What Theorems can I use to solve a problem of this type?

Question

Find All 17 real solutions $(w,x,y,z)$ to the following system of equalities: $$ 2w+w^2x=x $$ $$ 2x+x^2y=y $$ $$ 2y+y^2z=z $$ $$ -2z+z^2w=w $$

I was thinking of using something related to cyclic sums, but I couldn't figure it out since one of the signs is negative, so I get $$ \sum_{cyc} (x+x^2y)=4z $$ so finding the zeroes is no longer an easy process.

I also tried to find upper and lower bounds for the cyclic sum in the left hand side, using the AM-GM inequality but that really didn't help.

So I don't really know what to do, other than solving for each variable and substituting to end up with a monstrous polynomial, which makes me think there is an easier method I'm overlooking

Answer 1

You have $$x=\frac{2w}{1-w^2}$$ etc. Let's not worry about denominators being zero right now. This suggests setting $x_1=\arctan x$ etc. Then $x_1=2w_1$ or $x=2w_1\pm\pi$, so let's say $x_1\equiv 2w_1\pmod\pi$. One then gets $z_1\equiv8w_1\pmod\pi$.

The last equation is $$w=-\frac{2z}{1-z^2}$$ so $w_1\equiv-2z_1\equiv-16w_1\pmod\pi$. Therefore $w_1=k\pi/17$ where $k\in\Bbb Z$ so we have $$(w,x,y,z)=\left(\tan\frac{k\pi}{17},\tan\frac{2k\pi}{17}, \tan\frac{4k\pi}{17},\tan\frac{8k\pi}{17}\right).$$

I suppose one ought to check there are no solutions with any variable equal to $\pm1$.

Answer 2

All solutions are given by computing a resultant. We have the trivial solution $(x,y,z,w)=(0,0,0,0)$ and $16$ solutions parametrized by the $16$ solutions of $w$, which are the roots of the resultant polynomial $$ f(w)=w^{16} - 136w^{14} + 2380w^{12} - 12376w^{10} + 24310w^8 - 19448w^6 + 6188w^4 - 680w^2 + 17. $$ Note that all solutions are real. The there are exactly $17$ complex solutions, which are all real.