Solve the following equation in radicals.


I use Magma to verify that its Galois group is a solvable group.

R := RationalField(); 
R < x > := PolynomialRing(R); 
f := x^8-8*x^7+8*x^6+40*x^5-14*x^4-232*x^3+488*x^2-568*x+1; 
G := GaloisGroup(f); 
print G;
GroupName(G: TeX:=true);

The output of Magma(Online) is:

Permutation group G acting on a set of cardinality 8
Order = 16 = 2^4
    (2, 4)(6, 8)
    (1, 2, 3, 4)(5, 6, 7, 8)
    (1, 5)(2, 8)(3, 7)(4, 6)
C_2\times D_4

I also tried to calculate with PARI/GP(64-bit)v_2.13.3+GAP(64-bit)v_4.11.1, but failed.

gap> LoadPackage("radiroot");
gap> x := Indeterminate(Rationals,"x");;
gap> g := UnivariatePolynomial( Rationals, [1,-8,8,40,-14,-232,488,-568,1]);
gap>  RootsOfPolynomialAsRadicals(g, "latex");
Error,AL_EXECUTABLE,the executable for PARI/GP,has to be set at /proc/
cygdrive/C/gap-4.11.1/pkg/aInuth-3.1.2/gap/kantin.gi : 205 called from
  • $\begingroup$ True , the galois group is solvable and hence the polynomial solvable by radicals. But the expressions can still be extremely complicated , just look at the formulas for degree $3$ and $4$. Did you try it with Wolfram Alpha ? $\endgroup$
    – Peter
    Jan 25 at 11:22
  • 1
    $\begingroup$ Where does this come from? It seems not to be the 8th degree eqn you would get from constructing the regular 17-gon. $\endgroup$ Jan 25 at 11:46

3 Answers 3


Well, in honor of an old cartoon I'll say a miracle occurs. But can we get behind the curtain to see how the special effects are made?

If you take the square root of, let us say, $2358$ by the standard "long division" method, you get $48$ with a remainder of $54$, which may be interpreted as the equation


We can adapt this method to determining the square root of a polynomial, and for the one given in this problem we end with this:


If the remainder were a constant times a square then we would be able to render our octic polynomial in the form $a^2-b^2=(a+b)(a-b)$ or perhaps $a^2+b^2=(a+bi)(a-bi)$, getting a pair of quartic factors which would then be solvable by radicals in the usual way. Sadly, we can't do that because the remainder is a cubic polynomial. Nonetheless, the fact that the coefficients of this remainder have a common factor makes one go "hmmm...". What if there were a way to modify the remainder so that it has an even degree and could be a square quantity (or next best, a constant times one)?

I started by noting that the square root of $2358$ as determined by the standard method comes out as $48$ with a remainder of $54$. But did I really have to render the "quotient" as $48$? If I allow a negative remainder in the final stage maybe I could render the root as $49$ instead, in which case the remainder is indeed negative and we get an expression equally valid as the first one I quoted:

$2358=48^2+54$ but also


We might even say that the second form is superior because, with the absolutely smaller remainder, it renders the rounded value of $\sqrt{2358}$ (correctly) as $49$ instead of $48$.

Now what can we do with our polynomial square root? Let us say that, just as we rendered the last digit of the root as $9$ instead of $8$ when we extracted the square root of $2358$, we leave the constant term in our quartic expression as something other than $1$. We get


Can this remainder be a squared quantity, perhaps multiplied by a constant, for some value of $h$, presumably rational?

A necessary condition for this to occur in the quartic expression $ax^4+bx^3+cx^2+dx+e$ is $a/e=(b/d)^2$. Here we require



We turn this to a cubic polynomial equation for $h$, seek rational roots and discover $h=49$. We again go "hmmm...", for not only did we hit on a rational root but we incremented $h$ from its earlier value ($1$) by half the common factor of $96$ we saw in the earlier remainder.

We insert $h=49$ and obtain


If the bracketed quantity were to be a square, it would be $(x^2-x-5)^2$ to match the degree 4, degree 3, degree 1 and degree 0 terms (which our equation for $h$ was designed to do). But do we get the proper degree 2 term? In fact:


and we have hit on our squared remainder!

