Exercise 5.4.6 (Equation $x^3 + 2y^3 + 4z^3 = 6xyz$)

Question

Show that if $x^3 + 2y^3 + 4z^3 \equiv 6xyz \pmod 7$, then $x \equiv y \equiv z \equiv 0 \pmod 7$. Deduce that the equation $x^3 + 2y^3 + 4z^3 = 6xyz$ has no nontrivial integral solution.

Richard Ganaye · Accepted Answer

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

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

\begin{proof} 
$(\overline{0}, \overline{0}, \overline{0})$ is the only solution of $x^3 + 2y^3 + 4z^3 = 6xyz$ in $\Z/7\Z$. The $343$ verifications are done by Sagemath:
\begin{verbatim}
sage: l = []
....: for x in Integers(7):
....:     for y in Integers(7):
....:         for z in Integers(7):
....:             if x^3 + 2 * y^3 + 4 * z^3 == 6 * x * y * z:
....:                 l.append((x,y,z))
....: l
....: 
[(0, 0, 0)]
\end{verbatim}
 If $x^3 + 2y^3 + 4z^3 \equiv 6xyz \pmod 7$, then $x \equiv y \equiv z \equiv 0 \pmod 7$.
 
 \bigskip
 Suppose now that $(x,y,z) \in \Z^3 \setminus \{(0,0,0)\}$ satisfies  $x^3 + 2y^3 + 4z^3 = 6xyz$. Since $x,y,z$ are not all zero, $d = x \wedge y \wedge z 
e 0$. put $X = x/d, Y = y/d, Z = z/d$. Then
 $$X^3 + Y^3 + Z^3 = 6 XYZ,\qquad 	ext{where } X\wedge Y \wedge Z = 1.$$
 
 By the first part, $X \equiv Y \equiv Z \equiv 0 \pmod 7$. Therefore $7 \mid X\wedge Y \wedge Z = 1$, which is impossible. This contradiction shows that the equation $x^3 + 2y^3 + 4z^3 = 6xyz$ has no integral solution $(x,y,z) 
e (0,0,0)$.
\end{proof}

Exercise 5.4.6 (Equation $x^3 + 2y^3 + 4z^3 = 6xyz$)

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Exercise 5.4.6 (Equation $x^3 + 2y^3 + 4z^3 = 6xyz$)

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