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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 7"
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==Solution==
 
==Solution==
  
−
The roots of <math>P_n(x)</math> will be in the form <math>x=e^{\frac{2\pi k}{n+1}}</math> for <math>k=1,2,\cdots,n</math> with the only real solution when <math>n</math> is odd and <math>k=\frac{n+1}{2}</math> and the rest are complex.
+
The roots of <math>P_n(x)</math> will be in the form <math>x=e^{\frac{k}{n+1}2\pi i}</math> for <math>k=1,2,\cdots,n</math> with the only real solution when <math>n</math> is odd and <math>k=\frac{n+1}{2}</math> and the rest are complex.
  
 
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.
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But to get how many more '''distinct''' complex roots, we must subtract the complex roots of <math>P_{1004}</math> that can be found in <math>Q_{1003}</math>
 
But to get how many more '''distinct''' complex roots, we must subtract the complex roots of <math>P_{1004}</math> that can be found in <math>Q_{1003}</math>
  
−
complex roots of <math>P_{1004}:\; x=e^{\frac{1}{1005}2\pi},e^{\frac{2}{1005}2\pi},\cdots,e^{\frac{1004}{1005}2\pi}</math>
+
complex roots of <math>P_{1004}:\; x=e^{\frac{1}{1005}2\pi i},e^{\frac{2}{1005}2\pi i},\cdots,e^{\frac{1004}{1005}2\pi i}</math>
  
  

Revision as of 20:20, 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 $x=e^{\frac{k}{n+1}2\pi i}$ 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.

To get how many more complex roots does $Q_{1004}$ have than $Q_{1003}$ that will be the number of complex roots of $P_{1004}$.

But to get how many more distinct complex roots, we must subtract the complex roots of $P_{1004}$ that can be found in $Q_{1003}$

complex roots of $P_{1004}:\; x=e^{\frac{1}{1005}2\pi i},e^{\frac{2}{1005}2\pi i},\cdots,e^{\frac{1004}{1005}2\pi i}$



~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.

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