Five roots

By Vieta's formulae, we know $\sum{r_i} = 2$ $\sum{r_i r_j} = 0$

Rearranging the original equation: $\begin{aligned} x^5 - 2x^4 &= 1 \\ x^4 (x-2) &= 1 \\ x^{-8} &= (x-2)^2 \\ \sum{r_i^{-8}} &= \sum{(r_i-2)^2} \\ &= \sum{(r_i^2 - 4r_i + 4)} \\ &= \sum{r_i^2} - 4\sum{r_i} + 20 \\ &= (\sum{r_i})^2 - \sum{r_i r_j} - 4\sum{r_i} + 20 \\ &= 4 - 0 - 8 + 20 \\ &= \boxed{16} \end{aligned}$

According to Vieta's formula, we have:

$\sum_{i=1}^5 r_i=2$ .

$\sum_{i=1}^4 \sum_{j=i+1}^5r_i r_j=0$ .

$r_1 r_2 r_3 r_4 r_5=1$ .

We have: $(\sum_{i=1}^5 r_i)^2=\sum_{i=1}^5 r^2_i + 2 \sum_{i=1}^4 \sum_{j=i+1}^5r_i r_j$ .

Therefore, $\sum_{i=1}^5 r^2_i=4$ It is obvious that 0 is not a root of the equation.

Therefore, for each root r of the given equation, we have: $\frac{1}{r^4}=r-2$

Therefore, $\sum_{i=1}^5 \frac{1}{r^8_i}= \sum_{i=1}^5 (r_i-2)^2$

$=\sum_{i=1}^5 (r^2_i-4r_i+4)=4-4*2+4*5=16$

We can rewrite the equation as $x^4(x-2)=1$ so $x^4=\frac{1}{(x-2)}$ . Square both sides and take the reciprocal to get $\frac{1}{x^8}=(x-2)^2$ . Now consider the roots to this equation: $r_1,r_2,...,r_5$ . These roots satisfy the equation above. Now consider the polynomial $(x+2)^5-2(x+2)^4-1=x^5+8x^4+24x^3+32x^2+16x-1$ . This polynomial will have roots $(r_1-2),(r_2-2),...,(r_5-2)$ . Apply Newton sums or expand and we find that the sum of the squares of the roots for this polynomial is $8^2-2\cdot24=16$ . Notice that the sum of the squares of the roots of this polynomial is the desired sum that we want with the original polynomial so the answer is $\boxed{16}$ .

Since $x=0$ is not a root, we can divide the original equation by $x^3, x^4, x^8$ each to get $x^2-2x=\frac{1}{x^3}$ , $x-2=\frac{1}{x^4}$ and $\frac{1}{x^3}-\frac{2}{x^4}=\frac{1}{x^8}$ respectively. By substituting the first 2 terms in the third equation with the first two equations, we get that $\frac{1}{x^8}=(x^2-2x)-2(x-2)=x^2-4x+4$ for all roots $x$ . Furthermore, by Vieta's theorem, we obtain that $\sum_{i=1}^5 r_i =2$ and $\sum_{i=1}^5 r_i^2 = (\sum_{i=1}^5 r_i)^2-2(\sum_{1 \leq i < j \leq 5}{r_i r_j})=2^2-2(0)=4$ Therefore, $\sum_{i=1}^5 \frac{1}{r_i^8} = \sum_{i=1}^5 r_i^2 - 4 \sum_{i=1}^5 r_i + \sum_{i=1}^5 4 = 4-4(2)+5(4)=16$

Observe that: $r_i^4 (r_i - 2) = 1 \Rightarrow \frac{1}{r_i^4} = r_i - 2 \Rightarrow \frac{1}{r_i^8} = r_i^2 - 4r_i + 4$ .

Hence, $\sum_{i=1}^5 \frac{1}{r_i^8} = \sum r_i^2 - 4 \sum r_i + 4*5 = ( \sum r_i)^2 - 2 \sum r_ir_j - 4 * 2 + 20$ [By Vieta's Relations]

$= 2^2 - 2 * 0 - 4 * 2 + 20 =$ $16$

The application of Newton sums seems to be a promising route, but we must first transform a sum of negative powers of roots into a sum of positive powers. This can be accomplished by carefully rearranging the equation:

$\begin{aligned} x^5-2x^4-1=0 \\ x^5-2x^4=1 \\ x^4(x-2)=1 \\ x-2=\frac{1}{x^4} \\ (x-2)^2=\frac{1}{x^8} \end{aligned}$

We can thus rewrite the desired expression as $\sum_{i=1}^5 (r_i-2)^2 = \sum_{i=1}^5 r_i^2-4r_i+4$ From Vieta's formulas (or from Newton sums), we can find that the sum of the roots is 2. From Newton sums, we can find that the sum of the squares of the roots is twice this, or 4. Therefore, the desired expression is $4-4 \times 2+20=\textbf{16}$ .