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# Tangent Power Sum = Integer

Prove that \begin{aligned} \tan^6 \frac{\pi}{9}+\tan^6 \frac{2\pi}{9} +\tan^6 \frac{4\pi}{9} \end{aligned} is an integer, and find the value of this integer.

3 years, 6 months ago

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If $$x = \tan \dfrac{k\pi}{9}$$ for any $$k \in \overline{-4,4}$$. Then $$\arg(1+ix) = \dfrac{k\pi}{9}$$, and so, $$\arg[(1+ix)^9] = k\pi$$.

Thus, $$\text{Im}[(1+ix)^9] = 0$$. So, $$x = \tan \dfrac{k\pi}{9}$$ is a root of $$\text{Im}[(1+ix)^9] = 0$$ for $$k \in \overline{-4,4}$$.

Using the binomial theorem, $$P(x) := \text{Im}[(1+ix)^9] = x^9 - 36x^7 + 126x^5 - 84x^3 + 9x$$.

Since $$\tan 0 = 0$$ and $$\tan \dfrac{\pm3\pi}{9} = \pm \sqrt{3}$$ are roots of $$P(x)$$, we know $$x(x^2-3)$$ divides $$P(x)$$.

Factoring yields $$P(x) = x(x^2-3)(x^6-33x^4+27x^2-3)$$.

Therefore, $$\pm \tan \dfrac{\pi}{9}$$, $$\pm \tan \dfrac{2\pi}{9}$$, $$\pm \tan \dfrac{4\pi}{9}$$ are the roots of $$Q(x) := x^6-33x^4+27x^2-3$$.

Thus, $$\tan^2 \dfrac{\pi}{9}$$, $$\tan^2 \dfrac{2\pi}{9}$$, $$\tan^2 \dfrac{4\pi}{9}$$ are the roots of $$R(x) := Q(\sqrt{x}) = x^3-33x^2+27x-3$$.

Let $$S_n$$ be the sum of the $$n$$-th powers of the roots of $$R(x)$$. Now, we use Newton's Sums.

$$S_1 -33 \cdot 1 = 0 \leadsto S_1 = 33$$.

$$S_2 -33 \cdot S_1 + 27 \cdot 2 = 0 \leadsto S_2 = 1035$$.

$$S_3 -33 \cdot S_2 + 27 \cdot S_1 -3 \cdot 3 = 0 \leadsto S_3 = 33273$$.

Therefore, $$S_3 = \tan^6 \dfrac{\pi}{9} + \tan^6 \dfrac{2\pi}{9} + \tan^6 \dfrac{4\pi}{9} = \boxed{33273}$$, which is clearly an integer. · 3 years, 6 months ago

Lovely. · 1 year ago