2006 iTest Problems/Problem 35

Problem 35

Compute the $\textit{number}$ of ordered quadruples $(w,x,y,z)$ of complex numbers (not necessarily nonreal) such that the following system is satisfied: $\begin{align*} wxyz&=1\\ wxy^2 + wx^2z + w^2yz + xyz^2&=2\\ wx^2y + w^2y^2 + w^2xz + xy^2z + x^2z^2 + ywz^2 &= -3\\ w^2xy + x^2yz + wy^2z + wxz^2 &= -1 \end{align*}$

Solution

as we are given $xyzw=1$ , so from this we get second equation as $\frac{y}{z}+\frac{x}{y}+\frac{w}{x}+\frac{z}{w}=2$ . so say $a=\frac{y}{z} , b=\frac{x}{y}, \frac{w}{x}=c,\frac{z}{w}=d$ . so we get $a+b+c+d=2$ . from fourth equation we get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=-1$ . so we get $abc+abd+acd+bcd=-1$ . also from third equation we get $ab+bc+cd+ad+w^2y^2+x^2z^2=-3$ . notice we want $ac$ and $bd$ . so $ac=\frac{1}{x^2z^2}$ . so this gives $ab+bc+cd+ad+ac+bd=-3$ . and $abcd=1$ . so we get a equation $\alpha^{4}-2\alpha^{3}-3\alpha^2+\alpha+1=0$ whose roots are $a,b,c,d$ . so we get $(\alpha+1)(\alpha^3-3\alpha^2+1)=0$ . this gives $\alpha=-1$ . and three distinct complex ( not necessarily non real) solutions. so as $\alpha=-$ 1. we get any one pair say $\frac{x}{y}=-1$ . so $x=-y=k$ for some $k \in \mathbb{C}$ . so as $z,w$ , will be distinct we will get $4$ quadruples from $-k,k,w,z$ solution so we can have such $4 \cdot 4 =\boxed{16}$ quadruples.

~https://artofproblemsolving.com/community/q1h1745445p11362157

2006 iTest (Problems, Answer Key)
Preceded by: Problem 34	Followed by: Problem 36
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2006 iTest Problems/Problem 35

Problem 35

Solution

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