Let $x^2=x+1$, the solution is the golden ratio phi :$x=\frac{1\pm\sqrt{5}}{2}$.
Now I tried $x=x^2-1$ substituted in the equation itself : $$(x^2-1)^2=x^2\Rightarrow x^4-3x^2+1=0$$ $$\Rightarrow x=\pm\sqrt{\frac{3\pm\sqrt{5}}{2}}$$
Now we get 4 roots because one step is not iff giving back $x^2-1=\pm x$.
However we can recover the numerical values of the first equation solutions.
What happens now if I use $(x^2-1)^2=x+1\Rightarrow x(x^3-2x-1)=0$ ?
Could the numerical values be recovered in this case ? I do not think, so what step is used that is not reversible, only $x^2=x+1$ was used ?
When you square $M = W$ to $M^2 = W^2$ you add in the extraneous solutions of $M = -W$.
So $x = (x^2 -1)$ has two solutions. And $x^2 =(x^2 -1)^2$ will have $4$ solutions: the two solutions to $x=x^2-1$ as well as two new and extraneous solutions to $x =-(x^2-1)$.
$x = \sqrt{\frac {3 \pm \sqrt 5}2} = \frac {1 \pm \sqrt 5}2$ are the two solutions to $x = x^2-1$.
$x = -\sqrt{\frac {3 \pm \sqrt 5}2} = -\frac {1 \pm \sqrt 5}2$ are the two solutions to $x = -(x^2 -1)$.
Notice that $( \frac {1 \pm \sqrt 5}2)^2=\frac {3 \pm \sqrt 5}2$ so $ \sqrt{\frac {3 \pm \sqrt 5}2} = \frac {1 \pm \sqrt 5}2$