Solution to a non-linear simultaneous equation

Question

I'm trying to solve the non-linear, two-variable system of equations \begin{align*} y-x+x^5-\frac{y^4x}{2(1+x^2)^2}-\frac{x^3}{1+y^2} &= 0 \\ -x-y+y^5-\frac{x^4y}{2(1+y^2)^2}-\frac{y^3}{1+x^2} &= 0 \end{align*} with $x,y \in \mathbb{R}$. You might notice that the second equation is almost obtained from the first by swapping $x$ and $y$, except for the first term whose signs are opposed. I think the only solution is the origin $x=y=0$, and this is seems to be confirmed numerically on WolframAlpha, but I'm not sure, and I don't know how to prove it. You can also write the system as \begin{align*} y-x+x^5+\frac{1}{4}\partial_x\left(\frac{y^4}{1+x^2}-\frac{x^4}{1+y^2}\right) &= 0 \\ -x-y+y^5-\frac{1}{4}\partial_y\left(\frac{y^4}{1+x^2}-\frac{x^4}{1+y^2}\right) &= 0 \end{align*} Any hints/ideas?

Answer 1

I converted both equations into polynomials and put them into Singular to find all solutions.

LIB "solve.lib";
ring r=0,(x,y),dp;
poly p1=2x9y2+2x9+4x7y2+2x7-4x5+2x4y3+2x4y-4x3y2-6x3+4x2y3+4x2y-xy6-xy4-2xy2-2x+2y3+2y;
poly p2=-x6y-x4y-2x3y4-4x3y2-2x3+2x2y9+4x2y7-4x2y3-2x2y-2xy4-4xy2-2x+2y9+2y7-4y5-6y3-2y;
ideal i=p1,p2;
def T=solve(i,30);

There are $89$ distinct solutions, but $88$ of them involve a complex number.

This shows that the last solution, $x=y=0$, is the only real solution.

Answer 2

According to Maple, the resultant of the numerators of the left sides of the two equations with respect to $y$ is $$ 2\,{x}^{9} \left( 4096\,{x}^{80}+69632\,{x}^{78}+516096\,{x}^{76}+ 2084864\,{x}^{74}+4272128\,{x}^{72}-163840\,{x}^{70}-25825280\,{x}^{68 }-65273856\,{x}^{66}-38729728\,{x}^{64}+154712064\,{x}^{62}+382149632 \,{x}^{60}+144474112\,{x}^{58}-791342976\,{x}^{56}-1359673728\,{x}^{54 }+226609664\,{x}^{52}+3532913024\,{x}^{50}+4419130368\,{x}^{48}- 672814592\,{x}^{46}-8622736384\,{x}^{44}-11440427968\,{x}^{42}- 5943181984\,{x}^{40}+2080361568\,{x}^{38}+6219084784\,{x}^{36}+ 8104484032\,{x}^{34}+13320834677\,{x}^{32}+21495544525\,{x}^{30}+ 25938523882\,{x}^{28}+22524511317\,{x}^{26}+14404728925\,{x}^{24}+ 7384009128\,{x}^{22}+3941633218\,{x}^{20}+2832052128\,{x}^{18}+ 2199851588\,{x}^{16}+1418048480\,{x}^{14}+701995632\,{x}^{12}+ 263885888\,{x}^{10}+74884576\,{x}^{8}+15715584\,{x}^{6}+2310144\,{x}^{ 4}+208896\,{x}^{2}+8192 \right) \left( {x}^{2}+1 \right) ^{6} $$ The only real root of this is $x=0$, which implies $y=0$ from the first equation.