So now we just factor the octic polynomial as a difference of squares whose roots contain $\sqrt{96}$ or equivalently $\sqrt{6}$:


and we then solve each quartic factor by the usual method.

The roots, with all radicals defined as nonnegative real numbers, are





$1+\sqrt2-\sqrt3\pm i\sqrt[4]{3}$

$1-\sqrt2-\sqrt3\pm i\sqrt[4]{3}$

This set of roots conforms with the $C_2×D_4$ symmetry from the Galois group calculation.

  • 6
    $\begingroup$ I would write one word, “Wow”, but I think so short a comment is not allowed. $\endgroup$
    – Lubin
    Jan 26 at 3:27

Just to explain why the attempt to calculate in GAP failed. The error message was produced by the Alnuth GAP package on which RadiRoot depends, and Alnuth in its turn requires PARI/GP. If everything is installed, the code in question worked (also note the correct order of the coefficients):

gap> LoadPackage("radiroot");
gap> x := Indeterminate(Rationals,"x");;
gap> g := UnivariatePolynomial( Rationals, [1,-568,488,-232,-14,40,8,-8,1]);
gap> RootsOfPolynomialAsRadicals(g, "latex");

It wrote a temporary file and displayed its name. The file contains all commands needed to compile it with LaTeX, and I just show here the crucial part of the output (without modifying the content, so it has some redundant details) which says the following:

An expression by radicals for the roots of the polynomial $$x^{8} - 8x^{7} + 8x^{6} + 40x^{5} - 14x^{4} - 232x^{3} + 488x^{2} - 568x + 1$$ with the $n$-th root of unity $\zeta_n$ and $$\omega_1 = \sqrt[2]{2},$$ $$\omega_2 = \sqrt[2]{3},$$ $$\omega_3 = \sqrt[2]{-\omega_2},$$ $$\omega_4 = \sqrt[2]{\omega_2},$$ is: $$1-\omega_1-\omega_2-\omega_3$$

For other sources of help with GAP, especially with specific GAP packages, please see the GAP tag description.

  • $\begingroup$ You define $\zeta_n$ but I do not see that symbol in the formulas, is something missing? Also square roots are generally not written with the $2$ explicitly on the radical sign. $\endgroup$ Feb 22 at 18:48
  • $\begingroup$ @OscarLanzi that was autogenerated LaTeX produced by the RadiRoot package - could be superfluous, the idea is that the user adjusts it as they need. $\endgroup$ Feb 23 at 10:10
  • $\begingroup$ Understood. But now, I find you may have reversed the order of the coefficients from the original equation. Might want to check that and adjust the code as needed. $\endgroup$ Feb 23 at 10:53
  • 1
    $\begingroup$ @OscarLanzi thanks! I've just used the GAP input from the question, and did not pay attention to that. Curious how did that work in the OP, or if that GAP session was manually edited (e.g. aInuth should be alnuth). I have now updated my answer. $\endgroup$ Feb 23 at 12:38

A Pari computation shows that the discriminant of the degree $8$ number field $F$ generated by a root of the polynomial $f=𝑥^8−8𝑥^7+8𝑥^6+40𝑥^5−14𝑥^4−232𝑥^3+488𝑥^2−568𝑥+1$ is the product of a power of $2$ and a power of $3$. Therefore any quadratic subfield of $F$ is of the form ${\bf Q}(\sqrt{\pm d})$ with $d$ a divisor of $6$. A Pari computation shows that $f$ factors into a product of two degree $4$ irreducible factors over ${\bf Q}(\sqrt{2})$ and over ${\bf Q}(\sqrt{3})$. This implies that we have ${\bf Q}(\sqrt{2},\sqrt{3})\subset F$. A final Pari computation shows that $f$ is a product of four quadratic polynomials in ${\bf Q}(\sqrt{2},\sqrt{3})[X]$. Their roots are equal to $1+\sqrt{2}+{\root 4 \of 3}(1+{\root 4 \of 3})$ and its conjugates.

  • 3
    $\begingroup$ Many computer-assisted answers include the code and I/O used. May get more upvotes this way. $\endgroup$ Jan 29 at 0:47

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