Problem 5

For the case $y=-\tfrac{1}{2}$, $z=\tfrac{1}{4}$, $x=-\tfrac{1}{4}$: $y=-\tfrac{1}{2}$ means $1+2y=(a+b)^2+(c+d)^2=0$, so $b=-a$ and $d=-c$. Then $a^2+b^2+c^2+d^2=1$ becomes $a^2+c^2=\tfrac{1}{2}$, while $x=ac+bd=2ac=-\tfrac{1}{4}$ gives $ac=-\tfrac{1}{8}$ (which automatically gives $z=ad+bc=-2ac=\tfrac{1}{4}$). Solving $a^2+c^2=\tfrac{1}{2}$, $ac=-\tfrac{1}{8}$ yields $\{a,c\}=\left\{\tfrac{\sqrt3-1}{4},-\tfrac{\sqrt3+1}{4}\right\}$ up to overall sign flip. The other case $z=-\tfrac{1}{2}$, $y=\tfrac{1}{4}$ gives the odd cyclic shifts (where $d=-a$ and $c=-b$). In total this produces $8$ solutions: the $4$ cyclic shifts of $\left(\tfrac{\sqrt3-1}{4},\tfrac{1-\sqrt3}{4},-\tfrac{1+\sqrt3}{4},\tfrac{1+\sqrt3}{4}\right)$ and their negations.