<s>[INST] Solve the radical equation for all reals: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$ Question:

Solve the radical equation for all reals: $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$$

My approach:
$$1+\sqrt {1-x^2}=\frac {\sqrt {1+x^2}}{x}\\1+2\sqrt {1-x^2}+1-x^2=\frac{1+x^2}{x^2}\\4(1-x^2)=\left(\frac{1+x^2}{x^2}+x^2-2\right)^2\\4x^4(1-x^2)=(x^4-x^2+1)^2$$
I don't know how can I proceed from here.
Is there a way so that not using the complicated expansion of polynomials?

I'm looking for methods that doesn't use $4$ or higher degree polynomial expansions.

 [/INST] We quickly observe that $x>0$.
Letting $1-x^2=u^2, \,u≥0$  we have:
$$x^2(1+2u+u^2)=1+x^2$$
This implies that,
$$\begin{align}&(1-u^2)(u^2+2u)=1\\
\implies &u(u+1)(u+2)(u-1)=-1\\
\implies &(u^2+u)(u^2+u-2)=-1\end{align}$$
Finally letting $t=u^2+u$, we have:
$$(t-1)^2=0\implies t=1$$
This leads to:
$$\begin{align}&u^2+u-1=0,\,u>0\\
\implies &u=1-u^2\\
\implies &x=\sqrt u=\sqrt {\frac{-1+\sqrt 5}{2}}~~~.\end{align}$$
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