Solving for the roots of a solvable quintic

Question

In the Wikipedia page about quintics, there was a list of quintics that could be solved with trigonometric roots.

For example:$$x^5+x^4-4x^3-3x^2+3x+1\tag1$$ has roots of the form $2\cos \frac {2k\pi}{11}$ $$x^5+x^4-16x^3+5x^2+21x-9=0\tag2$$ has roots of the form $\sum_{k=0}^{7}e^{\frac {2\pi i 3^k}{41}}$

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Question: Is there a formula or underlying structure to find the roots of the quintics?

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Wikipedia says that the roots are the sums of the first $n$-th roots of unity with $n=10k+1$, so I'm guessing we first have to find $n$, or do we first have to find $k$?

Any help is appreciated.

Answer 1

ADDED: I remembered where I had seen this before, with lower degree; let $\alpha \neq 1$ be a seventh root of unity, $$ \alpha^7 = 1, $$ and take $$ \gamma = \alpha + \alpha^6, $$ $$ \gamma^2 = \alpha^2 + 2 + \alpha^5, $$ $$ \gamma^3 = \alpha^3 + 3 \alpha + 3 \alpha^6 + \alpha^4. $$ Therefore $$ \gamma^3 + \gamma^2 - 2 \gamma - 1 = \alpha^3 + \alpha^2 + \alpha + 1 + \alpha^6 + \alpha^5 + \alpha^4 = 0 $$ This is a cubic with three irrational roots, and is https://en.wikipedia.org/wiki/Casus_irreducibilis

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I simply checked the one that seems easiest. Cute. Let $\omega \neq 1$ be an 11th root of unity, $$ \omega^{11} = 1, $$ so that a root is $$ \beta = \omega + \omega^{10}$$ $$ \beta^2 = \omega^2 + 2 + \omega^9, $$ $$ \beta^3 = \omega^3 + 3 \omega + 3 \omega^{10} + \omega^8, $$ $$ \beta^4 = \omega^4 + 4 \omega^2 + 6 + 4 \omega^9 + \omega^7, $$ $$ \beta^5 = \omega^5 + 5 \omega^3 + 10 \omega + 10 \omega^{10} + 5 \omega^8 + \omega^6. $$ Then $$ 1 + 3 \beta - 3 \beta^2 - 4 \beta^3 + \omega^4 + \omega^5 = 1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 + \omega^7 + \omega^8 + \omega^9 + \omega^{10} = \frac{\omega^{11} - 1}{\omega - 1} = 0 $$

The quintic above and a few more are listed in Wiki Quintic. I cannot tell the source of these, how the list may be extended, or whether this family gives all such possibilities. I checked with gp-pari, for each prime $p = 10n+1,$ the discriminant of their $x^5 + x^4 - 4n x^3 + a x^2 + b x + c$ was $\Delta = w^2 p^4,$ with $w \neq 0 \pmod p.$ For the quintic above, $p = 11,$ and $w = 1.$

Here is an early post that discusses things, How to solve a cyclic quintic in radicals?

Easy enough to find cubic monics of the form $x^3 + x^2 + b x + c$ with no rational roots and square discriminant, so that the Galois group is $\mathbb Z_3.$ Not that there is no loss in having either that form or $x^3 + bx + c,$ as a pure (integer) translation changes the coefficient of $x^2$ by $3,$ and the differences between $x^3 + x^2 + b x + c$ and $x^3 - x^2 + b x - c$ are not important for this problem.

