Worksheet09: Dickson 041
Current Work (editable)
To define three periods, each of two terms, (16') R + R^{6}, R^{2} + R^{5}, R^{3} + R^{4}, we select the first term R of (21) and the third term R^{6} after it and add them, then the second term R^{3} and the third term R^{4} after it, and finally R^{2} and the third term R^{5} after it. We may also define two periods, each of three terms, z_{1} = R + R^{2} + R^{4}, z_{2} = R^{3} + R^{6} + R^{5}, by taking alternate terms in (21). Since z_{1} + z_{2} = -1, z_{1} z_{2} = 3 + R + ... + R^{6} = 2, z_{1} and z_{2} are the roots of z^{2} + z + 2 = 0. Then R, R^{2}, R^{4} are the roots of w^{3} - z_{1} w^{2} + z_{2}w - 1 = 0. 39. Regular Polygon of 17 Sides. Let R be a root $$ 1 of x^{17} = 1. Then R^{17} - 1 / R - 1 = R^{16} + R^{15} + ... + R + 1 = 0. As in §38, we may take g = 3 and arrange the roots R, ..., R^{16} so that each is the cube of its predecessor: R, R^{3}, R^{9}, R^{10}, R^{13}, R^{5}, R^{15}, R^{11}, R^{16}, R^{14}, R^{8}, R^{7}, R^{4}, R^{12}, R^{2}, R^{6}. Taking alternate terms, we get the two periods, each of eight terms, y_{1} = R + R^{9} + R^{13} + R^{15} + R^{16} + R^{8} + R^{4} + R^{2}, y_{2} = R^{3} + R^{10} + R^{5} + R^{11} + R^{14} + R^{7} + R^{12} + R^{6}. Hence y_{1} + y_{2} = -1. We find that y_{1} y_{2} = 4(R + ... + R^{16}) = -4. Thus (22) y_{1}, y_{2} satisfy y^{2} + y - 4 = 0. Taking alternate terms in y_{1}, we obtain the two periods z_{1} = R + R^{13} + R^{16} + R^{4}, z_{2} = R^{9} + R^{15} + R^{8} + R^{2}. Taking alternate terms in y_{2}, we get the two periods w_{1} = R^{3} + R^{5} + R^{14} + R^{12}, w_{2} = R^{10} + R^{11} + R^{7} + R^{6}. Thus z_{1} + z_{2} = y_{1}, w_{1} + w_{2} = y_{2}. We find that z_{1} z_{2} = w_{1} w_{2} = -1. Hence (23) z_{1}, z_{2} satisfy z^{2} - y_{1}z - 1 = 0, (24) w_{1}, w_{2} satisfy w^{2} - y_{2}w - 1 = 0.
Answer Key (non-editable)
To define three \emph{periods}, each of two terms, \[ \Tag{(16')} R + R^{6},\qquad R^{2} + R^{5},\qquad R^{3} + R^{4}, \] we select the first term $R$ of \Eq{(21)} and the third term $R^{6}$ after it and add them, then the second term $R^{3}$ and the third term $R^{4}$ after it, and finally $R^{2}$ and the third term $R^{5}$ after it. We may also define two periods, each of three terms, \[ z_{1} = R + R^{2} + R^{4},\qquad z_{2} = R^{3} + R^{6} + R^{5}, \] by taking alternate terms in \Eq{(2)1}. \begin{Remark} Since $z_{1} + z_{2} = -1$, $z_{1} z_{2} = 3 + R + \dots + R^{6} = 2$, $z_{1}$ and $z_{2}$ are the roots of $z^{2} + z + 2 = 0$. Then $R$, $R^{2}$, $R^{4}$ are the roots of $w^{3} - z_{1} w^{2} + z_{2}w - 1 = 0$. \end{Remark} \Par{39. Regular Polygon of $17$ Sides.} Let $R$ be a root $\neq 1$ of $x^{17} = 1$. Then \[ \frac{R^{17} - 1}{R - 1} = R^{16} + R^{15} + \dots + R + 1 = 0. \] As in §38, we may take $g = 3$ and arrange the roots $R$, \dots, $R^{16}$ so that each is the cube of its predecessor: \[ R,\ R^{3},\ R^{9},\ R^{10},\ R^{13},\ R^{5},\ R^{15},\ R^{11},\ R^{16},\ R^{14},\ R^{8},\ R^{7},\ R^{4},\ R^{12},\ R^{2},\ R^{6}. \] Taking alternate terms, we get the two periods, each of eight terms, \begin{align*} y_{1} &= R + R^{9} + R^{13} + R^{15} + R^{16} + R^{8} + R^{4} + R^{2}, \\ y_{2} &= R^{3} + R^{10} + R^{5} + R^{11} + R^{14} + R^{7} + R^{12} + R^{6}. \end{align*} Hence $y_{1} + y_{2} = -1$. We find that $y_{1} y_{2} = 4(R + \dots + R^{16}) = -4$. Thus \[ \Tag{(22)} y_{1},\ y_{2} \quad\text{satisfy}\quad y^{2} + y - 4 = 0. \] Taking alternate terms in $y_{1}$, we obtain the two periods \[ z_{1} = R + R^{13} + R^{16} + R^{4}, \qquad z_{2} = R^{9} + R^{15} + R^{8} + R^{2}. \] Taking alternate terms in $y_{2}$, we get the two periods \[ w_{1} = R^{3} + R^{5} + R^{14} + R^{12}, \qquad w_{2} = R^{10} + R^{11} + R^{7} + R^{6}. \] Thus $z_{1} + z_{2} = y_{1}$, $w_{1} + w_{2} = y_{2}$. We find that $z_{1} z_{2} = w_{1} w_{2} = -1$. Hence \begin{align*} \Tag{(23)} z_{1},\ z_{2} &\quad\text{satisfy}\quad z^{2} - y_{1}z - 1 = 0, \\ \Tag{(24)} w_{1},\ w_{2} &\quad\text{satisfy}\quad w^{2} - y_{2}w - 1 = 0. \end{align*}
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