How prove this $\tan^4{20^{\circ}}-4\sqrt{3}\tan^3{20^{\circ}}+6\tan^2{20^{\circ}}+4\sqrt{3}\tan{20^{\circ}}\in \mathbb{N}$

Show that $$\tan^4{20^{\circ}}-4\sqrt{3}\tan^3{20^{\circ}}+6\tan^2{20^{\circ}}+4\sqrt{3}\tan{20^{\circ}}\in \mathbb{N}.$$

This problem was created by me, and I have geometry methods. I hope see nice Algebra methods.

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Do you mean $20^\circ$ or $0$ power? –  Sigur Nov 22 '13 at 15:37
Have you tried to $\tan3\theta$ formula to eliminate cubic the term? –  lab bhattacharjee Nov 22 '13 at 15:38
@Sigur,is $20^{o}$ –  math110 Nov 22 '13 at 15:42

HINT:

As the minimum multiple of $\theta=20^\circ$ that produces a known value of $\tan\theta$ is $60^\circ,$

I start with $$\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\iff \tan^3\theta=3\tan\theta+3\tan3\theta\tan^2\theta-\tan3\theta$$

Putting $\theta=20^\circ,\tan3\theta=\tan60^\circ=\sqrt3$ $$\tan^3\theta=3\tan\theta+3\sqrt3\tan^2\theta-\sqrt3$$

$$\implies\tan^4\theta=\tan\theta(3\tan\theta+3\sqrt3\tan^2\theta-\sqrt3)=3\sqrt3\tan^3\theta+3\tan^2\theta-\sqrt3\tan\theta$$

$$=3\sqrt3(3\tan\theta+3\sqrt3\tan^2\theta-\sqrt3)+3\tan^2\theta-\sqrt3 \tan\theta$$

$$=30\tan^2\theta+8\sqrt3\tan\theta-27$$

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@math110, once we express $\tan^4\theta,\tan^3\theta$ in terms of $\tan^2\theta, \tan\theta;$ we can derive many such identities:) –  lab bhattacharjee Nov 22 '13 at 15:59
It's nice,Thank you +1 –  math110 Nov 22 '13 at 16:03
very nice solution...lab bhattacharjee.. –  juantheron Nov 22 '13 at 18:06
@juantheron, nice to hear that. Hope I could make the idea clear –  lab bhattacharjee Nov 24 '13 at 2:51