# Is it possible for sum of three sequences $x_n + y_n+z_n$ to equal $0$?

Define three sequences $$x_n, y_n, z_n$$ for $$n=1, 2, \dots,$$ by $$x_1 = 2$$, $$y_1 = 4$$, $$z_1 = \frac{6}{7}$$ and the recursion $$x_{n+1} = \frac{2x_n}{x_n^2-1}, \quad y_{n+1} = \frac{2y_n}{y_n^2-1}, \quad z_{n+1} = \frac{2z_n}{z_n^2-1}$$

Is it possible to have $$x_n + y_n + z_n = 0$$ for some $$n$$?

Looking at each sequence separately, they look quite messy and are not monotone and also don't seem to converge to a limit. I have tried an induction method, to start from $$x_{n+1} + y_{n+1} + z_{n+1}$$, expand out using the recursion, and try to show that it is non-zero provided that $$x_n+y_n+z_n$$ is non-zero, but I am not getting anywhere because it is too messy. Maybe it can be simplified to showing that $$x_n + y_n > z_n$$ for all $$n$$ or something along those lines which will be sufficient to prove the result.

Alternatively, I think it would be neater to find some invariant but I am really bad at finding invariants and I don't see any clue on how to start. The most obvious $$x_n +y_n + z_n$$ is not constant so that doesn't work.

Another thing maybe is to look at the sequence of numerators and denominators i.e. let $$a_n/b_n = x_n$$ where $$a_n$$ and $$b_n$$ are coprime and work from there but it also gets messy very quickly. I have $$\frac{a_{n+1}}{b_{n+1}} = \frac{2a_nb_n}{a_n^2 - b_n^2} = \frac{\sqrt{a_n^2b_n^2}}{\frac{a_n^2-b_n^2}{2}}$$ which doesn't seem to resemble anything.

How to approach this problem?

• Can you share the source of the problem? Jan 4 '20 at 8:24

Step 1: If $$x_n = \tan(\phi)$$ for some $$\phi$$ then $$x_{n+1} = -\frac{2 \tan(\phi)}{1- \tan^2(\phi)} = - \tan(2 \phi) = \tan(-2\phi) \, .$$ Therefore, if we define $$\alpha = \arctan(x_1), \, \beta = \arctan(y_1), \, \gamma = \arctan(z_1) \, ,$$ then $$x_n = \tan((-2)^{n-1} \alpha), \, y_n = \tan((-2)^{n-1} \beta), \, z_n = \tan((-2)^{n-1} \gamma)$$ for all $$n$$, as long as the sequences are defined, i.e. as long as no denominator in the sequence definitions becomes zero.
Step 2: Generally, $$\tan (u) + \tan (v) + \tan (w) = \tan (u) \tan(v) \tan(w)$$ holds if and only if $$u+v+w$$ is an integral multiple of $$\pi$$, see for example Prove the identity if x+y+z=xyz on AoPS.
Step 3: The initial values $$x_1 = 2$$, $$y_1 = 4$$, $$z_1 = 6/7$$ satisfy $$x_1 + y_1 + z_1 = x_1 y_1 z_1$$. It follows that $$\alpha + \beta + \gamma$$ is an integral multiple of $$\pi$$. But then also $$(-2)^{n-1} \alpha + (-2)^{n-1} \beta + (-2)^{n-1} \gamma = k \pi \text{ for some }k \in \Bbb Z$$ which in turn implies that $$x_n + y_n + z_n = x_n y_n z_n$$ as long as the sequences are defined. It follows directly from the sequence definitions that the right-hand side is never zero, therefore $$x_n + y_n + z_n$$ can not be zero either.
Remark: One can also verify directly from the recursive definition that $$x_{n+1} + y_{n+1} + z_{n+1} - x_{n+1} y_{n+1} z_{n+1} = \frac{2(1-x_n y_n - x_n z_n - y_n z_n)(x_n + y_n +z_n -x_ny_n z_n)}{(x_n^2-1)(y_n^2-1)(z_n^2-1)}$$ so that $$x_n + y_n +z_n = x_ny_n z_n \implies x_{n+1} + y_{n+1} + z_{n+1} = x_{n+1} y_{n+1} z_{n+1} \, .$$
• Ah okay so the idea here is to recognize the recurrence and use the trig substitution? Then the invariant $x_n + y_n + z_n - x_ny_nz_n = 0 \quad \forall n$ is less mysterious now. Thanks! Jan 4 '20 at 11:14
• $\alpha, \beta, \gamma$ are the angles of a triangle with sides $\sqrt{17}, \sqrt{20}, 3$. Jan 4 '20 at 13:06