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Is there a natural number $n$ such that the integral part of $(1+2\cos 20 ^{\circ})^n$ can be divided by $1000000$?

My thoughts. I know here is a well known propery Prove that $\cos 20^{\circ} + \cos 100^{\circ} + \cos {140^{\circ}} = 0$ but not sure how it could be applied

Or is there a better trick using the Euler formula?

Chen Chen
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1 Answers1

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It is a root of $x^3-3x^2+1=0$. So $y_n=x_1^n+x_2^n+x_3^n$ obeys the recurrence relation $$y_n=3y_{n-1}-y_{n-3}$$ and the other two roots of the cubic approach zero.

Empy2
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