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Is there any elementary way to solve the equation $ x^{13}= 1 $ by means of radicals? If not, how to get all the solutions?

Remark: The transcendental form of the solution by means of sines and cosines is not allowed, but only radicals, since this equation is solvable.

The solutions in terms of sines and cosines are $ \cos (k\pi/13)+i\sin(k\pi/13) $ for $ k = 0.1... 12$. So, what I say is that I am looking for an explicit solution in terms of radicals. The trivial answer $\sqrt[13]{1} $ is also excluded.

Tito Piezas III
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Jost Bürgi
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  • You mean stating all 13 complex solutions on the unit circle without using trig functions? – mvw May 20 '16 at 23:29
  • It has 12 complex and 1 real (=1) solutions, how do you want to find the complex ones by radicals ? – G Cab May 20 '16 at 23:34
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    Some confusion may be arising because you're using the word radical in a non-standard way. I think you want to allow only $n$th roots of real numbers. – Dan Piponi May 20 '16 at 23:57
  • Expressing $\cos\left(\frac{n\pi}{m}\right)$ in radicals for general integers $n,m$ is a hard problem. It might be that $m=13$ is small enough / has properties that makes it simpler (but I don't know this). Take a look at related questions of this type: http://math.stackexchange.com/questions/523077/radical-expression-for-cosine-formulas, http://math.stackexchange.com/questions/516142/how-does-cos2-pi-257-look-like-in-real-radicals?lq=1 ...and here is a method that works when $m$ is a Fermat prime. – Winther May 21 '16 at 00:02
  • See also wikipedia: Trigonometric constants expressed in real radicals. Unfortunately the denominator $13$ is not on the list. – Winther May 21 '16 at 00:08
  • Maybe you want to solve the equation by taking iterated square roots? – Noah Olander May 21 '16 at 00:09
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    This can be found by finding the roots of the Chebychev polynomial of order 13. Using symmetry, you can reduce it to a polynomial of the sixth degree $4096x^6-13312x^5+16640x^4-9984x^3+2912x^2-364x+13 $. I doubt that its roots can be found by radicals. This is probably provable using Galois theory. –  May 21 '16 at 00:19
  • @YvesDaoust Using Chebychev and try disproving radical solutions using Galois is a good idea. However I don't see how one can reduce it to degree $6$ (there should be $13$ different roots)? The polynomial you present does not seem to have the desired roots. – Winther May 21 '16 at 00:39
  • @Winther: probably a flaw in the derivation of the polynomial. That doesn't weaken the argument about solvability. –  May 24 '16 at 08:24
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    @Winther: By reducing the problem to just solving cubics and quartics, yes, we can satisfy the OP's request of "solving $x^{13}=1$ in radicals". – Tito Piezas III Dec 05 '16 at 17:04
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    @YvesDaoust: Yes, it can be solved in "radicals" by reducing the necessary equations to even less than a sextic. Kindly see my answer below. – Tito Piezas III Dec 05 '16 at 17:06

4 Answers4

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First, we establish some ground rules. All cubics and quartics can be solved by radicals, agreed? Thus, given,

$$\frac{x^{13}-1}{x-1}=x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=0\tag1$$

to solve for $x$ in radicals all we have to do is to "break-up" the $12$-deg into quadratics, cubics, or quartics.

I. Method 1

Equivalently, $(1)$ is the palindromic quartic,

$$F(x)=x^4 + u x^3 + (u^2 - u - 1)x^2 + u x + 1 =0\tag2$$

with its coefficients determined by,

$$u^3 - u^2 - 4u - 1 = 0\tag3$$

$F(x)=0\,$ has $4$ roots, but as there are $3$ choices of $u$, using each one will yield the $4\times3=12$ roots of $(1)$. As proof, eliminating $u$ between $(2),(3)$ will recover the $12$-deg. The WA command is,

Resultant[x^4 + u x^3 + (u^2 - u - 1)x^2 + u x + 1, u^3 - u^2 - 4u - 1, u]

II. Method 2

For primes $p=6n+1$, we can "factor" the cyclotomic polynomial into $n$ palindromic sextics with coefficients determined by an equation of degree $n$. But a palindromic sextic is just a cubic in disguise since given,

$$x^6 + a x^5 + b x^4 + c x^3 + b x^2 + a x + 1=0$$

then one can solve it as $x+1/x=y$ where $y$ are the roots of the cubic,

$$y^3+a y^2 + (b-3) y +(-2a+c)= 0$$

$p=13:$

Resultant[1 - u x + 2 x^2 - (1 + u) x^3 + 2 x^4 - u x^5 + x^6, -3 + u + u^2, u]

