Proof. Assume that the above equation has a solution in nonzero integers $x, y, z$. Since we may replace any number of the variables with their negatives without changing the validity of the equation, we may assume that $x, y, z$ are positive integers.

We may also suppose that $(x, y)=1$. To see this, let $(x, y)=d$. Then $x=d x_{1}$ and $y=d y_{1}$, with $\left(x_{1}, y_{1}\right)=1$, where $x_{1}$ and $y_{1}$ are positive integers. Since $x^{4}+y^{4}=z^{2}$, we have
$$\left(d x_{1}\right)^{4}+\left(d y_{1}\right)^{4}=z^{2}$$
so that
$$d^{4}\left(x_{1}^{4}+y_{1}^{4}\right)=z^{2}$$

Hence $d^{4} \mid z^{2}$, and, by problem 32 of Section 2.2 , we know that $d^{2} \mid z$. Therefore, $z=d^{2} z_{1}$, where $z_{1}$ is a positive integer. Thus,
$$d^{4}\left(x_{1}^{4}+y_{1}^{4}\right)=\left(d^{2} z_{1}\right)^{2}=d^{4} z_{1}^{2}$$
so that
$$x_{1}^{4}+y_{1}^{4}=z_{1}^{2}$$

This gives a solution of $x^{4}+y^{4}=z^{2}$ in positive integers $x=x_{1}, y=y_{1}, z=z_{1}$ with $\left(x_{1}, y_{1}\right)=1$.

So, suppose that $x=x_{0}, y=y_{0}, z=z_{0}$ is a solution of $x^{4}+y^{4}=z^{2}$, where $x_{0}, y_{0}$, and $z_{0}$ are positive integers with $\left(x_{0}, y_{0}\right)=1$. We will show that there is another solution in positive integers $x=x_{1}, y=y_{1}, z=z_{1}$ with $\left(x_{1}, y_{1}\right)=1$, such that $z_{1}<z_{0}$.

Since $x_{0}^{4}+y_{0}^{4}=z_{0}^{2}$, we have
$$\left(x_{0}^{2}\right)^{2}+\left(y_{0}^{2}\right)^{2}=z_{0}^{2}$$
so that $x_{0}^{2}, y_{0}^{2}, z_{0}$ is a Pythagorean triple. Furthermore, we have $\left(x_{0}^{2}, y_{0}^{2}\right)=1$, for if $p$ is a prime such that $p \mid x_{0}^{2}$ and $p \mid y_{0}^{2}$, then $p \mid x_{0}$ and $p \mid y_{0}$, contradicting the fact that $\left(x_{0}, y_{0}\right)=1$. Hence, $x_{0}^{2}, y_{0}^{2}, z_{0}$ is a primitive Pythagorean triple, and by Theorem 11.1, we know that there are positive integers $m$ and $n$ with $(m, n), m \not \equiv n(\bmod 2)$, and
$$\begin{aligned} x_{0}^{2} & =m^{2}-n^{2} \\ y_{0}^{2} & =2 m n \\ z_{0} & =m^{2}+n^{2}, \end{aligned}$$
where we have interchanged $x_{0}^{2}$ and $y_{0}^{2}$, if necessary, to make $y_{0}^{2}$ the even integer of this pair.
From the equation for $x_{0}^{2}$, we see that
$$x_{0}^{2}+n^{2}=m^{2}$$

Since $(m, n)=1$, it follows that $x_{0}, n, m$ is a primitive Pythagorean triple. Again using Theorem 11.1, we see that there are positive integers $r$ and $s$ with $(r, s)=1, r \not \equiv s(\bmod 2)$, and
$$\begin{aligned} x_{0} & =r^{2}-s^{2} \\ n & =2 r s \\ m & =r^{2}+s^{2} \end{aligned}$$

Since $m$ is odd and $(m, n)=1$, we know that $(m, 2 n)=1$. We note that because $y_{0}^{2}=(2 n) m$, Lemma 11.3 tells us that there are positive integers $z_{1}$ and $w$ with $m=z_{1}^{2}$ and $2 n=w^{2}$. Since $w$ is even, $w=2 v$ where $v$ is a positive integer, so that
$$v^{2}=n / 2=r s$$

Since $(r, s)=1$, Lemma 11.3 tells us that there are positive integers $x_{1}$ and $y_{1}$ such that $r=x_{1}^{2}$ and $s=y_{1}^{2}$. Note that since $(r, s)=1$, it easily follows that $\left(x_{1}, y_{1}\right)=1$. Hence,
$$x_{1}^{4}+y_{1}^{4}=z_{1}^{2}$$
where $x_{1}, y_{1}, z_{1}$ are positive integers with $\left(x_{1}, y_{1}\right)=1$. Moreover, we have $z_{1}<z_{0}$, because
$$z_{1} \leqslant z_{1}^{4}=m^{2}<m^{2}+n^{2}=z_{0} .$$

To complete the proof, assume that $x^{4}+y^{4}=z^{2}$ has at least one integral solution. By the well-ordering property, we know that among the solutions in positive integers, there is a solution with the smallest value $z_{0}$ of the variable $z$. However, we have shown that from this solution we can find another solution with a smaller value of the variable $z$, leading to a contradiction. This completes the proof by the method of infinite descent.