jmc

To find the two solutions, we use the quadratic formula. We can write our equation as $x^2-x-1=0$. Making the coefficients more visible, we have the equation $$(1)x^2+(-1)x+(-1)=0.$$The quadratic formula then gives $$x=\frac{-(-1)\pm \sqrt{(-1{)}^2-4(1)(-1)}}{2(1)}=\frac{1\pm \sqrt{5}}{2}.$$Letting $\Phi =\frac{1+\sqrt{5}}{2}$ and $\phi =\frac{1-\sqrt{5}}{2}$, we have $$\begin{array}{rl}\Phi -\phi & =\left(\frac{1+\sqrt{5}}{2}\right)-\left(\frac{1-\sqrt{5}}{2}\right)\\ & =\frac{1}{2}+\frac{\sqrt{5}}{2}-\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\\ & =\frac{1}{2}+\frac{\sqrt{5}}{2}-\frac{1}{2}+\frac{\sqrt{5}}{2}\\ & =\frac{\sqrt{5}}{2}+\frac{\sqrt{5}}{2}\\ & =\sqrt{5}.\end{array}$$The problem didn't tell us which solution was $\Phi$, but that doesn't matter: if $\Phi$ and $\phi$ are swapped, then $\Phi -\phi =-\sqrt{5}$, but either way, $(\Phi -\phi {)}^2=\boxed{5}$.