$p=19:$

Resultant[1 - u x - (2 - u^2) x^2 - (6 + 2 u - u^2) x^3 - (2 - u^2) x^4 - u x^5 + x^6, -7 - 6 u + u^2 + u^3, u]

and so on. Thus, you only need to solve $-3 + u + u^2=0$ and $-7 - 6 u + u^2 + u^3=0$, respectively. (Click link for more on $p=19$). For $p=31$, you'll have to tackle the quintic $-5 + u + 21 u^2 - 12 u^3 - u^4 + u^5=0$. But that's another story.

Tito Piezas III
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  • Is (2) the equation (1) cubed? Would be helpful to say so. – abnry Dec 06 '16 at 00:42
  • @abnry: No, $(2)$ is the decomposition of $(1)$. The $12$-deg is irreducible over the rationals, but you can "factor" it into three quartics with coefficients that are roots of cubics. That's what I did. – Tito Piezas III Dec 06 '16 at 03:00
  • @abnry: Specifically, the three factors are $F(x,,u_1)$, $F(x,,u_2)$, and $F(x,,u_3)$. – Tito Piezas III Dec 06 '16 at 03:13
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$x^{13}=1$ implies $x=\sqrt[13]{1}=\zeta_{13}^n$ for any integer $n$ where $\zeta_{13}$ is a primitive 13th root of unity.

The 13 distinct numbers $\zeta_{13}^n$ are the vertices of a regular 13-gon with one vertex at 1. Now you can find the internal angle of a regular 13-gon and apply trigonometry. Not sure if this is what your looking for, but I hope it helps.

user140776
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Why sine and cosine are not allowed? $sin(\pi/2)$ for example is algebraic. So the solutions are $$ x_k=\omega^k, \quad \omega=\cos(2\pi/13)+i\sin(2\pi/13) $$

guestDiego
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    cosinus? like a nasal sinus or a chicken sinus? – user140776 May 20 '16 at 23:41
  • @user140776 It's french :-). – YoTengoUnLCD May 20 '16 at 23:44
  • @user140776 The names of sine and cosine are sinus and cosinus in several European countries. – Eff May 20 '16 at 23:44
  • Better now? Thanks. However I take the occasion to emendate also the notation (i should represent only the imaginary unit) – guestDiego May 20 '16 at 23:44
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    Didn't mean to be offensive. It was a joke. The word sinus in latin means breast (apparently, according to http://www2.clarku.edu/~djoyce/trig/sines.html) . So I asked, like a nasal sinus or a chicken breast? I guess it wasn't funny :( – user140776 May 20 '16 at 23:58
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    My Latin dictionary says that “sinus” means many things, including bowl and bay. On the Moon, the Sinus Iridum is the Bay of (I think) Rainbows. Anyway, my understanding is that the Latin word sinus was chosen as the closest possible translation of the original Arabic word. – Lubin May 21 '16 at 01:25
  • Yes. Also according to to www2.clarku.edu/~djoyce/trig/sines.html it comes from Sanskrit via Arabic. Sanskrit jya-ardha (half-chord), shortened to jiva, transliterated to Arabic as jiba, but since vowels aren't written in Arabic this becomes jb, which gets misinterpreted as "jaib" (breast), hence the half chord (sine being half the chord of twice the angle) becomes "sinus" or breast in Latin. So my original comment was a joke about miscommunication...but now its less funny. – user140776 May 21 '16 at 07:33
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Here’s an unsatisfactory contribution to the discussion:
The Galois group of the extension $\Bbb Q(\zeta_{13})\supset\Bbb Q$ is cyclic of order $12$, and so has cyclic quotients of order three and four. So there are intermediate fields $K_3$ and $K_4$ of degree three and four over $\Bbb Q$, both cyclic. Our field does not contain the cube roots of unity, nor the fourth roots. But once you adjoin these, you can now express $K_3(\zeta_3)$ as gotten from $\Bbb Q(\zeta_3)$ by adjunction of a cube root, and you can express $K_4(i)$ as gotten from $\Bbb Q(i)$ by adjunction of a fourth root.

Lubin
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