Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 7"

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Therefore, each <math>P_n(x)</math> will have <math>n</math> distinct complex roots when <math>n</math> is even and <math>n-1</math> distinct complex roots when <math>n</math> is odd.
 
Therefore, each <math>P_n(x)</math> will have <math>n</math> distinct complex roots when <math>n</math> is even and <math>n-1</math> distinct complex roots when <math>n</math> is odd.
  
The roots of <math>Q_n(x)</math> will be all of the roots of <math>P_1,P_2,\cdots P_n</math> which will include several repeated roots.
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The roots of <math>Q_n(x)</math> will be all of the roots of <math>P_1,P_2,\cdots, P_n</math> which will include several repeated roots.
  
 
~Tomas Diaz. orders@tomasdiaz.com
 
~Tomas Diaz. orders@tomasdiaz.com
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 20:13, 26 November 2023

Problem

Let $P_n(x)=1+x+x^2+\cdots+x^n$ and $Q_n(x)=P_1\cdot P_2\cdots P_n$ for all integers $n\ge 1$. How many more distinct complex roots does $Q_{1004}$ have than $Q_{1003}$?

Solution

The roots of $P_n(x)$ will be in the form $x=e^{\frac{2\pi k}{n+1}}$ for $k=1,2,\cdots,n$ with the only real solution when $n$ is odd and $k=\frac{n+1}{2}$ and the rest are complex.

Therefore, each $P_n(x)$ will have $n$ distinct complex roots when $n$ is even and $n-1$ distinct complex roots when $n$ is odd.

The roots of $Q_n(x)$ will be all of the roots of $P_1,P_2,\cdots, P_n$ which will include several repeated roots.

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.