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I got this polynomial $$ f(x)=x^8+x^7-7x^6-6x^5+15x^4+10x^3-10x^2-4x+1 $$ by calculating $$ z^{17}=1 $$ and then $x=z+\overline{z}=2\operatorname{Re}z$. Now one root of $f$ is therefore of course $$ 2\cos\left(\frac{2\pi}{17}\right). $$ My question is:

Can someone go on from here with the polynomial $f$ and derive the expression $$ \small{2\cos \frac{2\pi}{17} = \frac{1}{8} \left( -1 + \sqrt{17} + \sqrt{ 2 \left(17 - \sqrt{17} \right)} + 2 \sqrt{ 17 + 3 \sqrt{17} - \sqrt{2 \left(17- \sqrt{17} \right)} - 2 \sqrt{2 \left(17+ \sqrt{17} \right)} } \right)} $$ that Gauss got while proving that the heptadecagon is constructible?

Note: I know the way he did it by using the primitive root $3$ mod $17$, a generator of $\mathbb{Z}_{17}^*$, but I wanted to get to the solution using $f$.

Update:

In a similar Stack Exchange question I found that $$ g(t)=t^4 + t^3 - 6t^2 - t + 1 $$ is related to $f$.

Is there a substitution to get $$ g(\text{substitution})=f(x) $$

calculatormathematical
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    The galois-group has only $8$ elements , so there should be an elegant exact solution. – Peter Aug 14 '23 at 17:06
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    Gauss used this to construct the $17$-gon. It is now well known that we can construct the $n$-gon iff it is the product $\ge 3$ of a power of $2$ and distinct Fermat prime numbers. – Peter Aug 14 '23 at 17:22
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    Hardy and Wright: Introduction to the Theory of Numbers (p57ff of the fourth edition) give explicit details of how to do this by solving four quadratics. Essentially one groups the powers of $z$ into convenient sets (using indices determined using a primitive root of $17$) and this controls the coefficients of the quadratics involved. In principle the same method could be used for the higher Fermat primes. – Mark Bennet Aug 14 '23 at 17:39
  • It is in this chapter of Cox, Galois Theory: http://zakuski.math.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf – Will Jagy Aug 14 '23 at 18:52
  • suggest you take a look at Reuschle for prime 17 https://archive.org/details/tafelncomplexer00unkngoog/page/18/mode/2up?view=theater where he shows polynomials for different choices of summing the roots. Also, in chapter 10, Cox gives more detail on 17, he mentions in the paragraph with formula (9.19). I guess I did not scan chapter 10 because I was primarily interested in polynomials of degrees 3,5, and later 7. Here is one of mine: http://math.stackexchange.com/questions/1996552/any-more-cyclic-quintics – Will Jagy Aug 14 '23 at 19:57
  • In W.Mathematica command Factor[x^8 + x^7 - 7 x^6 - 6 x^5 + 15 x^4 + 10 x^3 - 10 x^2 - 4 x + 1, Extension -> Cos[2 Pi/17]] return all roots. – Dmitry Ezhov Aug 15 '23 at 01:06

1 Answers1

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There are several ways to do it. One straightforward way is to factor the octic into two quartics over a square root extension, then use the quartic formula. Quartics have a cubic resolvent, but since the $17$th root of unity is construcible, then we know this cubic will also factor over a square root.

I. Factoring the octic

Given,

$$x^8+x^7-7x^6-6x^5+15x^4+10x^3-10x^2-4x+1= 0$$

Factor this into two quartics over $\sqrt{17}$ choosing,

$$x^4 + \frac{1 - \sqrt{17}}2x^3 - \frac{3 + \sqrt{17}}2x^2 + (2 + \sqrt{17})x - 1 = 0$$

such that $x = 2\cos \left(\frac{2\pi}{17} \right)$. Depress the quartic with the transformation,

$$x = y+\frac{-1 + \sqrt{17}}8$$

to get,

$$y^4 + \frac{-51 - 5\sqrt{17}}{16}y^2 + \frac{17 + 7\sqrt{17}}{16}y + \frac{221 + 43 \sqrt{17}}{512} = 0$$

II. Depressed quartic

Given,

$$y^4+py^2+qy+r = 0$$

then its cubic resolvent is simply,

$$ u^3 + 2p u^2 + (p^2 - 4r)u - q^2 =0$$

With any non-zero root $u$, then the depressed quartic's formula for the four roots are,

$$y_k = \frac12\left(\sqrt{u}\pm\sqrt{-(2p+u)-\frac{2q}{\sqrt{u}}}\right)$$

for $\pm\sqrt{u}$ and you are done.