    sqrt d         b         c                 d     sqrt d
     7           -142      -701                49       7
     7             -2        -1                49       7
    13             -4         1               169      13
    19             -6        -7               361      19
    37            -12        11              1369      37
    49            -16       -29              2401      49
    56             -9        -1              3136      56
    62            -10        -8              3844      62
    65            -30        53              4225      65
    65            -56      -181              4225      65
    79            -26        41              6241      79
    86            -14         8              7396      86
    91            -16        13              8281      91
    91            -44      -127              8281      91
    91            -86      -337              8281      91
    97            -32       -79              9409      97
   104            -17       -25             10816     104
   133            -82       259             17689     133
   139            -46       103             19321     139
   152            -25        31             23104     152
   163            -54      -169             26569     163
   172           -100       352             29584     172
   182            -30       -64             33124     182
   183            -20        -9             33489     183
   201            -22         5             40401     201
   203            -30        41             41209     203
   209            -44      -121             43681     209
   217            -72       209             47089     217
   219            -24       -27             47961     219
   247            -82      -311             61009     247
   248            -72      -256             61504     248
   254            -42        80             64516     254
   273            -30        27             74529     273
   287            -30       -43             82369     287
   296            -49      -137             87616     296
   301            -44        83             90601     301
   309            -34       -61             95481     309
   313           -104       371             97969     313
   325            -30       -25            105625     325
   349           -116      -517            121801     349
   392            -65       167            153664     392
   399            -44        69            159201     399
   427           -142       601            182329     427
   436            -36        -4            190096     436
   446            -74      -256            198916     446
   448            -37       -29            200704     448
   453            -50      -123            205209     453
   469           -156      -799            219961     469
   496            -41        23            246016     496
   520           -121      -545            270400     520
   532            -44       -64            283024     532
   566            -94       304            320356     566
   579            -64       143            335241     579
   581           -114       419            337561     581
   589            -44        -7            346921     589
   611            -82       235            373321     611
   628            -52        64            394384     628
   632           -105      -433            399424     632
   651            -72      -225            423801     651
   679            -86       251            461041     679
   688            -57      -121            473344     688
   728            -65      -169            529984     728
   776           -129       503            602176     776
   813            -90       261            660969     813
   832            -69       131            692224     832
   845            -56       -25            714025     845
   851            -86       233            724201     851
  sqrt d           b         c                d     sqrt d

Answer 2

Denoted \begin{align*} \qquad\left\{ \begin{aligned} \varepsilon_1&=\exp\left(\dfrac{\>\pi\,\!i\>}{5}\right)=\phantom{+}\frac{1+\sqrt{5}}{4}+i\sqrt{\frac{5-\sqrt{5}}{8}}\approx\phantom{+}0.809017 + 0.587785i\qquad\\ \varepsilon_2&=\exp\left(\dfrac{2\pi\,\!i}{5}\right)=\frac{-1+\sqrt{5}}{4}+i\sqrt{\frac{5+\sqrt{5}}{8}}\approx\phantom{+}0.309017 + 0.951057i\\ \varepsilon_3&=\exp\left(\dfrac{3\pi\,\!i}{5}\right)=\phantom{+}\frac{1-\sqrt{5}}{4}+i\sqrt{\frac{5+\sqrt{5}}{8}}\approx-0.309017 + 0.951057i\\ \varepsilon_4&=\exp\left(\dfrac{4\pi\,\!i}{5}\right)=-\frac{1+\sqrt{5}}{4}+i\sqrt{\frac{5-\sqrt{5}}{8}}\approx-0.809017+0.587785i\\ \end{aligned} \right.\qquad \end{align*} and \begin{gather*} \qquad\left\{ \begin{aligned} \alpha_1&=\sqrt[\underset{\,}{5}]{\frac{11}{4}\left(89+25\sqrt{5}+5i\sqrt{410-178\sqrt{5}}\right)}\approx3.315688+0.078842i\\ \alpha_2&=\sqrt[\underset{\,}{5}]{\frac{11}{4}\left(89-25\sqrt{5}+5i\sqrt{410+178\sqrt{5}}\right)}\approx3.197878+0.879531i\\ \alpha_3&=\sqrt[\underset{\,}{5}]{\frac{11}{4}\left(89-25\sqrt{5}-5i\sqrt{410+178\sqrt{5}}\right)}\approx3.197878-0.879531i\\ \alpha_4&=\sqrt[\underset{\,}{5}]{\frac{11}{4}\left(89+25\sqrt{5}-5i\sqrt{410-178\sqrt{5}}\right)}\approx3.315688-0.078842i \end{aligned} \right.\qquad\\ \end{gather*} Then, \begin{align*} \bbox[#EFF,15px,border:2px solid blue]{ \left\{\begin{aligned} x_1&=\dfrac{\phantom{+}\,\alpha_1\varepsilon_1+\alpha_2\varepsilon_1-\alpha_3\varepsilon_4-\alpha_4\varepsilon_4-1}{5}\\ x_2&=\dfrac{-\,\alpha_1\varepsilon_4-\alpha_2\varepsilon_2+\alpha_3\varepsilon_3+\alpha_4\varepsilon_1-1}{5}\\ x_3&=\dfrac{-\,\phantom{\varepsilon_0}\alpha_1-\alpha_2\varepsilon_4+\alpha_3\varepsilon_1-\phantom{\varepsilon_0}\alpha_4-1}{5}\\ x_4&=\dfrac{-\,\alpha_1\varepsilon_2+\alpha_2\varepsilon_3-\alpha_3\varepsilon_2+\alpha_4\varepsilon_3-1}{5}\\ x_5&=\dfrac{\phantom{+}\,\alpha_1\varepsilon_3-\phantom{\varepsilon_0}\alpha_2-\phantom{\varepsilon_0}\alpha_3-\alpha_4\varepsilon_2-1}{5} \end{aligned}\right. } \end{align*} satisfies the equation $x^5+x^4-4x^3-3x^2+3x+1=0$

