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.

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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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Was there a particular way you found these, or was it obvious to you? – Daniel Littlewood Nov 1 '12 at 22:18
@DanielLittlewood, I think the trick is that two +'s and two -'s will eliminate all the single variable terms on the LHS. Wish I thought of it! – sperners lemma Nov 1 '12 at 22:31

If you ever have a more complicated problem like this, systems of polynomial equations can be best solved with the Dixon Resultant. If you google that and my name, you will find lots of information.

Robert H. Lewis

Fordham University

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