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April 27th, 2018, 08:47 AM   #1
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Polynomial solutions

I'm not able to define the solutions.
How many solutions does the polynomial equation below have for $\displaystyle n$ natural? Note: (without complex solutions).
$\displaystyle \sum_{j=1}^{j=n} x^j =x^n+x^{n-1}+...+x^2+x=0$

Last edited by skipjack; April 27th, 2018 at 10:57 AM.
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April 27th, 2018, 09:01 AM   #2
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It appears that (excluding complex solutions)

$x = \begin{cases}0 &\text{n odd}\\\{0,-1\} &\text{n even} \end{cases}$

so the number of real solutions is

$N = \begin{cases}1 &\text{n odd}\\ 2 &\text{n even}\end{cases}$
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April 27th, 2018, 09:44 AM   #3
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Hint: multiply with x-1
Thanks from topsquark

Last edited by skipjack; April 27th, 2018 at 10:58 AM.
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