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 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. |
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com | ||
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 20:11, 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.
~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.
