Is the Galois group of this explicit family of irreducible Pisot polynomials always $S_n$?

Let $p$ be an odd prime and let $n \ge 3$. I have been considering the following family of polynomials.

Set

$$ r_n:=\frac{1}{2\sqrt{n-1}}, $$

and define

$$ B(p,n) := r_n+\frac{p+1}{r_n}+\frac{2p}{r_n^{n-1}}. $$

Let

$$ A(p,n) := 2+2p\left( \left\lfloor \frac{B(p,n)-2}{2p} \right\rfloor+1 \right). $$

Thus $A(p,n)$ is the smallest integer of the form $2+2pk$ which is strictly larger than $B(p,n)$.

Now define

$$ P_{p,n}(x) = x^n-A(p,n)x^{n-1} +(p+1)x^{n-2}+2p. $$

Since $A(p,n)$ is even and $p$ is odd, every nonleading coefficient is divisible by $2$, while $4\nmid 2p$. Hence $P_{p,n}(x)$ is Eisenstein at $2$, so it is irreducible over $\mathbb Q$.

By the choice of $A(p,n)$ and Rouché's theorem, $P_{p,n}$ has exactly $n-1$ roots in

$$ |z|<\frac{1}{2\sqrt{n-1}}, $$

and one remaining root $\alpha>1$. In particular, $\alpha$ is the unique root of largest modulus. Let

$$ f(p,n):=\alpha. $$

Thus $f(p,n)$ is a Pisot number of degree exactly $n$.

There are also some simple facts about the real roots. By Descartes' rule of signs, $P_{p,n}$ has exactly two positive real roots. Moreover,

$$ P_{p,n}(-x) = (-1)^n x^n-(-1)^{n-1}A(p,n)x^{n-1} +(-1)^{n-2}(p+1)x^{n-2}+2p. $$

It follows that:

• if $n$ is even, $P_{p,n}$ has exactly $2$ real roots;
• if $n$ is odd, $P_{p,n}$ has exactly $3$ real roots.

So for $n\ge 4$, the field $\mathbb Q(f(p,n))$ is generally not the splitting field.

Let $L_{p,n}$ denote the splitting field of $P_{p,n}$ over $\mathbb Q$, and set

$$ G_{p,n} := \operatorname{Gal}(L_{p,n}/\mathbb Q). $$

Since $P_{p,n}$ is irreducible of degree $n$,

$$ G_{p,n}\le S_n $$

is a transitive subgroup.

This led me to the following conjecture.

Conjecture

For every odd prime $p$ and every integer $n\ge 5$,

$$ \boxed{ \operatorname{Gal}(P_{p,n}/\mathbb Q)\cong S_n. } $$

If this is true, then since $S_n$ is not solvable for $n\ge 5$, none of the algebraic numbers $f(p,n)$ with $n\ge 5$ would be expressible by radicals over $\mathbb Q$.

There is some additional arithmetic structure which may or may not be useful. Since

$$ A(p,n)\equiv 2\pmod p, $$

we have

$$ P_{p,n}(x) \equiv x^{n-2}(x-1)^2 \pmod p. $$

Thus reduction modulo the prime indexing the family is highly non-generic.

Also, complex conjugation has a very specific cycle structure in $G_{p,n}$. Since the number of real roots is known exactly, complex conjugation has cycle type

$$ 1^2,2^{(n-2)/2} $$

when $n$ is even, and

$$ 1^3,2^{(n-3)/2} $$

when $n$ is odd.

My questions are:

1. Is the conjecture $G_{p,n}\cong S_n$ true for all odd primes $p$ and all $n\ge 5$?
2. If not, can the exceptional pairs $(p,n)$ be characterized?
3. Is there a systematic way to use factorizations of $P_{p,n}$ modulo auxiliary primes $q\ne p$ to produce enough cycle types to force $G_{p,n}=S_n$?
4. Can the discriminant of $P_{p,n}$ be analyzed well enough to rule out $G_{p,n}\subseteq A_n$?
5. Are there known results on Galois groups of sparse polynomials of the form $$ x^n-Ax^{n-1}+Bx^{n-2}+C $$ that would apply directly to this family?

I would also be interested in computational evidence for small $n$ and $p$, especially whether examples such as $P_{3,5},P_{5,5},P_{3,6}$ already have full symmetric Galois group.