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.