When you graph y=x^2-1 and x=y^2-1 you get 4 intersections, being: (-1;0), (0;-1), (-0.618...;-0.618...) and (1.618...; 1.618...) the last two intersections are equal to phi, can you explain why this happens?
I have tried to solve these two equation, but I can't manage to solve them.
On
To find the intersections, we have to solve this system of two equations: $$\left\{\begin{matrix} y=x^2-1 \\x=y^2-1 \end{matrix}\right.$$ We can substract the second equation to the first and we have: $$y-x=x^2-y^2\leftrightarrow (x-y)(x+y+1)=0 \leftrightarrow x=y \vee x=-(y+1)$$ Substituing, we have: $$y=y^2-1 \vee y=(y+1)^2-1$$ In the first case, we have: $$\left\{\begin{matrix} y=\frac{1\pm\sqrt{5}}{2}=\pm \phi \\x=y=\pm\phi \end{matrix}\right.$$ In the second case, we arrive at: $$\left\{\begin{matrix} y=0 \vee y=-1 \\x=-(y+1)=-1 \vee x=-(y+1)=0 \end{matrix}\right.$$
From the well-known relation $\phi+1=\phi^2$ (or numerically $1.618034\cdots^2=2.618034\cdots$), you draw that
$$\phi^2-1=\phi.$$
Hence $(\phi,\phi)$ belongs to both curves.
And dividing by $-\phi^2$,
$$\left(-\frac1\phi\right)^2-1=-\frac1\phi.$$
The explicit resolution is by means of
$$x=y^2-1=(x^2-1)^2-1$$ or
$$x^4-2x^2-x=x(x+1)(x^2-x-1)=0.$$