Given the polynomial $X^4+X^3+X^2+X+1=P(X)$. I know that $X^5=1$ and I need to find $x_1^{8}+x_2^{18}+x_3^{28}+x_4^{38}$ would this be equal to $4x^3$? What do I do next
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Are $x_1, \ldots, x_4$ the roots of $P$? – Travis Willse Jun 16 '17 at 16:30
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yes they are the roots – Lola Jun 16 '17 at 16:32
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1Since you know the roots satisfy $x^5 = 1$, we have, for example, $x_1^8 = x_1^5 x_1^3 = x_1^3$. – Travis Willse Jun 16 '17 at 16:36
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Related : https://math.stackexchange.com/questions/2323526/a-polynomial-is-divisible-with-another-one/2323530#2323530 – lab bhattacharjee Jun 16 '17 at 17:36
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Let $x_1,\ldots,x_4$ be the roots of $P$. Then $x_i^5=1$ and $x_i=x_1^i$ for all $i$, hence $$ x_1^{8}+x_2^{18}+x_3^{28}+x_4^{38}=\sum_{i=1}^4 x_i^{3}=\sum_{i=1}^4 x_1^{3i}=\sum_{i=1}^4 x_1^{i}=\sum_{i=1}^4 x_i=-1. $$ (Here, we can "remove" the $3$ because $\{3i: i=1,\ldots,4\}\equiv \{1,\ldots,4\}\bmod{5}$.)
Paolo Leonetti
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Since $x_k =e^{2k\pi i/5} $,
$\begin{array}\\ x_1^{8}+x_2^{18}+x_3^{28}+x_4^{38} &=e^{8\cdot 2\pi i/5}+e^{18\cdot 4\pi i/5}+e^{28\cdot 6\pi i/5}+e^{38\cdot 8\pi i/5}\\ &=e^{16\pi i/5}+e^{72\pi i/5}+e^{168\pi i/5}+e^{304\pi i/5}\\ &=e^{6\pi i/5}+e^{2\pi i/5}+e^{8\pi i/5}+e^{4\pi i/5}\\ &=x_1+x_2+x_3+x_4\\ &=\text{ the sum of the roots}\\ \end{array} $
Since $\prod_{k=1}^4 (x-x_k) =x^4-x^3\sum_{k=1}^4 {x_k}+... =x^4+x^3+... $, the sum of the roots is $-1$.
marty cohen
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