3

Prove that one root of the Equation $${ \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $$

is

$$x=2\left(\cos\frac{4\pi}{19}+\cos\frac{6\pi}{19}+\cos\frac{10\pi}{19}\right) $$

My progress:

After simplyfying the given Equation I got $$ x^8-16x^6+88x^4-192x^2+x+140=0$$

Then I tried to factorise it, got this

$ x^8-16x^6+88x^4-192x^2+x+140=(x^3+x^2-6x-7)(x^5-x^4-9x^3+10x^2+17x-20)$

$ x^8-16x^6+88x^4-192x^2+x+140=(x^3+x^2-6x-7)(x^3-2x^2-3x+5)(x^2+x-4)$ Should I now solve cubic?

Makar
  • 2,267
  • see page 26 https://books.google.com/books?id=wt7lgfeYqMQC&pg=PR1&lpg=PR1&dq=reuschle++tafeln+complexer+primzahlen&source=bl&ots=VGZFPrfUBn&sig=MlQ667PqXaQ9rAvLWkG3_F1rwsk&hl=en&sa=X&ved=0ahUKEwiIwtSvm9TQAhUJ-2MKHXJIA_kQ6AEIODAE#v=onepage&q=reuschle%20%20tafeln%20complexer%20primzahlen&f=false – Will Jagy Feb 09 '18 at 18:44
  • @MathLover see http://math.stackexchange.com/questions/2022216/on-the-trigonometric-roots-of-a-cubic/2022887#2022887 for my list of these...the method of Gauss is written up nicely in Galois Theory by Cox. Maybe I can find it: one of the earliest times Tito asked about these it was in exactly this format, square roots http://math.stackexchange.com/questions/1065862/something-strange-about-sqrt-4-sqrt-4-sqrt-4-x-x-and-its-friends – Will Jagy Feb 09 '18 at 18:51
  • @ Will Jagy Google seems to jump pages 21 to 32? – Narasimham Feb 09 '18 at 22:41

2 Answers2

3

$\pm 2,\pm 3,\pm 5$ are precisely twice the cubic residues $\!\pmod{19}$. Let $\omega=\exp\left(\frac{2\pi i}{19}\right)$. $\omega$ is al algebraic number over $\mathbb{Q}$ with degree $18$, hence by Galois theory both $$ \sum_{k\in\{\pm 1,\pm 7,\pm 8\}}\omega^k,\qquad \sum_{k\in\{\pm 2,\pm 3,\pm 5\}}\omega^k $$ (related to the Kummer sums $\!\!\pmod{19}$) are algebraic numbers over $\mathbb{Q}$ with degree $3$. Indeed, the minimal polynomial of the second sum is $z^3+z^2-6z-7$, and the second sum is exactly $2\left(\cos\frac{4\pi}{19}+\cos\frac{6\pi}{19}+\cos\frac{10\pi}{19}\right)$.

Jack D'Aurizio
  • 353,855
  • 1
    Gauss's simple method for constructing polynomials with such roots, in the Disquisitiones, is given a pleasant modern treatment in Galois Theory by D. A. Cox. I put lists of such cubics at http://math.stackexchange.com/questions/2022216/on-the-trigonometric-roots-of-a-cubic/2022887#2022887 and quintics and septics at http://math.stackexchange.com/questions/1996552/any-more-cyclic-quintics I notice Wikipedia does not give any of these at https://en.wikipedia.org/wiki/Septic_equation , I would like to put in a half dozen septics as they have for quintics – Will Jagy Feb 09 '18 at 20:07
1

I created this equation by the following way.

$$3^n\equiv1,3,9,8,5,15,7,2,6,...\!\pmod{19}$$ Now, let $$x_1=2\left(\cos\frac{2\cdot1\pi}{19}+\cos\frac{2\cdot8\pi}{19}+\cos\frac{2\cdot7\pi}{19}\right),$$ $$x_2=2\left(\cos\frac{2\cdot3\pi}{19}+\cos\frac{2\cdot5\pi}{19}+\cos\frac{2\cdot2\pi}{19}\right)$$ and $$x_3=2\left(\cos\frac{2\cdot9\pi}{19}+\cos\frac{2\cdot15\pi}{19}+\cos\frac{2\cdot6\pi}{19}\right).$$ Thus, $x_1$, $x_2$ and $x_3$ they are roots of the cubic equation with rational coefficients.

Easy to show that it's $$x^3+x^2-6x-7=0.$$

Also, easy to show that $$-x_1=\sqrt{4-x_2},$$ $$x_2=\sqrt{4-x_3}$$ and $$-x_3=\sqrt{4-x_1}$$ which gives an equation with radicals.

I posted this equation here: https://artofproblemsolving.com/community/c6h69475

Michael Rozenberg
  • 194,933