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How to solve the equation $x^6-2\varphi^5x^5+2\varphi x+\varphi^6=0$ in radicals using Mathematica?

where $\varphi$ is the golden ratio.

Solution is Ramanujan's Class Invariant $G_{125}$, Class Invariant is always algebraic, therefore this equation is solvable in radicals.

I try to use "Solve and Reduce ", but there is no "Radical" output

  t = (1 + Sqrt[5])/2  

  Solve[x^6 - 2 t^5 x^5 + 2 t x + t^6 == 0, x, Reals]  

out

  {{x -> 1}, {x -> Root[1 - 20 #1 - 45 #1^2 - 70 #1^3 - 95 #1^4 - 118 #1^5 -   95 #1^6 - 70 #1^7 - 45 #1^8 - 20 #1^9 + #1^10 &, 2]}} 
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  • $\begingroup$ @Dr.belisarius Solution is Ramanujan's Class Invariant $G_{125}$. Class Invariant is always algebraic $\endgroup$
    – vito
    Jan 7, 2016 at 16:40
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    $\begingroup$ But not every algebraic number has a radical representation. $\endgroup$
    – John Doty
    Jan 7, 2016 at 17:19
  • $\begingroup$ @JohnDoty are you sure? could you give me an example? $\endgroup$
    – vito
    Jan 7, 2016 at 17:27
  • $\begingroup$ See the "polynomial roots" bullet at en.wikipedia.org/wiki/Algebraic_number $\endgroup$
    – John Doty
    Jan 7, 2016 at 17:33
  • $\begingroup$ @JohnDoty aaah of cource.. Galois theory.. :)) $\endgroup$
    – vito
    Jan 7, 2016 at 17:45

2 Answers 2

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Only a partial answer here, but I believe your Root object is indeed solvable in radicals. I have no idea how to find the solution, but here's how I came to my conclusion.

First let's define the polynomial we're working with:

p[x_] := 1 - 20 x - 45 x^2 - 70 x^3 - 95 x^4 - 118 x^5 - 95 x^6 - 70 x^7 - 45 x^8 - 20 x^9 + x^10

Get its roots:

roots = x /. Solve[p[x] == 0, x];

Now we find a generator for the root's splitting field:

θ = ToNumberField[roots, All][[1, 1]];

Find the degree of the splitting field extension:

Exponent[θ[[1]][x], x]
(* 20 *)

From this we can find the order of the Galois group: $$ \left|\text{Gal}(\mathbb{Q}(\theta)/\mathbb{Q})\right| = [\mathbb{Q}(\theta) : \mathbb{Q}] = 20. $$

Now there's only 5 groups of order 20 and they're all solvable, so we can stop here, i.e. we can conclude $p(x)$ can be solved in radicals.

p.s. Here's info on all groups of order 20. It mentions they are all solvable.

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  • $\begingroup$ Maybe cross-list this question with math.stackexchange.com? $\endgroup$
    – QuantumDot
    Jan 7, 2016 at 23:03
  • $\begingroup$ @QuantumDot how do I cross list a question? $\endgroup$
    – Greg Hurst
    Jan 8, 2016 at 3:29
  • $\begingroup$ I have no idea. I thought you would know... $\endgroup$
    – QuantumDot
    Jan 8, 2016 at 14:03
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    $\begingroup$ @ChipHurst I think something is wrong here. As can be seen either from above or Factor[x^6 - 2 t^5 x^5 + 2 t x + t^6 /. t -> GoldenRatio, Extension -> GoldenRatio], one factor is of degree 5. If this splits then that split involves algebraic conjugates, which means it splits into subfactors all of the same degree over Q. Or so I believe. With 5 being prime this of course means there can be no such split. (But maybe I'm seeing this all wrong, what with the fog and all.) $\endgroup$ Jan 8, 2016 at 16:27
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    $\begingroup$ By the way I like that method of figuring the extension degree (so I upvoted). And wipe the L off your forehead, it makes you look gloomy. $\endgroup$ Jan 8, 2016 at 18:15
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t = (1 + Sqrt[5])/2;

p=x^6 - 2 t^5 x^5 + 2 t x + t^6;

Not really an answer (there is already an answer here ) just mentioning something rather interesting that has an underlying lesson to be wary of generalizations from finite samples. I am also completing the answer in the previous link to explain how one can obtain a fully radical solution.