III. Cubic resolvent

Given $(p,q,r)$ we find the resolvent as,

$$u^3 + \frac{-51 - 5 \sqrt{17}}8u^2 + \frac{323 + 53 \sqrt{17}}{32}u - \left(\frac{17 + 7\sqrt{17}}{16}\right)^2 = 0$$

this has the root,

$$u = \frac{17-\sqrt{17}}{8}$$

Using the formula and reversing the transformation, we find,

$$x = \frac{-1+\sqrt{17}}8+\frac12\sqrt{u}+\frac12\sqrt{\sqrt{17}+2u-\sqrt{\frac{17}{u}}-\sqrt{u}}$$

After some minor algebraic manipulation, we end up with the original form,

$$\small{x_1 = 2\cos \left(\frac{2\pi}{17} \right) = \frac{1}{8} \left( -1 + \sqrt{17} + \sqrt{ 2 \left(17 - \sqrt{17} \right)} + 2 \sqrt{ 17 + 3 \sqrt{17} - \sqrt{2 \left(17- \sqrt{17} \right)} - 2 \sqrt{2 \left(17+ \sqrt{17} \right)} } \right)}$$

IV. Conjugate

Using one of its conjugates,

$$\small{x_2 = 2\cos \left(\frac{8\pi}{17} \right) = \frac{1}{8} \left( -1 + \sqrt{17} + \sqrt{ 2 \left(17 - \sqrt{17} \right)} \color{red}{-} 2 \sqrt{ 17 + 3 \sqrt{17} - \sqrt{2 \left(17- \sqrt{17} \right)} - 2 \sqrt{2 \left(17+ \sqrt{17} \right)} } \right)}$$

then,

$$t=x_1+x_2 = 2\cos \left(\frac{2\pi}{17} \right)+2\cos \left(\frac{8\pi}{17} \right) = \frac{1}{4} \left( -1 + \sqrt{17} + \sqrt{ 2 \left(17 - \sqrt{17} \right)}\right)$$

a root of,

$$t^4 + t^3 - 6t^2 - t + 1 = 0$$

which was the second question of the OP.

Tito Piezas III
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  • Thank you! Is there an easy way to find the coefficients of the two quartics or did you use an engine? – calculatormathematical Aug 15 '23 at 20:32
  • @calculatormathematical I used an engine (Mathematica). However, if you want to do it the classical way then assume, $$\big(x^4+(a+b\sqrt{17})x^3+(c+d\sqrt{17})x^2+(e+f\sqrt{17})x+(g+h\sqrt{17})\big)\ \big(x^4+(a-b\sqrt{17})x^3+(c-d\sqrt{17})x^2+(e-f\sqrt{17})x+(g-h\sqrt{17})\big)=0$$ expand, and compare coefficients with the octic. This will give you a system of 8 equations in 8 unknowns. (This can be simplified by setting $g=1$ and $h=0$.) – Tito Piezas III Aug 16 '23 at 03:49
  • I tried to solve a similar system but I don't get why you can express the coefficients as $d\pm\sqrt{17}t$ for numbers $d,t$, this is the last thing in this whole procedure that I don't understand. – calculatormathematical Aug 16 '23 at 09:43
  • @calculatormathematical I really don't understand your question. The octic factors over $\sqrt{17}$. Therefore the two quartics will have coefficients of form $a_k\pm b_k\sqrt{17}$ for some rational $(a,b)$. P.S. And I meant $\color{blue}{g=-1}$. – Tito Piezas III Aug 16 '23 at 10:01
  • I was unclear, let me improve my question: When you see a random octic, can you see directly over which square root it factors? – calculatormathematical Aug 16 '23 at 10:50
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    @calculatormathematical Oh, I understand now. No, you have to look at its discriminant. The WolframAlpha command is FactorInteger[Discriminant[P(x),x]]. If you try it on the octic, you'll see it is $d=17^7$ so that's your clue. But octics are tricky: you may need the higher quartic extension like in this MSE post or even over a $7$th deg extension like this MO post. – Tito Piezas III Aug 16 '23 at 11:02