Solvability by radicals of an equation of prime degree

Question

For which prime $p$ the equation $x^p-p^px+p=0$ is solvable by radicals?

I don't know how to solve this for primes $p\neq 2$, so any help is welcome.

Answer 1

Suppose $p > 3$.

Clearly $f(x) = x^p-p^px+p$ is irreducible (over $\Bbb{Q}$) by Eisenstein's criterion.

Now $f'(x) =p x^{p-1} - p^{p} = p (x^{p-1} - p^{p-1})$ has exactly two real roots $\pm p$, and since $f(-p) > 0$ and $f(p) < 0$, it follows that $f$ has exactly $3$ real roots, and $p - 3 > 0$ non-real ones, which come in conjugate pairs.

So the Galois group of $f(x)$ contains a $p$-cycle, which may be taken in the form $$g = (1 2 \dots p),$$ and an element of the form $$ h = (a_{1} a_{2}) (a_{3} a_{4}) \dots (a_{p-4}, a_{p-3}). $$

---

Addendum

Now, as noted by @JackSchmidt in his comment below, a result of Galois applies. This states that if $p$ is an odd prime, and $G$ is a transitive subgroup of $S_{p}$, then the following are equivalent:

1. $G$ has a unique $p$-Sylow subgroup, which is then normal in $G$;
2. $G$ is soluble;
3. $G$ is (similar to) a subgroup of the affine group on the line over the field with $p$ elements;
4. $G$ is a Frobenius group, that is, the only element of $G$ fixing two distinct points is the identity.

Here I am quoting from B. Huppert's Endliche Gruppen I, Satz II.3.6.

Of interest to us is the fact that if $G$ is as in (1), or (4), then, as noted by @JackSchmidt, the involution $h$ must normalize $\langle g \rangle$. But involutions doing this are $$ (2, p) (3, p-1) \dots \left(\dfrac{p+1}{2}, \dfrac{p+3}{2}\right), $$ and its conjugates under powers of $g$. These involutions invert $g$, as $g^{-1} = (1, p, p-1, \dots, 3, 2)$.

Thus an involution normalizing $\langle g \rangle$ is a product of $(p-1)/2$ disjoint transpositions, unlike $h$, which is a product of $(p-3)/2$ disjoint transpositions. Therefore (1) is not satisfied, hence (2) is not satisfied, and $G$ is not soluble.