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This largest solution to this gorgeous equation is the first local extremum on a function related to the Fibonacci sequence:

$$x^2 \cdot \sin \left(\frac{2\pi}{x+1} \right) \cdot \left(3+2 \cos \left(\frac{2\pi}{x} \right) \right) = (x+1)^2 \cdot \sin \left(\frac{2\pi}{x} \right) \cdot \left(3+2 \cos \left(\frac{2\pi}{x+1} \right) \right)$$

This is as simplified as I could get it. The largest solution to this equation is around $x = 2.1392.$

It appears there is no closed-form solution for this; is there any way to prove if the solution is algebraic or transcendental?

P.S. Can anyone approximate this constant to more decimal places? ANSWERED

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  • $\begingroup$ You should try Mathematica, Maple or Matlab. $\endgroup$
    – markvs
    Mar 1, 2021 at 7:04
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    $\begingroup$ Mathematica to 100 digits yields 2.139212753707616852225181645744445960113284555577699941769231477262384484929696716416193938704042851 $\endgroup$ Mar 1, 2021 at 7:24
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    $\begingroup$ It may at least be possible to show that $x$ is irrational. Conway and Jones gave an algorithm for transforming a trigonometric Diophantine equation into an ordinary Diophantine equation. So one could try running their algorithm on your equation for $x$, and with luck, one might be able to show that the resulting Diophantine equation has no rational solutions. $\endgroup$ Mar 2, 2021 at 1:56
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    $\begingroup$ @Tim, I don't think OP's equation meets the Conway-Jones definition of a trigonometric Diophantine equation. $\endgroup$ Mar 2, 2021 at 4:07
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    $\begingroup$ @GerryMyerson Can't we take $l=1$, $n=4$, trig$_1 = \sin$, $\theta_1(x) = 2/(x+1)$, trig$_2 = \cos$, $\theta_2(x) = 2/x$, trig$_3 = \sin$, $\theta_3(x)=2/x$, trig$_4=\cos$, $\theta_4(x) = 2/(x+1)$, and $R(A,B,C,D,E) = A^2\cdot B \cdot(3 + 2C) - (A+1)^2\cdot D \cdot(3+2E)$? $\endgroup$ Mar 2, 2021 at 4:57

1 Answer 1

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It should be possible to show that $x$ is irrational using Theorem 7 of Trigonometric diophantine equations (On vanishing sums of roots of unity) by J. H. Conway and A. J. Jones, Acta Arithmetica 30 (1976), 229–240, although I have not carried out the full calculation. By letting $\alpha = 2\pi/(x+1)$ and $\beta = 2\pi/x$ and using standard trig identities, we can rewrite the given equation as $$3x^2\sin \alpha -3(x+1)^2\sin\beta - (2x+1)\sin(\alpha+\beta) + (2x^2+2x+1)\sin(\alpha-\beta) = 0.$$ We can convert from sines to cosines via $\sin \gamma \equiv \cos(\pi/2 - \gamma)$. Then Theorem 7 of Conway and Jones tells us that there are only a few "primitive" ways of getting a rational linear combination of four cosines of rational multiples of $\pi$ to vanish: $$\eqalign{{1\over2} &= \cos{\pi\over 3}\cr 0 &= -\cos\phi + \cos\biggl({\pi\over 3}-\phi\biggr) + \cos\biggl({\pi\over 3}+\phi\biggr)\cr {1\over2} &= \cos{\pi\over 5} - \cos{2\pi\over 5}\cr {1\over2} &= \cos{\pi\over 7} - \cos{2\pi\over 7} + \cos{3\pi \over 7}\cr {1\over2} &= \cos{\pi\over 5} - \cos{\pi\over 15} + \cos{4\pi\over 15}\cr {1\over2} &= -\cos{2\pi\over 5} + \cos{2\pi\over 15}-\cos{7\pi\over 15}\cr {1\over2} &= \cos{\pi\over 7} + \cos{3\pi\over 7} - \cos{\pi\over 21} + \cos{8\pi\over21}\cr {1\over2} &= \cos{\pi\over 7} -\cos{2\pi\over7}+\cos{2\pi\over21}-\cos{5\pi\over 21}\cr {1\over2} &= -\cos{2\pi\over 7} + \cos{3\pi\over 7}+ \cos{4\pi\over21} + \cos{10\pi\over 21}\cr {1\over2} &= -\cos{\pi\over 15}+\cos{2\pi\over15}+\cos{4\pi\over 15}-\cos{7\pi\over 15}\cr }$$ One should be able to go through this list and check case by case that no rational value of $x$ can yield the desired equation.

I do not know how to show that $x$ must be transcendental.

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