Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

(The question is at the bottom of the post.) Here's a "natural" solvable 17-th deg eqn with small coefficients:

$$\begin{align*} x^{17}-6 x^{16}&-24 x^{15}-42 x^{14}-31 x^{13}-23 x^{12}-7 x^{11}-x^{10}\\ &\quad-4 x^9-11 x^8-7 x^7-13 x^6-x^5+x^3+x^2+x-1 = 0 \quad\text{(eq.1)} \end{align*}$$

Its unique real root is exactly given as $x = \zeta_{48} \eta(\tau)/(\sqrt{2}\,\eta(2\tau)) = 9.1630942 \dots$ where $\eta(\tau)$ is the Dedekind eta function, the root of unity $\zeta_{48} = \exp(2\pi i/48)$, $\tau = (1+\sqrt{-d})/2$, and $d = 383$. This $d$ has class number $h(-d) = 17$.

To solve this, depress eq.1 (get rid of its $x^{n-1}$ term), by letting $x = (y+6)/17$ to get, $$\begin{align*} y^{17}&-11832 y^{15}-1124346 y^{14}-55393735 y^{13}-1784741617 y^{12}\\ &\quad-41171464807 y^{11}-711423456455 y^{10}-9455898295636 y^9-99724287747103 y^8\\ &\quad 887992943070295 y^7-7665207188897171 y^6-70479807472769473 y^5\\ &\quad -592167373130143650 y^4-3496187093606980919 y^3-8695712981307573757 y^2\\ &\quad +68265051092799270505 y-427806967360317821039 = 0 \qquad \text{(eq.2)} \end{align*}$$ Its 16-deg resolvent, a polynomial with INTEGER coefficients, call this $R_{16}$, has roots,

$$\begin{align*} z_k &= [(y_1 + w^k y_2 + w^{2k} y_8 + w^{3k} y_7 + w^{4k} y_{16} + w^{5k} y_4 + w^{6k} y_{12} + \\ &\qquad + w^{7k} y_{15} + w^{8k} y_{11} + w^{9k} y_{10} + w^{10k} y_{14} + w^{11k} y_{13} + w^{12k} y_5 +\\ &\qquad + w^{13k} y_{17} + w^{14k} y_6 + w^{15k} y_9 + w^{16k} y_3)/17]^{17} \end{align*}$$

for $k = 1,\dots,16$ where w is any complex 17th root of unity.

Note the specific arrangement of the $y_n$. There are $16! \approx 2 x 10^{13}$ possible permutations of the $y_n$, and out of that huge number, there are only 16 such that $R_{16}$ has integer coefficients, and we have given one of them. Of course, I used a short cut to find it, because even if your computer can check a million permutations a second, it would still take about 8 months to go through them all. The short cut took less than two hours to find R_16.

The $y_n$ follows the root object Root[poly, n] ordering in Mathematica. Approximately, these are,

$$\begin{align*} y_1 &= 149.7726\\ \{y_2, y_3\} &= -27.62 \mp 18.49i\\ \{y_4, y_5\} &= -21.61 \mp 7.52i\\ \{y_6, y_7\} &= -16.58 \mp 6.34i\\ \{y_8, y_9\} &= -10.57 \mp 15.32i\\ \{y_{10}, y_{11}\} &= -5.02 \mp 13.71i\\ \{y_{12}, y_{13}\} &= -2.34 \mp 13.15i\\ \{y_{14}, y_{15}\} &= 2.57 \mp 2.60i\\ \{y_{16}, y_{17}\} &= 6.31 \mp 7.04i \end{align*}$$

$R_{16}$ has extremely large integer coefficients, with the largest being the 248-digit constant term $429534618434587^{17}$ which, naturally enough, is a 17th power. (Note: $R_{16}$ can easily be formed using 500-digit precision or more on the $y_n$, and multiplying the 16 factors together to form the polynomial.)

The polynomial $R_{16}$ can be factored into two octics over the radical extension $\sqrt{17}$. This, in turn, can be factored into 2 quartics over $\sqrt{2(17+\sqrt{17})}$. This can be factored further into 2 quadratics using an expression involved in the 17th root of unity. Apparently, to solve $R_{16} = 0$, only square roots of square roots of square roots, etc, are needed.

The real root of eq.2 in radicals is then,

$$y_1 = {z_1}^{1/17} + {z_2}^{1/17} + {z_3}^{1/17} + \dots + {z_{16}}^{1/17} = 149.7726 \dots$$

Problem: Express the roots of this particular $R_{16}$ purely in terms of the complex 17th root of unity. (If anyone knows how to contact the mathematician Peter-Lawrence Montgomery, he probably will know how, since he has done something similar with a septic root and the 29th root of unity.)

share|improve this question
    
Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI looks like that is his email –  FiniteA Dec 9 '11 at 16:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.