Exercise 5.3.10 ($x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z) = 1$)

Question

Prove that $x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z) = 1$.

Richard Ganaye · Accepted Answer

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

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

\begin{proof} 
We search only the solutions $(x,y,z)$ such that $x$ is odd. By Problem 7,
$$x^2 = \left(\frac{x^2+1}{2} ight)^2 - \left(\frac{x^2 -1}{2} ight)^2.$$ 
We obtain solutions of $x^2  + y^2= z^4$ (where $y = \frac{x^2 -1}{2}$) if $\frac{x^2+1}{2} = z^2$, that is if 
\begin{align}
x^2 - 2 z^2 = -1.
\end{align}
Note that $ z^2 -y = \frac{x^2+1}{2} - \frac{x^2-1}{2} = 1$, thus $y \wedge z = 1$, a fortiori $x\wedge y \wedge z = 1$.

It remains to prove that (1) has infinitely many solutions such that $x$ is odd.

First $(1,1)$ is such a solution. If $(u,v)$ is a solution of the Pell-Fermat equation 
\begin{align}
u^2 - 2 v^2 = 1,
\end{align}
then
\begin{align*}
-1 &= (1^2 - 2\cdot 1^2)(u^2 - 2 v^2)\
&= [(1 - \sqrt{2})(1 + \sqrt{2})][(u - v \sqrt{2})(1 + v \sqrt{2})]\
&=  [(1 - \sqrt{2})(u - v \sqrt{2})][(1 + \sqrt{2})(1 + v \sqrt{2})]\
&=[(u + 2v) - \sqrt{2}(u+v)][(u + 2v) + \sqrt{2}(u+v)]\
&= (u+2v)^2 - 2 (u+v)^2.
\end{align*}
This shows that $(x,z) = (u + 2v, u+v)$ is also a solution of (1), and $x$ is odd if and only if $u$ is odd.

Note that $(u_0,v_0) = (3,2)$ is a solution of (2). If $(u,v)$ is a solution of (2), with $u$ odd, we have similarly
\begin{align*}
1 &= (3^2 - 2\cdot 2^2)(u^2 - 2 v^2)\
&= [(3 - 2\sqrt{2})(3 + 2\sqrt{2})][(u - v \sqrt{2})(u + v \sqrt{2})]\
&=  [(3 - 2 \sqrt{2})(u - v \sqrt{2})][(3+ 2\sqrt{2})(u + v \sqrt{2})]\
&=[(3u + 4v) - \sqrt{2}(2u+3v)][(3u + 4v) + \sqrt{2}(2u + 3v)]\
&= (3u + 4v)^2 - 2 (2u+3v)^2,
\end{align*}
so $(3u + 4v, 2u + 3v)$ is a solution of (2), where $3u + 4v$ is odd. Starting from the solution $(u_0, v_0) = (1, 0)$ of (2), we obtain  recursively the solutions $(u_n, v_n)$ given by
$$(u_{n+1}, v_{n+1}) = (3u_n + 4 v_n, 2 u_n + 3v_n).$$
Note that by easy induction $u_n > 0, v_n > 0$ for all $n \in \N^*$, thus $u_{n+1} > 3 u_n$ for all $n \in \N$. This prove that these solutions are distinct.

Finally, we obtain  the solution $(x_n, z_n)$ of (1), given by
$$(x_n, z_n) = (u_n + 2v_n, u_n + v_n).$$
Since $z_{n+1} = u_{n+1} + v_{n+1} = (3u_n + 4 v_n) + ( 2 u_n + 3v_n) = 5 u_n + 7 v_n > u_n + v_n = z_n$, these solutions are distinct. There are infinitely many solutions $(x_n, z_n)$ of (1) such that $x_n$ is odd, therefore the equation $x^2 + y^2 = z^4$ has infinitely many solutions with $x \wedge y \wedge z = 1$.
\end{proof}
Verification with Python:
\begin{verbatim}
def solution(n):
    (u, v) = (1, 0)
    for i in range(n):
        x = u + 2 * v
        y = (x * x - 1) // 2
        z = u + v
        (u, v ) = (3 * u + 4 * v, 2 * u + 3 * v)
        yield (x, y, z)
        
for sol in solution(10):
    (x, y, z) = sol
    print(sol, '=>',  x ** 2 + y ** 2 - z ** 4)    
  
(1, 0, 1) => 0
(7, 24, 5) => 0
(41, 840, 29) => 0
(239, 28560, 169) => 0
(1393, 970224, 985) => 0
(8119, 32959080, 5741) => 0
(47321, 1119638520, 33461) => 0
(275807, 38034750624, 195025) => 0
(1607521, 1292061882720, 1136689) => 0
(9369319, 43892069261880, 6625109) => 0
\end{verbatim}

Exercise 5.3.10 ($x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z) = 1$)

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Exercise 5.3.10 ($x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z) = 1$)

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