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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). $\displaystyle \sum_{j=1}^{j=n} x^j =x^n+x^{n1}+...+x^2+x=0$ 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 x1
Last edited by skipjack; April 27th, 2018 at 10:58 AM. 

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