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Solve the following equations on $\mathbb{R}$(the result must be expressed by elementary function) \begin{eqnarray} \begin{cases} x^5-5x^3y+5x^2+5xy^2-5y+1=0 \\ y^5-5xy^3+5y^2+5x^2y-5x+2=0 \ \end{cases} \end{eqnarray}

(I have input it into wolfram alpha only get the approximate solution.This problem is provided by a chinese tieba(one kind of BBS) user "我不知道我不知道”）

(tips: let $x=a+b+c,y=1/a+1/b+1/c$ and $abc=1$)

chenyueyue
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    Quintics in general do not have closed form solutions using radicals. Is this question guaranteed to have one? – Henry Jul 22 '22 at 10:10
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    "I don't know I don't know" Tne number of $7$ in the approximation is quite impressive – Claude Leibovici Jul 22 '22 at 10:56
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    Here's how to ask a good question. Follow these guidelines to get help in this forum. It's particularly important that you show your work and thoughts on the problem. – jjagmath Jul 24 '22 at 10:19
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    I think that you do not understand that, on this site, we try to help you. Hiding the hints (that you more than likely have from the start) is less than fair play. Do not rely on me for any further question from you. – Claude Leibovici Jul 24 '22 at 13:36

2 Answers2

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In $\mathbb{R}$, there is only one solution.

From the first equation, we have $$y=\frac{5 x^3+5+\sqrt{5} \sqrt{x^6-10 x^3-4 x+5}}{10 x}$$ Plugging in the second, we need to solve for $x$ $$x^{25}-50 x^{22}+5 x^{20}+875 x^{19}-75 x^{17}-6750 x^{16}+1260 x^{15}+125 x^{14}+23125 x^{13}-$$ $$17425 x^{12}+1625 x^{11}-31240 x^{10}+63250 x^9+11875 x^8+24550 x^7-$$ $$29625 x^6-16245 x^5+32750 x^4-11250 x^3+26325 x^2-22375 x+6876=0$$ which does not show any rational solution.

Its closest approximation is $$x=-\sin ^{\frac{24}{7}}(e \pi ) (-\cos (e \pi ))^{3/7} \sec ^6(e \pi )\,\pi ^{\frac{3(9 e-2)}{7} }\,\exp\Bigg[-\frac{10 e+8 \pi -9 e \pi -12 e^2 \pi +34 e \pi ^2}{7 e \pi } \Bigg]$$ which is in an error of $3.89\times 10^{-18}$

Plugging these $(x,y)$ in the equations, the rhs are $0$ (for sure) and $-6.456\times 10^{-17}$.

Claude Leibovici
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After a hard search, I contacted the author of the problem. He did it in this way

Let $x=a+b+c,y=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ and $abc=1$

$(a+b+c)^5-5(a+b+c)^3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+5(a+b+c)^2+5(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2-5(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+1=0 :A$

$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^5-5(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^3+5(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2+5(a+b+c)^2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-5(a+b+c)+2=0 :B$

Through complex operations on $A$ and $B$ and $abc=1$,we find $a^5,b^5,c^5$ is the three different roots of equation $t^3+t^2-2t-1=0$

Then let $t=m+\frac{1}{m}$,we get $m^7=1$ and $m\neq1$.

let $ω_1=e^\frac{2πi}{7}$,$ω_2=e^\frac{4πi}{7}$,$ω_3=e^\frac{6πi}{7}$

then $a=(ω_1+\frac{1}{ω_1})^{0.2}$ ,$b=(ω_2+\frac{1}{ω_2})^{0.2}$ ,$c=(ω_3+\frac{1}{ω_3})^{0.2}$

and $x=a+b+c,y=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

chenyueyue
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