Quadratic System of Equations

First, we factor each of the LHS to get the system $\left\{\begin{array}{l}2(x-1)^2=y\\ 2(y-1)^2=z\\ 2(z-1)^2=x\end{array}\right.$

To make our lives easier, let $x'=x-1$ , $y'=y-1$ , and $z'=z-1$ . Thus our new variables $x',y',z'\in [-1,1]$ and our equations are $\left\{\begin{array}{l}2x'^2=y'+1\\ 2y'^2=z'+1\\ 2z'^2=x'+1\end{array}\right.$

Isolating the $x',y',z'$ variables on the RHS again, we get $\left\{\begin{array}{l}2x'^2-1=y'\\ 2y'^2-1=z'\\ 2z'^2-1=x'\end{array}\right.$

Wait a second! We recognize $2x'^2-1=y'$ : it is the same as the cosine double-angle formula! In addition, the range of $x'$ is $[-1,1]$ , which is the same as the range for $\cos\theta$ ! This means we should plug in $x'=\cos\theta$ . Thus, $y'=\cos2\theta$ . Since $2y'^2-1=z'$ , we have $z'=\cos 4\theta$ . Finally, since $2z'^2-1=x'$ , we have $x'=\cos 8\theta$ .

However, we already know that $x'=\cos\theta$ . Thus, $\cos\theta=\cos 8\theta$ .

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We now have two cases to consider:

Case 1

We let $\theta+2\pi n = 8\theta$ . This gives $\theta=\dfrac{2\pi}{7}n$ . We can vary $n=0\to 6$ to get different values of $\theta$ ; however, since $\cos\dfrac{2\pi}{7}n=\cos\left(2\pi -\dfrac{2\pi}{7}n\right)$ , only $n=0\to 3$ give unique values.

Case 2

We let $\theta+2\pi n=-8\theta$ . This gives $\theta=\dfrac{2\pi}{9}n$ . We can vary $n=0\to 8$ , but like a similar argument as before, only $n=0\to 4$ give unique values.

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Our solutions for $(\theta, 2\theta, 4\theta)$ are then as follows: $\left(0,0,0\right)$ $\left(\dfrac{2\pi}{7},\dfrac{4\pi}{7},\dfrac{8\pi}{7}\right)$ $\left(\dfrac{4\pi}{7},\dfrac{8\pi}{7},\dfrac{2\pi}{7}\right)$ $\left(\dfrac{8\pi}{7},\dfrac{2\pi}{7},\dfrac{4\pi}{7}\right)$ $\left(\dfrac{2\pi}{9},\dfrac{4\pi}{9},\dfrac{8\pi}{9}\right)$ $\left(\dfrac{4\pi}{9},\dfrac{8\pi}{9},\dfrac{2\pi}{9}\right)$ $\left(\dfrac{8\pi}{9},\dfrac{2\pi}{9},\dfrac{4\pi}{9}\right)$ $\left(\dfrac{2\pi}{3},\dfrac{2\pi}{3},\dfrac{2\pi}{3}\right)$

However, we want unordered pairs , and the 2nd, 3rd, 4th, and 5th, 6th, 7th are cyclic permutations of each other. Thus, they only count once towards our total, so our total number of unordered pairs will be $\boxed{4}$ .

If any two of x, y and z are the same, the other should be the same. That is, except for the trivial solutions for x = y = z, the other solutions, if any, should be x != y != z != x.

The two trivial solutions are x = y = z = 0.5 or 2.

The system of equations has 6 non-trivial solutions and they have cyclic symmetry under the conversion (x, y, z) to (y, z, x) or (z, x, y). That is, the system of equasions has 6 / 3 = 2 unordered triplets.

Therefore, the answer is 2 (trivial) + 2 (non-trivial) = 4.