Notice:

p // Factor

$$-\left((x-1) \left(-x^5+5 \sqrt{5} x^4+10 x^4+5 \sqrt{5} x^3+10 x^3+5 \sqrt{5} x^2+10 x^2+5 \sqrt{5} x+10 x+4 \sqrt{5}+9\right)\right)$$

So that one may focus on the degree 5 irreducible factor:

(9 + 4 Sqrt[5] + 10 x + 5 Sqrt[5] x + 10 x^2 + 5 Sqrt[5] x^2 + 10 x^3 + 5 Sqrt[5] x^3 + 10 x^4 + 5 Sqrt[5] x^4 - x^5) //Simplify

$$-x^5+5 \left(\sqrt{5}+2\right) x^4+5 \left(\sqrt{5}+2\right) x^3+5 \left(\sqrt{5}+2\right) x^2+5 \left(\sqrt{5}+2\right) x+4 \sqrt{5}+9$$

I used AskConstants to guess a formula for the unique real root from a numerical approximation. Once that root is factored out, one is left with a degree 4 polynomial which can be solved by radicals.

This is the solution I got from AskConstants:

1/(2^(1/4) DedekindEta[(25 I)/2])

This solution has a problem which is that one expects a ratio of Dedekind functions. In particular, that is needed to removed the usual $\Gamma(1/4)$ that accompanies these Dedekind eta functions evaluated at imaginary rationals (see here for the case of integers (one of the comments mentions that the formula for $\eta(30i)$ there is false)).

This approximation has a relative error with the exact solution of the order of 10^(-68) when computed with N[#,100] ! That is, the approximation has more than 50 correct decimals and the formula is still false !

The lesson/moral of the story part is over. The next section completes the answers provided in the link above


The answer provided in the link for the real root of the quintic factor is:

$$\frac{2^{-1 / 4} \eta^2(\tau)}{\eta\left(\frac{\tau}{2}\right) \eta(2 \tau)}$$

With $\tau=i25=i\sqrt{625}$.

Notice how close AskConstants was. Th missing factor is very close to 1:

(η[τ])^2/η[2 τ] /. η -> 
DedekindEta /. τ -> Sqrt[-25^2] // N[#, 80] &

(* 0.999999999999999999999999999999999999999999999999999999999999999999987915955843352 *)

$\eta(i25)$ can be found in the particular values section of MathematicalFunctionData[DedekindEta]["Dataset"].

$\eta(2\times 25 i)$ can be found in the answer here (I recall that there is a comment that there is a mistake for $\eta(i 30)$ so maybe verify numerically first (I did not check)). The main issue is $\eta(i 25/2)$ which I did not find in my search. This can be derived form the reasoning below.

Denoting $R$ the Rogers-Ramunajan continued fraction and $q=e^{2i\pi\tau}$, the following formula from wikipedia 1 (I also searched for the source referenced for the formula):

$$\frac{1}{R(q)}-R(q) = \frac{\eta(\frac{\tau}{5})}{\eta(5\tau)}+1$$

can be used.

Indeed, for $\tau=\frac{5i}{2}$ one has:

$$\frac{1}{R(e^{-5\pi})}-R(e^{-5\pi}) = \frac{\eta(\frac{i}{2})}{\eta(\frac{25i}{2})}+1$$

The formula for $R(e^{-5\pi})$ is given in the same wikipedia page and involves the golden ratio. $\eta(\frac{i}{2})$ is given in this wikipedia page: https://en.wikipedia.org/wiki/Dedekind_eta_function#Special_values.

The final result with all 6 roots might have a rather interesting computational graph given both the complexity and simplicity of the result.

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  • $\begingroup$ You do understand algebraic numbers may be roots of a quintic and that does not have (generally, that is, some do have) radical representation? $\endgroup$ Dec 25, 2022 at 6:27
  • $\begingroup$ @ВалерийЗаподовников yes why ? The question mentions "therefore this equation is solvable in radicals" the other answer says that it is solvable by radicals, and the title mentions that they would like a solution in radicals. $\endgroup$ Dec 26, 2022 at 0:04
  • $\begingroup$ "Class Invariant is always algebraic, therefore this equation is solvable in radicals." That is false. Abel–Ruffini. $\endgroup$ Dec 26, 2022 at 3:35
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    $\begingroup$ @ВалерийЗаподовников indeed I did not bother much with the deduction given in the question but in any case the other answer mentioned that the solution is indeed solvable by radicals. Still I am not sure why you wrote your initial comment given that the answer provided here gives the steps for constructing the roots by radicals and the other answer already confirmed that the roots can be found by radicals. $\endgroup$ Dec 26, 2022 at 5:28
  • $\begingroup$ That answer just says "It's too long to describe step-by-step the method to solve solvable quintics." And other answer here just says "but I believe your Root object is indeed solvable in radicals". $\endgroup$ Dec 26, 2022 at 6:48

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