Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 7"
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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{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. | ||
| − | 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. | ||
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com | ||
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 19:12, 26 November 2023
Problem
Let 
and 
for all integers 
. How many more distinct complex roots does 
have than 
?
Solution
The roots of 
will be in the form 
for 
with the only real solution when 
is odd and 
and the rest are complex.
Therefore, each will have distinct complex roots when is even and 
distinct complex roots when is odd.
The roots of 
will be all of the roots of 
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.
