Exercise 5.5.4 (Equation $5x^2 + 7y^2 + 9y^2 + 11xy + 13 yz + 15 zx = 0$)

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