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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

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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 $\pm1$.

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  • $\begingroup$ This is a really elegant solution, I can see the right hand side resembling the double angle formula for tangent; what if I had a slightly different but similar problem it seems I won't be able to use such an identity. Is there any approach through algebraic principles? I tried something similar to this brilliant.org/practice/amc-am-gm-inequality/?p=8 but I only got the trivial solution from it $\endgroup$ – Mahoma Mar 19 '18 at 12:00
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    $\begingroup$ @Mahoma But surely this problem was designed to be solved with this method? $\endgroup$ – Lord Shark the Unknown Mar 19 '18 at 12:29
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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.

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  • $\begingroup$ That looks like the monstrous polynomial I mentioned. After a few steps on that direction I gave up on it. Do you have a quick method to obtain the polynomial? (other than mix up the equations until we have only one variable) At this point though, the only options I see are synthetic division, which does not look pleasant; or get a new system of equations in the roots using Newton Sums or Vieta's... maybe since quite a few of the terms are zero this could yield something "fast" $\endgroup$ – Mahoma Mar 19 '18 at 11:41
  • $\begingroup$ Oh, this is not "monstrous" at all, when dealing with systems of polynomial equations. For a reference, see Bernd Sturmfeld's book Solving Systems of Polynomial Equations. $\endgroup$ – Dietrich Burde Mar 19 '18 at 12:07
  • $\begingroup$ @Mahoma Just set $-w=f(f(f(f(w))))$ where $f(u)=2u/(1-u^2)$. Expanding and simplifying will give Dietreich's formula. Of course there is a nice formula for the iterates of $f$, coming from the fact that $f$ is the duplication formula for the tangent function. $\endgroup$ – Lord Shark the Unknown Mar 19 '18 at 12:35

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