Euler's derivation of Lemniscate addition theorem In the notes "a Brief history of elliptic integral addition theorems", the author states at the very beginning of Chapter 4 that Euler found a general solution of the equation
$$
\frac{dx}{\sqrt{1-x^4}}=\frac{dy}{\sqrt{1-y^4}}\tag{1}
$$
to be
$$
x=\frac{c\sqrt{1-y^4}+y\sqrt{1-c^4}}{1+x^2c^2}\tag{2}
$$
I don't have a problem proving that this is a solution, but how could Euler have possibly arrived at such solution?
 A: The following comments are summarized from Sec2 Chapter 1 in Topic In Complex Function Theory Vol.1. (By C.L. Siegel; translated by A. Shenitzer and D. Solitar) by my own understanding. In fact, at the begining of this section, named The Euler addition theorem, the author states that "In this section we shall attempt to reconstruct Euler's train of thought".

It may date back to Fagnano, who attempted to rationalize the following integrand, which now is referred to as the lemniscatic integral,
\begin{equation}
 s=s(r)=\int_{0}^{r} \frac{{\rm d}r}{\sqrt{1-r^4}}.
\end{equation}
It is well-known that an analogous formula for the arc of a circle
\begin{equation}\label{eq-2}
 \arcsin r=\int_{0}^{r} 
 \frac{{\rm d} r}{\sqrt{1-r^2}} \tag{1}
\end{equation}
can be rationalized by taking the follow substitution
$$r=\frac{2t}{1+t^2}, \qquad (0\leq t\leq 1.)$$
Now in the above lemniscatic integral we have $\sqrt{1-r^4}$ instead of $\sqrt{1-r^2}$. This suggests the use of the analogous substitution
$$r^2=\frac{2t^2}{1+t^4}, \qquad (0\leq t\leq 1.)$$
Then one can obtain
\begin{equation}
 \int_{0}^{r} \frac{d r}{\sqrt{1-r^{4}}}=\sqrt{2} \int_{0}^{t} \frac{d t}{\sqrt{1+t^{4}}}, \quad  \quad(0 \leq r \leq 1) \tag{2} \label{eq-1}
\end{equation}
If we treat the right-hand of of the above  integrand in an analogous manner by using the substitution
$$t^{2}=\frac{2 u^{2}}{1-u^{4}},$$
then we can get that
$$\int_{0}^{t} \frac{d t}{\sqrt{1+t^{4}}}=\sqrt{2} \int_{0}^{u} \frac{d u}{\sqrt{1-u^{4}}}.$$
In view of \eqref{eq-1}, we now have the doubling of a lemniscatic arc, that is,
$$\int_{0}^{r} \frac{d r}{\sqrt{1-r^{4}}}=2 \int_{0}^{u} \frac{d u}{\sqrt{1-u^{4}}},$$
where
$$r^{2}=\frac{4 u^{2}\left(1-u^{4}\right)}{(1+u^{4})^{2}}, \qquad (0\leq u \leq 1).$$
Now we go back to \eqref{eq-2}. In fact,the doubling of a circular arc states that
$$\int_{0}^{r} \frac{{\rm d} r}{\sqrt{1-r^{2}}}=2 \int_{0}^{u} \frac{{\rm d} u}{\sqrt{1-u^{2}}},$$
where
$$r=2u \sqrt{1-u^2}.$$
This result can be obtained by using the transcendental substitution
$u=\sin x$. This result includes the relation
$$\sin (2x)=2\sin x \cos x,$$
which is a special case of the trigonometric addition theorem:
$$\sin(x+y)=\sin x \cos y +\sin y \cos x.$$
The latter can be obtained by means of the substitutions
$$u=\sin x, \quad v=\sin y, \quad u \sqrt{1-v^{2}}+v \sqrt{1-u^{2}}=\sin (x+y),$$
and the formation of inverse functions and takes the form
$$\int_{0}^{u} \frac{{\rm d} u}{\sqrt{1-u^2}}+\int_{0}^{v} \frac{{\rm d} v}{\sqrt{1-v^2}}=\int_{0}^{r} \frac{{\rm d} r}{\sqrt{1-r^2}}, \quad r=u \sqrt{1-v^{2}}+v \sqrt{1-u^{2}}. \tag{4}\label{eq-4}$$
If we view $v$ as constant, then  the above equality  suggests that
$r=u \sqrt{1-v^{2}}+v \sqrt{1-u^{2}}$ is a general solution of
$$\frac{{\rm du}}{\sqrt{1-u^2}}=\frac{{\rm d}r}{\sqrt{1-r^2}}.$$
Note that \eqref{eq-4} is obtained by replacing $r=2 u \sqrt{1-u^2}$ by  $r=u \sqrt{1-v^{2}}+v \sqrt{1-u^{2}}$.
Recall that the substitution, which gives us the doubling of lemniscatic arc,
$$r=\frac{2u \sqrt{1-u^4}}{1+u^4}.$$
It is plausible to replace  the numerator $2u \sqrt{1-u^4}$ to $u\sqrt{1-v^4}+v\sqrt{1-u^2}$. As denominator we choose the simplest symmetric function of $u$ and $v$ which becomes $1+u^4$ for $u=v$, namely, $1+u^2 v^2$. We thus arrive at the substitution
$$r=\frac{u\sqrt{1-v^4}+v\sqrt{1-u^2}}{1+u^2 v^2}.$$
Taking this substitution, it  can be proved that
$$\int_{0}^{r}\frac{{\rm d}r}{\sqrt{1-r^4}}=\int_{0}^{u}\frac{{\rm d}u}{\sqrt{1-u^4}}+\int_{0}^{v}\frac{{\rm d}v}{\sqrt{1-v^4}}.$$
Then one obtains, by viewing $v$ as constant, that
$$r=\frac{u\sqrt{1-v^4}+v\sqrt{1-u^2}}{1+u^2 v^2}$$
is a general solution of
$$\frac{{\rm d} u}{\sqrt{1-u^4}}=\frac{{\rm d} r}{\sqrt{1-r^4}}.$$
