Consider the equations:
$ab+c+d=3$, $bc+d+a=5$, $cd+a+b=2$ and $da+c+b=6$
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.