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 April 27th, 2018, 08:47 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 248 Thanks: 27 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.
 April 27th, 2018, 09:01 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,124 Thanks: 1103 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 $N = \begin{cases}1 &\text{n odd}\\ 2 &\text{n even}\end{cases}$
 April 27th, 2018, 09:44 AM #3 Senior Member   Joined: Oct 2009 Posts: 562 Thanks: 179 Hint: multiply with x-1 Thanks from topsquark Last edited by skipjack; April 27th, 2018 at 10:58 AM.

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