The answer is 6.

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This is a correct solution, and well written too! Hint for a more elegant one: tan(2t)=...

In response to the Challenge Master:

I see now! The funny thing is that I did have a good look into trying to apply a substitution featuring trigonometric or hyperbolic functions, but only after I had made my substitution to $a$ , $b$ and $c$ , because that felt so "natural".

Until someone posts an "official" solution, let me give one:

We can rewrite the equations to:

$3x = \frac{2z}{1-z^2} \\ 3y = \frac{2(3x)}{1-(3x)^2} \\ z = \frac{2(3y)}{1-(3y)^2}$

Note that this has introduced the invalid solution $x=y=z=0$ . Substitution time:

$3x = \tan \alpha \qquad 3y = \tan \beta \qquad z = \tan \gamma$

This gives:

$\tan \alpha = \tan 2\gamma \iff \alpha \equiv 2\gamma \text{ (mod } \pi \text{)} \\ \tan \beta = \tan 2\alpha \iff \beta \equiv 2\alpha \text{ (mod } \pi \text{)} \\ \tan \gamma = \tan 2\beta \iff \gamma \equiv 2\beta \text{ (mod } \pi \text{)}$

Combining all three:

$\alpha \equiv 8\alpha \text{ (mod } \pi \text{)} \iff \alpha = \frac{k\pi}7, \ k \in \mathbb{Z}$

Note that these solutions provide only 7 different solutions for $x = \frac13 \tan \alpha$ , e.g. for $0 \leq k \leq 6$ , and that $k=0$ yields the forbidden solution $x=0$ . This leaves 6 solutions, which can all be easily verified to give genuine solutions for $(x,y,z)$ .

Thomas Beuman
- 7 years, 10 months ago

Not a particularly elegant solution I admit, but at the time of writing this is the only one, so it's better than nothing. I've tried to think of something smart, even after solving it, but I failed (so far).

Thomas Beuman
- 7 years, 10 months ago

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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.

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At the start, you should start that $x, y, z$ are non-zero from the conditions, which allows us to multiply throughout. Otherwise, it is not true that $x - z^2x = \frac {2z}{3} \Leftrightarrow \frac {x}{z} - zx = \frac {2}{3}$ .

This solution can be simplified slightly by working modulo $\pi$ , to show that $8A \equiv 4B \equiv 2C \equiv A \pmod{\pi}$ , hence $7A \equiv 0 \pmod{\pi}$ .

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.

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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:

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 $(x,y,z)$ .