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