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Let $f(n,t) = \sum_{k=0}^{r-1} d_k t^k$ where $D_n = \{d_0=1,d_1,\cdots,d_{r-1}\}$ are all divisors of $n$.

For instance

$$f(28,t) = 28 t^{5} + 14 t^{4} + 7 t^{3} + 4 t^{2} + 2 t + 1$$

For even perfect numbers $n = 2^{p-1}(2^p-1)$ this polynomial always factors as:

$$f(n,t) = ((2^p-1)t^p+1)(1+2t +2^2t^2+\cdots+2^{p-1}t^{p-1})$$

Furthermore the two irreducible (?) factors seem to have interesting Galois groups:

6 (2*t + 1) * (3*t^2 + 1)
6 (2*t + 1, 1) Galois group PARI group [1, 1, 1, "S1"] of degree 1 of the Number Field in t with defining polynomial 2*t + 1
6 (3*t^2 + 1, 1) Galois group PARI group [2, -1, 1, "S2"] of degree 2 of the Number Field in t with defining polynomial 3*t^2 + 1
28 (4*t^2 + 2*t + 1) * (7*t^3 + 1)
28 (4*t^2 + 2*t + 1, 1) Galois group PARI group [2, -1, 1, "S2"] of degree 2 of the Number Field in t with defining polynomial 4*t^2 + 2*t + 1
28 (7*t^3 + 1, 1) Galois group PARI group [6, -1, 2, "S3"] of degree 3 of the Number Field in t with defining polynomial 7*t^3 + 1
496 (16*t^4 + 8*t^3 + 4*t^2 + 2*t + 1) * (31*t^5 + 1)
496 (16*t^4 + 8*t^3 + 4*t^2 + 2*t + 1, 1) Galois group PARI group [4, -1, 1, "C(4) = 4"] of degree 4 of the Number Field in t with defining polynomial 16*t^4 + 8*t^3 + 4*t^2 + 2*t + 1
496 (31*t^5 + 1, 1) Galois group PARI group [20, -1, 3, "F(5) = 5:4"] of degree 5 of the Number Field in t with defining polynomial 31*t^5 + 1
8128 (64*t^6 + 32*t^5 + 16*t^4 + 8*t^3 + 4*t^2 + 2*t + 1) * (127*t^7 + 1)
8128 (64*t^6 + 32*t^5 + 16*t^4 + 8*t^3 + 4*t^2 + 2*t + 1, 1) Galois group PARI group [6, -1, 1, "C(6) = 6 = 3[x]2"] of degree 6 of the Number Field in t with defining polynomial 64*t^6 + 32*t^5 + 16*t^4 + 8*t^3 + 4*t^2 + 2*t + 1
8128 (127*t^7 + 1, 1) Galois group PARI group [42, -1, 4, "F_42(7) = 7:6"] of degree 7 of the Number Field in t with defining polynomial 127*t^7 + 1
  1. What are the Galois groups of these irreducible (?) factors?
  2. Whats is the Galois group of $f(n,t)$ for $n$ an even perfect numbers?
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The second factor is $P(2t)$ where $P=X^{p-1}+\cdots+X+1$, the $p$-th cyclotomic polynomial. Hence the Galois group of $P(2t)$ is the same as the Galois group of $P(t)$, which is simply $(\mathbb{Z}/p\mathbb{Z})^\times$, which is cyclic of order $p-1$.

The other factor has the form $t^p -a$. The splitting field is $\mathbb{Q}(\zeta_p, a^{1/p})$, and it is known that the Galois group is $(\mathbb{Z}/p\mathbb{Z})\rtimes (\mathbb{Z}/p\mathbb{Z})^\times$ (with the obvious action).

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