Exercise 5.4.3 (Equation $(x^2 + y^2)^2 - 2(3x^2-5y^2)^2 = z^2$ )

Question

Show that the equation $(x^2 + y^2)^2 - 2(3x^2-5y^2)^2 = z^2$ has no integral solution.

\bigskip
Hint: Remove powers of $2$ common to $x$ and $y$, then argue $\pmod{16}$.

Richard Ganaye · Accepted Answer

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

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

\begin{proof} 
Suppose for the sake of contradiction that the integers $x,y$ satisfy $(x^2 + y^2)^2 - 2(3x^2-5y^2)^2 = z^2$.

Let $k$ be the largest integer such that $2^k \mid x$ and $2^k \mid y$, so that $x= 2^k u, y = 2^k v$ for suitable integers $u,v$. Then $2^{k+1} 
mid x$ or $2^{k+1} 
mid y$, thus $2 
mid u$ or $2 
mid v$, so $u$ or $v$ is odd (perhaps both).

Since $2^k \mid x$ and $2^k \mid y$, then $2^{4k} \mid z^2$, hence $2^{2k} \mid z$. Thus $z = 2^{2k}w$ for some integer $w$. This gives
$$((2^k u)^2 + (2^k v)^2)^2 - 2 (3 (2^k u)^2 - 5 (2^kv)^2)^2 = (2^{2k}w)^2.$$
Simplifying by $2^{4k}$, we obtain
$$(u^2 + v^2)^2 - 2(3u^2-5v^2)^2 = w^2,$$
where $u$ or $v$ is odd.

For all integers $v$,
$$v^2 \equiv 0, 1,4, 9 \pmod{16},$$
and if $u$ is odd,
$$u^2 \equiv 1,9 \pmod{16}.$$
\begin{itemize}
\item If $u$ is odd, 
\begin{align}
(u^2 + v^2)^2 - 2(3u^2-5v^2)^2 \equiv 7,12,15 \pmod {16}
\end{align}
\item If $v$ is odd,
\begin{align}
(u^2 + v^2)^2 - 2(3u^2-5v^2)^2 \equiv 7,12,15 \pmod {16}
\end{align}
\end{itemize}
But $7, 12, 15$ are not squares modulo $16$, hence the equation $(x^2 + y^2)^2 - 2(3x^2-5y^2)^2 = z^2$ has no integral solution.
\end{proof}
With Sagemath:
\begin{verbatim}
A = set()
for v in range(16):
        for u in range(16):
            if u % 2 == 1 or v % 2 == 1:
                A.add((Mod(u,16)^2 + Mod(v, 16)^2)^2 
                	     -2*(3 * Mod(u,16)^2 - 5 * Mod(v, 16)^2)^2)
A

{7, 12, 15}
\end{verbatim}

Exercise 5.4.3 (Equation $(x^2 + y^2)^2 - 2(3x^2-5y^2)^2 = z^2$ )

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Exercise 5.4.3 (Equation $(x^2 + y^2)^2 - 2(3x^2-5y^2)^2 = z^2$ )

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