Exercise 5.5.5 (Equation $x^2 + 3y^2 + 5z^2 + 2xy + 4 yz + 6zx = 0$)

Question

Determine whether the equation
$$x^2 + 3y^2 + 5z^2 + 2xy + 4 yz + 6zx = 0.$$
has a nontrivial integral solution.

Richard Ganaye · Accepted Answer

\begin{proof} 
Put
$$f(x,y) =x^2 + 3y^2 + 5z^2 + 2xy + 4 yz + 6zx.$$
Using formula (1) in Problem 3, we obtain
\begin{align*}
2 f(x,y) &= 2 \, {\left(x + y + 3 \, zight)}^{2} + {\left(2 \, y - zight)}^{2} - 9 \, z^{2}\
&= 2 \, {\left(x + y + 3 \, zight)}^{2} + {\left(2 \, y - zight)}^{2} -  \, (3z)^{2}\
\end{align*}
Put
$$g(x,y) = 2 X^2 + Y^2 -Z^2.$$
Then $f$ has nontrivial integral solutions if and only if $g$ has nontrivial integral solutions.

Since every integer is a quadratic residue modulo $1$ or modulo $2$, by Legendre's theorem, $g$ has nontrivial integral solutions (for instance $(2,1,3)$).

Therefore, the equation $x^2 + 3y^2 + 5z^2 + 2xy + 4 yz + 6zx = 0$ has a nontrivial integral solution.
\end{proof}
Note: The solution $(2,1,3)$ of $g(X,Y,Z) = 0$ is corresponding to the solution $(-2,1,1)$ of $f(x,y,z) = 0$:
$$2^2 + 3 \cdot 1^2 + 5 \cdot 1^2 - 2 \cdot 2 \cdot 1  + 4 \cdot 1 \cdot 1 - 6 \cdot 1 \cdot 2  =0.$$
\begin{verbatim}
sage: A = matrix([[1,1,3],[0,2,-1],[0,0,3]])
sage: A^(-1) * matrix([[2],[1],[3]])
[-2]
[ 1]
[ 1]
\end{verbatim}

Exercise 5.5.5 (Equation $x^2 + 3y^2 + 5z^2 + 2xy + 4 yz + 6zx = 0$)

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Exercise 5.5.5 (Equation $x^2 + 3y^2 + 5z^2 + 2xy + 4 yz + 6zx = 0$)

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