Can you solve this system of equations with four variables?

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The forum discussion focuses on solving a system of equations with four variables: \(a + b + c + d = 5\), \(ab + bc + cd + da = 4\), \(abc + bcd + cda + dab = 3\), and \(abcd = -1\). The solution involves applying algebraic techniques to find the values of \(a\), \(b\), \(c\), and \(d\) that satisfy all four equations simultaneously. Participants in the discussion shared various methods, including substitution and polynomial factorization, to derive the solutions effectively.

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anemone
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Find all real solutions to the following system of equations:

$a+b+c+d=5$

$ab+bc+cd+da=4$

$abc+bcd+cda+dab=3$

$abcd=-1$
 
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Solution that I saw somewhere online:

We're given

$a+b+c+d=5$

$ab+bc+cd+da=4$

$abc+bcd+cda+dab=3$

$abcd=-1$

Let $X=a+c$ and $Y=b+d$. Then the system of equations is equivalent to

$x+Y=5$

$XY=4$

$Xbd+Yac=3$

$(Xbd)(Yac)=-4$

The first two of these equations imply $(X,\,Y)=(1,\,4)$ and the last two give $(Xbd,\,Yac)=(4,\,-1)$.

And this yields:

[TABLE="class: grid, width: 500"]
[TR]
[TD]$X$[/TD]
[TD]$Y$[/TD]
[TD]$Xbd$[/TD]
[TD]$Yac$[/TD]
[TD]$(a,\,c)$[/TD]
[TD]$(b,\,d)$[/TD]
[/TR]
[TR]
[TD]1[/TD]
[TD]4[/TD]
[TD]4[/TD]
[TD]-1[/TD]
[TD]$\dfrac{1\pm 2}{2}$[/TD]
[TD]2[/TD]
[/TR]
[TR]
[TD]1[/TD]
[TD]4[/TD]
[TD]-1[/TD]
[TD]4[/TD]
[TD]-[/TD]
[TD]-[/TD]
[/TR]
[TR]
[TD]4[/TD]
[TD]1[/TD]
[TD]4[/TD]
[TD]-1[/TD]
[TD]-[/TD]
[TD]-[/TD]
[/TR]
[TR]
[TD]4[/TD]
[TD]1[/TD]
[TD]-1[/TD]
[TD]4[/TD]
[TD]2[/TD]
[TD]$\dfrac{1\pm 2}{2}$[/TD]
[/TR]
[/TABLE]
 

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