9 times around

First, let us apply the following change of variables:

$a = 3zx \qquad b = 9xy \qquad c = 3yz$

The system of equations then transforms into:

$\frac{b}{c} - a = \frac{c}{a} - b = \frac{a}{b} - c = 2$

We can already see a solution: $a=b=c=-1$ . However, this corresponds to the two imaginary solutions $x = y = \pm\frac{i}3$ , $z = \pm i$ . In order for a solution $(a,b,c)$ to give a real solution $(x,y,z)$ , we need to have $abc > 0$ . Also note that any solution $(a,b,c)$ corresponds to two solutions: if $(x,y,z)$ is one, then $(-x,-y,-z)$ is the other.

We rewrite the equations:

1. $b = c(a+2)$
2. $c = a(b+2)$
3. $a = b(c+2)$

Entering (1) into (2) gives:

$c = a(c(a+2)+2)$

Solving for $c$ :

$c = \frac{2a}{1-2a-a^2}$

And thus:

$b = \frac{2a(a+2)}{1-2a-a^2}$

Putting the expressions for $b$ and $c$ into (3) leads to:

$a(1-2a-a^2)^2 = 2a(a+2)(2a + 2(1-2a-a^2))$

This can be expanded and rearranged to:

$0 = a(a^4+8a^3+14a^2-7)$

Obviously $a=0$ is no good, and we know that $a=-1$ is not a solution we're interested in, so we can divide by $a$ and $a+1$ :

$0 = a^3+7a^2+7a-7 = f(a)$

This has no rational roots, but from noting that $f(-1000) < 0$ , $f(-4) > 0$ , $f(0) < 0$ and $f(1000) > 0$ we can deduce that there are three real roots, two of them negative and one positive. From the symmetry in $a$ , $b$ and $c$ , it follows that $b$ and $c$ must also each be equal to one of these roots. Furthermore, no two of them can be equal, since e.g. $a=b$ implies $c=-1$ , which does not work. Therefore, for all three possible values of $a$ , the values of $b$ and $c$ are precisely the two other roots, and they obey $abc > 0$ . We thus have three relevant solutions for $(a,b,c)$ , and thus $\boxed{6}$ real solutions for $(x,y,z)$ .

Let $r=3x,\ s=3y,\ t=z$ , then the original system becomes $\left\{\begin{array}{l} &\frac{r}{t}-rt=2\\ &\frac{s}{r}-sr=2\\ &\frac{t}{s}-ts=2 \end{array}\right.$ Now, let $t=\tan\theta$ , where $\theta\in[0,\pi)$ . Substitute into the first equation and simplify to get $r=\frac{2t}{1-t^2}=\frac{2\tan\theta}{1-\tan^2\theta}=\tan2\theta$ Similarly, $s=\tan4\theta$ and $t=\tan8\theta$ . Now, $\tan8\theta=\tan\theta\Leftrightarrow \sin8\theta\cos\theta=\cos8\theta\sin\theta\Leftrightarrow\sin7\theta=0$ The equation has 7 solutions, but the solution $\theta=0$ gives $r=s=t=0$ , which makes the equations undefined, so there are 6 solutions.

Let $3x=\tan A, 3y=\tan B, z=\ tanC$

From the 1st, 2nd and 3rd equations respectively we get the following 3 equations:

$\begin{aligned} \tan A & = \tan2C &-(1)\\ \tan B &= \tan 2A & -(2)\\ \tan C& =\tan 2B & -(3)\\ \end{aligned}$

So, $A=n\pi+2C \quad \forall n \in \mathbb{I}$ -(4)

$B=m\pi+2n\pi+4C \quad \forall m \in \mathbb{I}$ -(5)

$C=k\pi+(2m+4n)\pi+8C \quad \forall k \in \mathbb{I}$

$\implies C=\frac{-(k+2m+4n)\pi}{7}$

Substituting the value of C obtained above in (4) and (5) we get,

$B=\frac{-(4k+m+2n)\pi}{7}$

$A=\frac{-(2k+4m+n)\pi}{7}$

We set $A,B,C \in (0,\pi)$ . We are allowed to do so since all values of $x$ can be represented by using $\tan \theta$ for $\theta \in(0,\pi)$

Suppose we have got a solution for x. Observe (2). As value of x is fixed the value of y is also fixed(as (2) becomes a linear equation in y). Now observe (3). As the value of y is fixed the value of z is fixed (same reason as the previous one). So, we have now proved that for a particular x we get a unique y and for a particular pair of x and y we get a unique z. -(6)

This gives us 7 solutions, of the form $x = \tan \frac {k \pi}{7}$ for $k=0$ to 6. However, we must exclude the case when $k=0$ , as that leads to $x=0$ . Hence, there are 6 solutions in all.

x/z - zx = 2/3 => x = 2/3 { z/(1-z^2)}

y/x - 9xy =2 => y = 2 { x/(1-9x^2)}

z/y - 9yz =6 => z = 6 {y/(1-9y^2)} Now, Let 3x = tana , 3y = tanb and z=tanc we get, tana = tan2c=>a =2c +n1* 180 tanb = tan2a=>b =2a+n2* 180 tanc = tan2b=>c =2b+n3 180 where n1,n2,n3 any integer, Also, c = 2(2a+n2 *180) +n3 180 c =4a + 2n2 180+n3 180 c =4(2c+n1 180)+2n2 180+n3 180 c =8c + (4n1+2n2+n3) 180 7c= - (4n1+2n2+n3)*180 c= - (4n1 +2n2 +n3)/7 *180 Since n1 , n2 and n3 can be any integers (4n1 +2n2+n3) can also be any integer. Thus, c= n/7 *180 , where n is any integer Again, z=tanc = tan(n/7 * 180) Since the period of tan(theta) is 180. we have 7 different solutions for z out of which one is 0. But z cannot be 0 so we reject it. Now, from every other non zero solutions of z we can find one non zero x for each z as x = 2/3 { z/(1-z^2)} and from every this non zero x values we can find one y for each x as y = 2 { x/(1-9x^2)}. Thus, we have 6 set of distinct real solutions.