## ganeshie8 one year ago find a third degree polynomial with integer coefficients whose roots are $\cot^2\frac{\pi}{7},~\cot^2\frac{2\pi}{7},~\cot^2\frac{3\pi}{7}$

1. anonymous

Consider the polynomial equation: $$x^6 +x^5 + x^4 + x^3 + x^2 + x + 1 = 0$$ It's roots are $$e^{i2k\pi/7}$$ where $$k = 1$$ to $$6$$ Divide the equation by $$x^3$$ , then we have, $$x^3 + x^2 + x + 1 + 1/x + 1/x^2 + 1/x^3 = 0$$ $$(x+1/x)^3 -3(x+1/x) + (x+1/x)^2 - 2 + (x+1/x) + 1 = 0$$ When x = $$e^{i2k\pi/7}$$ , $$(x+1/x) = 2\cos(2k\pi/7)$$ So, $$y^3 + y^2 -2y -1 = 0$$ has the roots $$2\cos(2\pi/7), 2\cos(4\pi/7) , 2\cos(6\pi/7)$$ Now, note that $$\cos(2k\pi/7) = \cos^2(k\pi/7) - \sin^2(k\pi/7)$$ $$= \frac{1}{\sec^2(k\pi/7)} - \frac{\tan^2(k\pi/7)}{\sec^2(k\pi/7)}$$ $$= \frac{1-\tan^2(k\pi/7)}{1 + \tan^2(k\pi/7)}$$ $$= \frac{\cot^2(k\pi/7) - 1}{\cot^2(k \pi/7) + 1}$$ So, if $$\cot^2(k\pi/7) = t$$ , $$2\cos(2k\pi/7) = 2\frac{(t-1)}{(t+1)}$$ Therefore $$\cot^2(k\pi/7)$$ satisfies the following equation $$(2\frac{(t-1)}{(t+1)})^3 + (2\frac{(t-1)}{(t+1)})^2 -2(2\frac{(t-1)}{(t+1)}) - 1 = 0$$ Multiplying both sides by $$(t+1)^3$$ , $$(2t-2)^3 + (t+1)(2t-2)^2 -2(t+1)^2(2t-2) - (t+1)^3 = 0$$ Therefore, $$(2t-2)^3 + (t+1)(2t-2)^2 -2(t+1)^2(2t-2) - (t+1)^3$$ is clearly a polynomial with integer coefficients and has as its roots $$\cot^2(k\pi/7)$$ for $$k = 1,2,3$$

2. anonymous

Also, simplifying the final expression, the final polynomial turns out to be $$7t^3 - 35t^2 + 21t-1$$

3. ganeshie8

beautiful! it seems there is no question i know which you cannot solve !

4. ikram002p

Andre thats make me cry :\ @adxpoi lol

5. anonymous

Haha, nah I'm sure there's stuff I can't solve :P

6. ikram002p

extremely interesting question hmmm and answer :O

7. ParthKohli

Hey I'm pretty sure I've asked this one before

8. ganeshie8

I think @adxpoi 's method generalizes to any arbitrary degree polynomial : Find a polynomial of $$n$$th degree whose roots are $\cot^2\frac{\pi}{2n+1},~\cot^2\frac{2\pi}{2n+1},~\cot^2\frac{3\pi}{2n+1},\cdots, \cot^2\frac{n\pi}{2n+1}$

9. ganeshie8

Yes @ParthKohli I remember, we had solved it using Vieta's formulas or something adhoc... :)

10. ParthKohli

No, complex numbers.

11. ParthKohli

In fact just like how adx has solved it.

12. ganeshie8

Ahh okay, can't tryst my memory lol knw how to solve the general case for arbitrary degree ?

13. ParthKohli

My book (KD Joshi) has this, yeah.

14. ganeshie8

Ohk.. I'm hoping we can simply mimic the adx's method, but I haven't tried it yet... so not completely sure yet..

15. ParthKohli

Where did you find this question?

16. ganeshie8

IMO shortlisted practice problems

17. ganeshie8

@praxer has been asking these recently

18. ParthKohli

Hmm, this one is pretty standard for an IMO problem if you've read about it.

19. ganeshie8

I read and forget most of what I read, so I must have encountered this problem before.. but I'm pretty sure I never had this observation before : $x^6+x^5+\cdots+x+1 = \dfrac{x^7-1}{x-1}$ so $$e^{i2k\pi/7}$$ is a root

20. ParthKohli

Exactly.

21. ganeshie8

Is there any other way to see that $$e^{i2k\pi/7}$$ satisfies that polynomial equation ? (w/o using geometric series)

22. ParthKohli

23. ParthKohli

No, not really.

24. ParthKohli

Pages 246-247, yes.

25. ganeshie8

slightly different problem, but the methods are similar yeah https://i.gyazo.com/27c20454519da07d956f53137adc3ee6.png