Exercise 5.3.11 (Solutions of $x^2 + y^2 = 2 z^2$)

Question

Using Theorem 5.5, determine all solutions of the equation $x^2 + y^2 = 2z^2$.

\bigskip
Hint. Write the equation in the form $(x+y)^2 + (x- y)^2 = (2z)^2$.

Richard Ganaye · Accepted Answer

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

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

We use Theorem 5.5 in the form 
\begin{align*}
x^2 + y^2 = z^2 \iff &\exists (\lambda, r, s) \in \Z^3,\  r \wedge s = 1, \
&\left\{
\begin{array}{ll}
x &= \lambda (r^2 -s^2),\
y &= \lambda \, 2 r s,\
z&= \lambda (r^2 + s^2),
\end{array}
ight.
	ext{ or } 
\left\{
\begin{array}{ll}
x &= \lambda \, 2 r s,\
y &= \lambda (r^2 -s^2),\
z&= \lambda (r^2 + s^2).
\end{array}
ight.
\end{align*}
\begin{proof} 
The equation $x^2 + y^2 = 2z^2$ is equivalent to  
\begin{align}
(x+y)^2 + (x- y)^2 = (2z)^2.
\end{align}

Using Theorem 5.5, $(x+y)^2 + (x- y)^2 = (2z)^2$ if and only if there exist integers $\lambda, r,s$ such that
\begin{align}
\left\{
\begin{array}{ll}
x + y &= \lambda (r^2 -s^2),\
x - y &= \lambda \, 2 r s,\
2 z &= \lambda (r^2 + s^2),
\end{array}
ight.
\qquad 	ext{ or } \qquad 
\left\{
\begin{array}{ll}
x  - y &= \lambda (r^2 -s^2),\
x + y &= \lambda \, 2 r s,\
2 z &= \lambda (r^2 + s^2),
\end{array}
ight.
\end{align}
that is
\begin{align}
\left\{
\begin{array}{ll}
x + \varepsilon  y &= \lambda (r^2 -s^2),\
x -  \varepsilon y &= \lambda \, 2 r s,\
2 z &= \lambda (r^2 + s^2).
\end{array}
ight.
\end{align}
where $r \wedge s = 1$ and $\varepsilon = \pm 1$.

Then (3) becomes
 \begin{align}
 \left\{
\begin{array}{ll}
2x  &= \lambda (r^2 -s^2 +  2rs),\
2 \varepsilon y &= \lambda (r^2 - s^2 - 2 r s),\
2 z &= \lambda (r^2 + s^2),
\end{array}
ight.
\end{align}
where $r \wedge s = 1$.

If $r$ and $s$ have not same parity, then $r^2 + s^2$ is odd. Then $2z = \lambda (r^2 + s^2)$ implies that $\lambda$ is even, so $\lambda = 2\mu$ for some integer $\mu$.

If $r$ and $s$ have same parity, then $r$ and $s$ are odd (because $r \wedge s = 1$), and $r^2 -s^2, r^2 + s^2$ are even.

In conclusion, $x^2 + y^2 = 2z^2$ if and only if there are integers $r,s, \varepsilon$ such that
 \begin{align*}
 \left\{
\begin{array}{ll}
x  &=\frac{1}{2} \lambda (r^2 -s^2 +  2rs),\
 y &=\frac{\varepsilon }{2} \lambda (r^2 - s^2 - 2 r s),\
 z &= \frac{1}{2}\lambda (r^2 + s^2),
\end{array}
ight.
\end{align*}
where $r \wedge s = 1$ and $\varepsilon = \pm 1$.

\end{proof}
For instance, the solution $(1,7,5)$ is given by 
$$r = 1, \quad s = 2, \quad\lambda = 1, \quad\varepsilon = -1,$$
and the solution $(7, 1,5)$ by
$$ r = 3,\quad s = 1, \quad\lambda = 1, \quad\varepsilon = 1.$$

Exercise 5.3.11 (Solutions of $x^2 + y^2 = 2 z^2$)

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Exercise 5.3.11 (Solutions of $x^2 + y^2 = 2 z^2$)

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