References

Dummit, D. S. "Solving Solvable Quintics."Math. Comput.57, 387-401, 1991. https://www.ams.org/journals/mcom/1991-57-195/S0025-5718-1991-1079014-X/S0025-5718-1991-1079014-X.pdf

Answer 3

I am not sure if there is any simple formula for the root structure of polynomials having a solvable polynomial. The only thing is that the subnormal series tells you the order in which you adjoin radicals. It does not tell you which number in each field formed you take a radical of to make the next field, but if you make the correct choices, then the final field is the splitting field, and the root of the polynomial will be just linear combinations over the base field, of the elements of the splitting field.

For example: $1\triangleleft G_2\triangleleft V\triangleleft A_4\triangleleft S_4$, where $G_2$ is any order two group containing a double-transposition has the following.

$\text{ord}(S_4/A_4)=2$

$\text{ord}(A_4/V)=3$

$\text{ord}(V/G_2)=2$

$\text{ord}(G_2/1)=2$

Thus the general quartic's splitting field is constructed by these steps:

Step 1: Pick a number in the base field and adjoin it's square root.

Step 2: Pick a number in the new field and adjoin it's cube root and $e^{2\pi i/3}$.

Step 3: Pick a number in the new field and adjoin it's square root.

Step 4: Pick a number in the new field and adjoin it's square root, thus producing the splitting field.

By the way, for your two quintics, we know that: $$e^{\pm 2\pi i/5}=\frac{-1+\sqrt{5}\pm i\sqrt{10+2\sqrt{5}}}{4}$$ $$e^{\pm 4\pi i/5}=\frac{-1-\sqrt{5}\pm i\sqrt{10-2\sqrt{5}}}{4}$$

For $x^5+x^4-4x^3-3x^2+3x+1=0$:

If $(a,b,k)$ is any of $(-4,4,2)$, $(2,-4,4)$, $(4,0,6)$, $(-2,-2,8)$, $(0,2,10)$ then \begin{align*} 2\cos \frac{2k\pi}{11}=-\frac{1}{5} &-\frac{1}{10}e^{a\pi i/5}\sqrt[5]{7832-2200\sqrt{5}+(1320+880\sqrt{5})i\sqrt{10+2\sqrt{5}}}\\ &-\frac{1}{10}e^{-a\pi i/5}\sqrt[5]{7832-2200\sqrt{5}-(1320+880\sqrt{5})i\sqrt{10+2\sqrt{5}}}\\ &-\frac{1}{10}e^{b\pi i/5}\sqrt[5]{7832+2200\sqrt{5}+(1320-880\sqrt{5})i\sqrt{10-2\sqrt{5}}}\\ &-\frac{1}{10}e^{-b\pi i/5}\sqrt[5]{7832+2200\sqrt{5}-(1320-880\sqrt{5})i\sqrt{10-2\sqrt{5}}} \end{align*}

For $x^5+x^4-16x^3+5x^2+21x-9=0$:

If $(a,b,n)$ is any of $(0,-2,1)$, $(-4,0,2)$, $(2,2,4)$, $(-2,4,8)$, $(4,-4,16)$ then

\begin{align*} \sum_{k=0}^7e^{2n\pi i3^k/41}=-\frac{1}{5} &+\frac{1}{10}e^{a\pi i/5}\sqrt[5]{321768-8200\sqrt{5}+(9840+14760\sqrt{5})i\sqrt{10+2\sqrt{5}}}\\ &+\frac{1}{10}e^{-a\pi i/5}\sqrt[5]{321768-8200\sqrt{5}-(9840+14760\sqrt{5})i\sqrt{10+2\sqrt{5}}}\\ &+\frac{1}{10}e^{b\pi i/5}\sqrt[5]{321768+8200\sqrt{5}+(9840-14760\sqrt{5})i\sqrt{10-2\sqrt{5}}}\\ &+\frac{1}{10}e^{-b\pi i/5}\sqrt[5]{321768+8200\sqrt{5}-(9840-14760\sqrt{5})i\sqrt{10-2\sqrt{5}}} \end{align*}