(restored slightly simplified version of earlier post). How about:
$1- \zeta^{p-1} = -\zeta^{-1}(1-\zeta)$. Doing likewise for $\zeta^{2},\ldots,\zeta^{p-1}$,
\zeta^{2},\ldots,\zeta^{\frac{p-1}{2}}$,
and setting $\alpha = \prod_{i=1}^{\frac{p-1}{2}} (1- \zeta^{i}),$ we see that
$\bar{\alpha} = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2-1}{8}} \alpha.$
Hence we have $$p = \prod_{i=1}^{p-1}(1- \zeta^i) = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2-1}{8}}\prod_{i=1}^{\frac{p-1}{2}} (1- \zeta^{i})^{2}.$$
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7 | typo | ||
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6 | restored simplified version of earlier post | ||
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Post deleted (restored slightly simplified version of earlier post). How about:
$1- \zeta^{p-1} = -now redundant\zeta^{-1}(1-\zeta)$. Doing likewise for $\zeta^{2},\ldots,\zeta^{p-1}$,
and setting $\alpha = \prod_{i=1}^{\frac{p-1}{2}} (1- \zeta^{i}),$ we see that
$\bar{\alpha} = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2-1}{8}} \alpha.$
Hence we have |
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5 | removed post as redundant | ||
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How about $\frac{1- \zeta^{p-1}}{1-\zeta} = (1 + \zeta + \ldots + \zeta^{p-2}) = -\zeta^{p-1}.$
The same applies to $\zeta^2,\ldots,\zeta^{\frac{p-1}{2}}.$ Hence we obtain
Post deleted- \zeta^{i})^{2}.$$now redundant |
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