Exercise 5.5.1 (Equation $2x^2 + 5y^2 - 7z^2 = 0$ with Legendre's Theorem.)

Question

Use Theorem 5.11 to show that the equation $2x^2 + 5y^2 - 7z^2 = 0$ has a nontrivial solution.

Richard Ganaye · Accepted Answer

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

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

\begin{proof} Put $a= 2, b=5,c = -7$. Since $abc$ is square-free, and $a,b,c$ do not have the same sign, we may apply Theorem 5.11.
\begin{itemize}
\item $-bc = 35\equiv 1^2 \pmod 2$ is a quadratic residue modulo $a = 2$.
\item $-ac = 14 \equiv 4 \equiv 2^2 \pmod 5 $ is a quadratic residue modulo $b = 5$.
\item $-ab = -10\equiv 4 \equiv 2^2 \pmod 7$ is a quadratic residue modulo $|c| = 7$.
\end{itemize}
This  shows that the equation $2x^2 + 5y^2 - 7z^2 = 0$ has a nontrivial solution.
\end{proof}
Note: Without Theorem 5.11, $(1,1,1)$ is a solution. More generally, $(\pm a,\pm a, \pm a)$ is a solution for every integer $a$. There are other solutions, for instance $(97, 13, 53)$:
$$2\cdot 97^2 + 5\cdot 13^2 = 19663 = 7 \cdot 53^2.$$

Exercise 5.5.1 (Equation $2x^2 + 5y^2 - 7z^2 = 0$ with Legendre's Theorem.)

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Exercise 5.5.1 (Equation $2x^2 + 5y^2 - 7z^2 = 0$ with Legendre's Theorem.)

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