Exercise 5.4.2 (Equation $x^2 + 2y^2 = 8z+5$ )

Question

Show that the equation $x^2 + 2y^2 = 8z+5$ has no integral solution.

Richard Ganaye · Accepted Answer

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

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

\begin{proof} 
If we reduce this equation modulo 8, we obtain no solution: for all integers $x,y$,
$$x^2 \equiv 0,1,4 \pmod 8,\qquad 2 y^2 \equiv 0, 2 \pmod 8,$$
thus
$$x^2 + 2y^2 \equiv 0,1,2,3,4,6 \pmod 8.$$
Therefore 
$$x^2 +2y^2 
ot \equiv 5 \pmod 8.$$
This shows that the equation
$$x^2 + 2y^2 = 8z + 5$$
has no integral solution.
\end{proof}
With Sagemath:
\begin{verbatim}
sage: A = set()
sage: for x in Integers(8):
....:     for y in Integers(8):
....:         A.add(x^2 + 2 * y^2)
....:         
sage: A
{0, 1, 2, 3, 4, 6}
\end{verbatim}

Exercise 5.4.2 (Equation $x^2 + 2y^2 = 8z+5$ )

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Exercise 5.4.2 (Equation $x^2 + 2y^2 = 8z+5$ )

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