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.

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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.

Jimmy Kariznov - 7 years, 9 months ago

Yes

U Z - 6 years, 5 months ago

Lovely.

Jun Arro Estrella - 5 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$