Permutated Peculiar Powers

If we let $S_n = x^n + y^n + z^n$ and suppose that $S_2 = a$ , we have $S_0=3$ , $S_1 = 5$ , $S_2 = a$ , $S_3 = -19$ and $S_5 = -235$ . Playing with elementary symmetric polynomials (expressing $x+y+z$ , $xy+xz+yz$ and $xyz$ in terms of $S_1$ , $S_2$ and $S_3$ ) tells us that $x,y,z$ are the roots of the cubic $X^3 - 5X^2 + \tfrac12(25-a)X - \tfrac12(29-5a) \; = \; 0$ and hence $S_{n+3} - 5S_{n+2} + \tfrac12(25-a)S_{n+1} - \tfrac12(29-5a)S_n \; = \; 0$ for any integer $n$ . Then $S_4 \; =\; \tfrac12(a^2 - 50a - 45)$ and the equation $S_5 - 5S_4 + \tfrac12(25-a)S_3 - \tfrac12(29-5a)S_2 = 0$ collapses to a linear equation for $a$ , which can be solved to obtain $a=3$ . We now easily calculate $S_6 = -285$ and $S_7 = 509$ .

Let $X^3 + TX^2 + UX + V = 0$ be the cubic equation that has roots $x,y,z$ , then

By Vieta's formula, $T = -5$

Apply the algebraic identity

$\begin{aligned} x^3 + y^3 + z^3 -3xyz & = & (x+y+z)(x^2+y^2+z^2 -xy-xz-yz) \\ -19 + 3V & = & (x+y+z) \left ( (x+y+z)^2 -3(xy+xz+yz) \right ) \\ & = & 5 ( 5^2 -3U ) \\ V & = & -5U + 48 \\ \end{aligned}$

Consider expanding the following expression

$\begin{aligned} (x^2+y^2+z^2)(x^3 + y^3 + z^3) & = & (x^5 + y^5 + z^5) + (xy)^2 (x+y) + (xz)^2 (x+z) + (yz)^2 (y+z) \\ (25 - 2U)(-19) & = & -235 + (xy)^2 (5-z) + (xz)^2 (5-y) + (yz)^2 (5-x) \\ (25-2U)(-19)+235 & = & 5 \left ( (xy)^2 + (xz)^2 + (yz)^2 \right ) - xyz(xy + xz + yz) \\ 38U - 240 & = & 5 \left ( (xy+xz+yz)^2 - 2xyz(x+y+z) \right ) - xyz (xy + xz + yz ) \\ & = & 5 \left ( U^2 - 2(-V)(5) \right ) - (-V) (U ) \\ & = & 5U^2 + 50V + UV \\ & = & 5U^2 + V(50+ U) \\ & = & 5U^2 + (-5U+48)(U+50) \\ \end{aligned}$

Solve for $U$ and $V$ simultaneously gives $U = 11, V = -7$ . Thus the equation $X^3 - 5X^2 + 11X - 7 = 0$ has roots $x,y,z$ . By rational root theorem, $(X-1)$ is a root of the equation. Factorize it gives $(X-1)(X^2 - 4X + 7) = 0$ . Quadratic formula gives the other two roots $2 \pm i \sqrt3$

Hence, the answer is $1^7 + (2 + i \sqrt3 )^7 + (2 - i \sqrt3 )^7 = \boxed{509}$