# Systems of polynomial equations involving sums of equal powers

Given the following system of polynomial equations: $$\left\{\begin{array}{lclclcr} x & + & y & + & z & = & 1 \\ x^{2} & + & y^{2} & + & z^{2} & = & 14 \\ x^{3} & + & y^{3} & + & z^{3} & = & 36 \end{array}\right.$$ What is $$x^{5} + y^{5} + z^{5}\ {\large ?}$$ .

How should I approach this? Is there a general formula for this kind of system?

• Try using this approach by blackpenredpen. – Andrew Chin Aug 10 at 21:38

Let $$p=x+y+z$$, $$q=xy+yz+zx$$, $$r=xyz$$. So we have $$\begin{cases} p=1\\p^2-2q=14\\p^3-3pq+3r=36 \end{cases}$$ $$\begin{cases} p = 1\\q = -\frac{13}{2}\\ r = \frac{31}{6} \end{cases}$$ By consecutive eliminating the highest powers terms we get $$x^5+y^5+z^5-(x+y+z)^5+5(xy+yz+xz)(x+y+z)^3-5(xy+yz+xz)^2(x+y+z)-5(x+y+z)^2xyz+5(xy+yz+xz)xyz=0$$ In other words, $$x^5+y^5+z^5=p^5-5qp^3+5q^2p+5p^2r-5qr$$ $$=\frac{877}{2}$$
The requested exponent does not matter that much. Each of the three is a root of the same $$6 t^3 - 6 t^2 - 39 t - 31,$$ one real and two complex conjugates, but we don't need them. Each also obeys $$6 t^{n+3} - 6 t^{n+2} - 39 t^{n+1} - 31 t^n,$$ so that $$a_n=x^n + y^n + z^n$$ is a sequence with linear recurrence $$a_{n+3} = a_{n+2} + \frac{13}{2} a_{n+1} + \frac{31}{6} a_n$$ or $$a_1 = 1, \; a_2 = 14, \; a_3 = 36, \; a_4 = \frac{793}{6}, \;$$ $$a_5 = \frac{877}{2}, \; a_6 = \frac{17803}{12}, \; a_7 = \frac{180601}{36}, \; a_8 = \frac{1218641}{72}, \;$$