How to find the $x^2+y^2=?$ Suppose that $ x$ and $ y$ are complex numbers, such $x+y=1$ and $x^{20}+y^{20}=20$, Find the sum of all possible values $x^2+y^2$ 
I find this problem answer is $-90$, But my methods is very very ugly, and I think this Problem have some nice methods.Thank you everyone.
 A: Let $xy=c$
So, $x^2+y^2=(x+y)^2-2xy=1-2c$
$x^3+y^3=(x+y)^3-3xy(x+y)=1-3c$
On multiplication, $(x^2+y^2)(x^3+y^3)=(1-2c)(1-3c)$
$\implies x^5+y^5+x^2y^2(x+y)=1-5c+6c^2\implies x^5+y^5=1-5c+5c^2$
On squaring, $ x^{10}+y^{10}+2(xy)^5=(1-5c+5c^2)^2$
$\implies  x^{10}+y^{10}=1-10c+35c^2-50c^3+25c^4-2c^5$
On squaring, $ x^{20}+y^{20}+2(xy)^{10}=(1-10c+35c^2-50c^3+25c^4-2c^5)^2$
$\implies  x^{20}+y^{20}=4c^{10}-100c^9+\cdots+1-2c^{10}$
$\implies  2c^{10}-100c^9+\cdots+1=20\implies 2c^{10}-100c^9+\cdots-19=0$
This is a $10$ degree equation in $c=xy$
Using Vieta's formula, $\sum xy=\frac{100}2=50$
$\implies \sum x^2+y^2=10-2\sum xy=10-2\cdot 50=-90$
A: This is quite doable without calculus if you can be clever with the polynomials. Rewrite $x^2 + y^2 = (x+y)^2  - 2xy = 1 - xy,$ and note that for any pair $(x,y)$ satisfying the relations, we must have that $x$ is a root of the polynomial $p(x) = x^{20} + (1-x)^{20} - 20 = 2x^{20} - 20 x^{19} + 190 x^{18} - ... -19 .$ Since $y$ is uniquely determined by $x,$ we know that there exist exactly $20$ such pairs. Summing over all $20$ pairs reduces to $20 - 2\displaystyle\sum_{i=1}^{20} x_i y_i = 20 - 2\displaystyle\sum_{i=1}^{20} x_i (1-x_i) = 20 - 2\displaystyle\sum_{i=1}^{20} x_i  + 2\displaystyle\sum_{i=1}^{20} x_i^2$ The first sum can be calculated with the coefficients of $p$ using the first of Vieta's Formulae  - these are obtained by factoring the polynomial and comparing coefficients. Its value is $- \frac{-20}{2} = 10$ - cancellation reduces the computation to only the last sum. 
Note $\displaystyle\sum_{i=1}^{20} x_i^2 = [\displaystyle\sum_{i=1}^{20} x_i ]^2 - 2 \displaystyle\sum_{i < j} x_i x_j = 10^2 - 2 \cdot (190/2) = -90 ,$ where I've now used the first and second Vieta formulae. This last step is also the first nontrivial Newton Identity.
