Timeline for Integer solutions of $2\cos\left(\frac{p\pi}n\right)+2\cos\left(\frac{q\pi}n\right)+4\cos\left(\frac{p\pi}n\right)\cos\left(\frac{q\pi}n\right)=1$
Current License: CC BY-SA 4.0
8 events
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| Sep 24, 2023 at 9:47 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor Math Jaxing (formula hyperlinking)
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| Aug 31, 2023 at 6:13 | comment | added | Gerry Myerson | @Ilya, Conway & Jones did that in the paper I cite in my comment on OP's question. | |
| Aug 30, 2023 at 23:42 | comment | added | Ilya Bogdanov | The equation obtained in the beginning claims that the sum of 9 roots of unity is 0. It is possible to (elementarily) describe all such 9-tuples, and then to check which of them are of the required form; that would be elementary, but I will not do that without a computer aid… | |
| Aug 30, 2023 at 17:48 | comment | added | David E Speyer |
I personally did it using the Mathematica command MinimalPolynomial[E^(-2 Pi I/12) + 1+ E^(2 Pi I/12), x] and then, when I saw that I got a quadratic, solving that quadratic. But a more reasonable method would be to remember that $\cos(\pm \pi/6) = \sqrt{3}/2$ and $\cos(\pm 5 \pi/6) = -\sqrt{3}/2$ :).
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| Aug 30, 2023 at 17:34 | comment | added | user506548 | Just one question... when you wrote "In this case, $\alpha$ and $\beta$ should be primitive $12$-th roots of unity. This means that $\alpha+1+\alpha^{-1}$, and $\beta+1+\beta^{-1}$ should be $1 \pm \sqrt{3}$.", how did you calculate $\alpha+1+\alpha^{-1}$? Is it just substituting the cosine values, or is there a neater trick? | |
| Aug 30, 2023 at 17:32 | comment | added | Fedor Petrov | There seems to be no need to open the brackets and close them again (in the beginning) | |
| Aug 30, 2023 at 17:31 | comment | added | user506548 | This is a beautiful solution (+1)! I will wait for two more days to see if there are any more solutions using more elementary methods, and then accept it. | |
| Aug 30, 2023 at 15:40 | history | answered | David E Speyer | CC BY-SA 4.0 |