Find one set of solutions for the following system: Find one set of solutions for the following system:
\begin{cases}
1+a^2+d^2=3+b^2+e^2=3+c^2+f^2 \\
1+ab+de=0 \\
ac+df=0 \\
bc+ef=0 \\
\end{cases}
 A: One way is to follow the mathematics of wishful thinking. Here, I wishfully think that I can choose many things to be zero. For instance, if $d = e = c = 0$, then the last two equations hold trivially and we are instead left with the system 
$$\begin{cases} 1 + a^2 = 3 + b^2 = 3 + f^2 \\
1 + ab = 0
\end{cases}$$
If $f = b$, then $f$ is covered. So really, we just have the pair $1 + ab = 0$ and $1 + a^2 = 3 + b^2$. Fortunately, the first means that $a = -1/b$. Plugging into the other gives 
$$1 + \frac{1}{b^2} = 3 + b^2.$$
Rearranging and calling $B = b^2$ gives
$$B^2 + 2B - 1 = 0,$$
with solutions $B = -1 \pm \sqrt 2$. Let's choose the positive square root, $B = 1 + \sqrt 2$. Then $b = \sqrt{1 + \sqrt 2}$, and $a = -1/b$.
So in total,
$$\begin{cases}
a = \frac{-1}{\sqrt{1 + \sqrt 2}}\\
b = \sqrt{1 + \sqrt{2}} \\
c = 0 \\
d = 0 \\
e = 0 \\
f = \sqrt{1 + \sqrt{2}}
\end{cases}$$
is a solution. Of course, this is not linear-algebra, nor algebra-precalculus, but I liked the challenge of finding a single solution to an otherwise untenable problem.
A: Setting $a = -d$ and $c = f$ and $ b = -e $, we are left with the top 2 equations.
Subbing in some values, these equations become, 
$1 + 2d^2 = 3 + 2e^2 = 3 + 2c^2$ and $1 + 2de = 0$.
From this, $2de = -1 \iff d = \frac{-1}{2e}$.
Thus, we have $1 + \frac{1}{2e^2} = 3 + 2e^2 = 3 + 2c^2$.
Then, $1 = 4e^2 + 4e^4 \iff e^2 = \frac{1}{\sqrt{2}} - \frac{1}{2} \iff e = \pm\sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}}$.
Now we can solve for $d$ and $c$. But we already have $d$ and $c = e$ from inspection.
We have $$\begin{equation} \begin{aligned} e = \pm\sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}} \\ d = \frac{-1}{2e} \\ b = -e \\ a = -d \\ c = f = e\end{aligned} \end{equation} .$$
