# Roots of this equation in x

I am examining the roots of the equation in $$x$$, $$\sum_{q=0}^{2k-1} (-1)^{q} {2k+1 \choose q+1} x^{2k-q} m^{q}+r=0$$ where $$m$$ and $$r$$ are positive integers.

I want to know whether the roots of this can always be fully expressed in radicals (which I suspect that it does), and if so, what can be said regarding the discriminants of the roots. I need as much info about these discriminants as possible. Does anyone know how to properly go about this?

• I guess it is common to say "solution" to an equation, vs "zeros/roots" of a polynomial.
– YCor
May 30 at 15:10

Upon dividing the given polynomial by $$m^{2k}$$ and replacing $$x$$ with $$mx+m$$, it assumes the simpler form $$\begin{equation} s+(x+1)^{2k+1}-x^{2k+1} \end{equation}$$ for some $$s$$.

Generically its Galois group is the wreath product $$C_2^k\rtimes S_k$$. This follow from Hilbert's irreducibility theorem together with the fact that its Galois group over $$\mathbb C(s)$$, where we consider $$s$$ as a transcendental over the complex numbers, is $$C_2^k\rtimes S_k$$, so in general the roots cannot be expressed in radicals once $$k\ge5$$.

Write $$n=2k$$, set $$f(x)=(x+1)^{n+1}-x^{n+1}$$ and consider the branched cover $$a\mapsto f(a)$$. First note that $$f(x-1/2)$$ is an even function, so the monodromy group $$G$$ of this cover is a subgroup of $$C_2^k\rtimes S_k$$.

The critical points of the cover are the roots of $$f'(x)$$. Let $$\gamma$$ be such a root, so $$(\gamma+1)^n-\gamma^n=0$$ and therefore $$f(\gamma)=\gamma^n$$. The possibilities for $$\gamma$$ are $$\gamma=\frac{1}{\zeta-1}$$, where $$\zeta$$ is an $$n$$-th root of unity different from $$1$$. Note that $$-\gamma$$ corresponds to $$1/\zeta$$. Looking at absolute values we see that each $$f(\gamma_1)=f(\gamma_2)$$ if and only if $$\gamma_1$$ and $$\gamma_2$$ are complex conjugate. Note that $$\gamma=-1/2$$ corresponds to $$\zeta=-1$$. (Recall that $$n$$ is even.) So we see that the finite generators of the monodromy group of the cover is one transposition and $$(n-2)/2$$ double transpositions. The double transpositions induce transpositions on the blocks, so the action on the blocks is $$S_k$$. The transposition fixes all blocks and flips the two elements in one block. This together yields the claimed structure of the group.

• This is probably a stupid question but wouldn't $s$ just be an integer, so that it is not reasonable to pretend it is an indeterminate/transcendental over $\Bbb C$? May 31 at 10:49
• @FShrike Let $f(x)$ be as above and $s$ be a transcendental over $\mathbb C$. By Hilbert's Irreducibility Theorem, $\text{Gal}(f(X)-s_0, \mathbb Q)=\text{Gal}(f(X)-s, \mathbb Q(s))$ for "most" rational $s_0$. Furthermore, $\text{Gal}(f(X)-s, \mathbb C(s))\le\text{Gal}(f(X)-s, \mathbb Q(s))$. May 31 at 11:06
• (I know little, am just a student) so if I understand correctly: you know $\operatorname{Gal}(f(X)-s,\Bbb C(s))\cong C_2^k\rtimes S_k=:K$ for transcendent $s$ and you conclude $\operatorname{Gal}(f(X)-s,\Bbb Q(s))$ contains a copy of $K$. For “most” rational $s_0$ you also know: $G:=\operatorname{Gal}(s_0+X^{n+1}-X^n,\Bbb Q)\cong\operatorname{Gal}(f(X)-s,\Bbb Q(s))$ so you conclude that the group of interest, $G$, contains $K$ most of the time and is consequently not solvable. But for the OP’s specific polynomial that doesn’t prove it because the integer $s_0$ might not be included in ‘most’. May 31 at 11:34
• I hope that’s right May 31 at 11:34
• @FShrike Let $f_{m,r}(x)$ be the polynomial in the question, and $t$ be a transcendental over $\mathbb C$. My argument shows the following: Given any rational $m\ne0$, then the Galois group of $f_{m,t}$ over $\mathbb Q(t)$ is not solvable if $k\ge5$. Now by HIT, this Galois group is the same as the one of $f_{m,r}$ over $\mathbb Q$ for infinitely many integers $r$. May 31 at 15:02

Q: Can the roots of this sum always be fully expressed in radicals?

A: No. I note that the sum can be carried in closed form: $$r+\sum_{q=0}^{2k-1} (-1)^{q} {2k+1 \choose q+1} x^{2k-q} m^{q}=$$ $$\qquad=r-(-1)^{2 k} m^{2 k}+m^{-1}x^{2 k} \left(m \left(1-\frac{m}{x}\right)^{2 k}-x \left(1-\frac{m}{x}\right)^{2 k}+x\right).$$ The roots are radicals for $$k\leq 4$$, but for $$k=5$$, for example, the roots involve the solution of a quintic equation.

Specific example, $$r=1,m=1,k=5$$:

$$1 - 11 x + 55 x^2 - 165 x^3 + 330 x^4 - 462 x^5 + 462 x^6 - 330 x^7 + 165 x^8 - 55 x^9 + 11 x^{10}$$ does not seem to have roots that can be expressed as radicals, it needs the roots of $$1+11x+44x^2+77x^3+55x^4+11x^5$$.

• According to Maple, the Galois group of $1+11x+44x^2+77x^3+55x^4+11x^5$ is cyclic of order 5. May 30 at 15:23