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What are the solutions of this system of equations, where $x,y \in \mathbb{R}$?

$\begin{cases} \frac{1}{x} + \frac{1}{2y} = (x^2+3y^2)(3x^2+y^2)\\ \frac{1}{x} - \frac{1}{2y} = 2(y^4-x^4) \end{cases}$

First I rewrote the equations as

$\begin{cases} \frac{1}{x} + \frac{1}{2y} = 3x^4+3y^4+10x^2y^2\\ \frac{1}{x} - \frac{1}{2y} = 2(y-x)(y+x)(y^2+x^2) \end{cases}$

then

$\begin{cases} \frac{1}{x} + \frac{1}{2y} = 3(x^2+y^2)^2+4x^2y^2\\ \frac{1}{x} - \frac{1}{2y} = 2(y-x)(y+x)(y^2+x^2) \end{cases}$

I tried to add the two equations and I got:

$\frac{2}{x}=2(y-x)(y+x)(y^2+x^2)+3(x^2+y^2)^2+4x^2y^2$

we can rewrite this as

$\frac{2}{x}=(y^2+x^2)[2(y-x)(y+x)+3(x^2+y^2)]+4x^2y^2$,

$\frac{2}{x}=(y^2+x^2)[2y^2-2x^2+3x^2+3y^2]+4x^2y^2$

$\frac{2}{x}=(y^2+x^2)(5y^2-x^2)+4x^2y^2$

Im stuck from here, how should I proceed? Or is there any trick to solving this system?

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  • $\begingroup$ z could be anything $\endgroup$
    – user121330
    Dec 4, 2023 at 19:36

3 Answers 3

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Given,

\begin{align} \frac{1}{x} + \frac{1}{2y} &- (x^2+3y^2)(3x^2+y^2)=0\\ \frac{1}{x} - \frac{1}{2y} &- 2(y^4-x^4)=0 \end{align}

Do the substitution,

\begin{align} x &= \frac{p+1}2\\ y &= \frac{p-1}2 \end{align}

and your two equations simplify as,

\begin{align} \frac{p\,(p^5-3)}{(p^2-1)}=0\\ \frac{(p^5-3)}{(p^2-1)}=0 \end{align}

So your real solution is,

\begin{align} x &= \frac{3^{1/5}+1}2 = 1.1228654698\dots\\ y &= \frac{3^{1/5}-1}2 = 0.1228654698\dots \end{align}

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  • $\begingroup$ What's the motivation for this particular substitution ? It restricts $x,y$ to a specific line $y = x -1$. So shouldn't you also check that there are no other solutions outside of this line. $\endgroup$
    – Digitallis
    Dec 4, 2023 at 12:24
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    $\begingroup$ @Digitallis For quick results with non-linear systems, I normally use resultants. I ended up with a quintic in $x$ which wasn't obviously solvable in radicals. But checking its Galois group, it turned out to be solvable. Depressing the quintic, it simplified to binomial form $z^5-3=0$, so I then knew what substitutions to use. $\endgroup$ Dec 4, 2023 at 12:29
  • $\begingroup$ I'm afraid the time when I understood Galois theory is long gone :'( $\endgroup$
    – Digitallis
    Dec 4, 2023 at 12:35
  • $\begingroup$ Finding the resultant, checking its Galois group and depressing the quintic seem like they should be part of the answer - that substitution will not appear on line two to most people. $\endgroup$
    – user121330
    Dec 4, 2023 at 20:00
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Numerically, by plotting the contour lines for both equations, we can guess that there is a single real solution. Observing the first plot, there is an apparent solution at $(0,0)$ that we must rule out, and an actual solution in the first quadrant. The second plot provides a zoom in. The approximate value of the solution is $(x,y)=(1.1228654698077587,0.12286546980788067)$.

enter image description here

enter image description here

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  • $\begingroup$ What software did you use? Your $y$ value is off by the last few digits, starting at the red numbers. \begin{align} x&\approx1.12286546980775=\frac{3^{1/5}+1}2\\ y&\approx0.12286546980775=\frac{3^{1/5}-1}2\\y&\approx0.122865469807\color{red}{88}\dots \end{align} $\endgroup$ Dec 4, 2023 at 12:23
  • $\begingroup$ @TitoPiezasIII it was Wolfram Mathematica. $\endgroup$ Dec 4, 2023 at 13:01
  • $\begingroup$ I guess Wolfram Mathematica may assume certain levels of accuracy. But it is strange it was more accurate with $x$ but not with $y$. $\endgroup$ Dec 4, 2023 at 13:07
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Using polar coordinates

$ x = r \cos \phi $, $y = r \sin \phi $

Let $ c = \cos \phi , s = \sin \phi $ , then the two equations become

$ \dfrac{1}{c} + \dfrac{1}{2 s} = r^5 ( 3 c^4 + 3 s^4 + 10 c^2 s^2 ) $

$ \dfrac{1}{c} - \dfrac{1}{2 s} = r^5 ( 2 s^4 - 2 c^4 ) $

Dividing the two equations eliminates $r$:

$ \dfrac{ 2 s + c } {2 s - c } = \dfrac{ 3 c^4 + 3 s^4 + 10 c^2 s^2 }{ 2 s^4 - 2 c^4 } $

Cross multiplication leads to

$ (2 s + c) (2 s^4 - 2 c^4) = (2 s - c) ( 3 c^4 +3 s^4 + 10 c^2 s^2 ) $

Divide by $c^5$ then

$ (2 t + 1 ) (2 t^4 - 2 ) = (2 t - 1) (3 + 3 t^4 + 10 t^2 ) $

where $t = \tan \phi$

Expanding and re-arranging gives

$ 2 t^5 - 5 t^4 + 20 t^3 - 10 t^2 + 10 t - 1 = 0 $

This quintic polynomial has only one real root

$ t = 0.109421363 $

The corresponding $\phi$ is

$ \phi = 0.108987771 $

Now $r$ can be computed from the above equations, and then

$ x = r \cos \phi = 1.12286547 $

$ y = r \sin \phi = 0.12286547 $

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    $\begingroup$ Your quintic is solvable in radicals and is given by, $$t=\frac{1+3^{1/5}-3^{2/5}+3^{3/5}-3^{4/5}}2 = 0.109421\dots$$ $\endgroup$ Dec 6, 2023 at 16:46

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