<s>[INST] Solving 4 Quadratic Simultaneous Equations 
Consider the equations:
$$\begin{cases}
ab+c+d=3\\
bc+d+a=5\\
cd+a+b=2\\
da+c+b=6
\end{cases}$$
  Where $a,b,c,d \in \mathbb{R}$. How can we find $a,b,c,d$?

The furthest I've got is by adding the first two equations and the last two equations together to get:
$ab+bc+c+d+d+a=(b+1)(a+c)+2d=8$ and $(d+1)(a+c)+2b=8$
We can rearrange these (assuming $b,d \neq -1$) to get $$\frac{8-2d}{b+1}=\frac{8-2b}{d+1} \implies -d^{2}+3d+4=-b^{2}+3b+4$$
Which leads us further to
$$b^{2}-d^{2}-3(b-d)=0 \implies (b-d)(b+d-3)=0$$
Now since $b=d$ gives us an absurdity (equations 1 and 4 reduce to 3=6), we must have $b+d=3$. This is as far as I can get unfortunately, since trying the same approach with the other pairs doesn't work in the same way.
Help would be much appreciated; the problem is from the BMO1 2004 paper.
 [/INST] Take $\#1 - \#2 + \#3 - \#4$ to get $(b-d)(a-c)=-6$ 
$\#1 + \#2 - \#3 - \#4$ gives $(b-d)(a+c-2) = 0$
$\#1 - \#2 - \#3 + \#4$ gives $(b+d-2)(a-c)=2$.
So now $a+c-2 = 0$, and $$\frac{1}{a-c} = \frac{b-d}{-6} = \frac{b+d-2}{2}$$
That gives us $d=3-2b$ and $a-c = \dfrac{-2}{b-1}$ (in particular, $b \ne 1$).
Together with $a+c-2=0$ we get
$a = \dfrac{b-2}{b-1}$, $c = \dfrac{b}{b-1}$.  Finally, substitute into $\#1$ to get
$-b+3=3$, or $b=0$, and thus $a=2$, $c=0$, $d=3$.
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