I found the proof by Kronecker that the expression 


$$X = p^e + p^{e-1} + ... + p^2 +p + 1$$ 


is irreducible. Four people translated the German, two in Detroit and two in Kiel, Germany. See [Kronecker's proof of irreducibility](http://oddperfectnumber.org/docs/KroneckerProofOfIrreducibility.pdf).


Kronecker also proved that the numeric factors of $X$ have these forms: 


$(e + 1) | | X,$ and $(k(e + 1) + 1) | X.$


I need the proof of the form of the factors of $\sigma(p^e)$ for a paper on a new result in the odd perfect number problem. I would like the proof in these ways:


1. An exact citation to "Leopold Kronecker's Werke" available on Google Books. 
The article name, the page number, the equation number, etc. to allow me to translate a page or two to understand his proof.
2. A proof by a reader who would like to share it with us. There are usually multiple ways to prove any theorem. Please share them all.
3. A citation of an article, preferably in English, showing the proof.


My knowledge of German is to count to 10, my French is a little better, and my Latin went out with Vatican II. However, I gained some insight into the proof by studying it.


Let $X = \sigma(p^e) = p^e + p^{e-1} + ... + p^2 +p + 1$, and $p = (j(e + 1) + a)$, where $p$ and $(e + 1)$ are prime. Then 


$$Y = \sigma(p^e)\bmod (e + 1) = a^e + a^{e-1} + ... + a^2 +a + 1.$$


For $a = 1$, then $Y \bmod (e + 1) = e + 1$. Removing the factor $(e + 1)$ leaves $1$, not $0$. Therefore, when $p = j(e + 1) + 1$, then $(e + 1) || X$.


For $a = 0 \bmod (e + 1)$, then $p$ is composite and not a prime.
For $a = 2, 3, ..., e$, then the following holds:


$$Y \bmod (e + 1) = a (a^{e-1} + a^{e-2} + ... + a^2 +a + 1) + 1$$


The complicated term in the parentheses is reducible into several algebraic terms.
Suppose $a = f\_1^{e1} f\_2^{e2} ... f\_m^{em}$. Then 


$$Y = a \times \sigma(a^{f\_1-1}) \times \sigma(a^{f\_2-1}) \times
 ... \times \sigma(a^{f\_m-1}) \times
 \sigma((-a)^{f\_1-1}) \times \sigma((-a)^{f\_2-1}) \times ... \times \sigma((-a)^{f\_m-1}) + 1$$


I think, maybe, perhaps. For every value of $a$, one of the terms is zero. Thus, for $p$ and $a$ as above, $X \bmod (e + 1) = 1$. One thing for sure, Kronecker proved it better.


Can someone please help me find a proof and the citation? The paper will be scrutinized by the best mathematicians and harshest critics, you.



Proof that the factors of sigma(p^e) have two forms
It's well-known that if $a$ is an integer then a prime factor of the number
$\Phi\_n(a)$ is either a factor of $n$ or congruent to $1$ modulo $n$.
Here $\Phi\_n$ is the $n$-th cyclotomic polynomial. The reason is that if
$p$ divdes $\Phi\_n(a)$ but not $n$ then $a$ has order exactly $n$
in the multiplicative group $(\mathbb Z/p\mathbb Z)^\*$. By Lagrange's
theorem then $n\mid(p-1)$.


When $e+1$ is prime, then $\Phi\_{e+1}(X)=X^e+X^{e-1}+\cdots+X+1$,
so in this case a prime factor of any $\Phi\_{e+1}(a)$ is either $e+1$
or congruent to $1$ modulo $e+1$. In general though $X^e+X^{e-1}+\cdots+X+1$
is the product of the $\Phi\_m(X)$ over the factors $m>1$ of $e+1$,
so a prime divisor of $a^e+a^{e-1}+\cdots+a+1$ is either a divisor of $e+1$
of congruent to $1$ modulo some prime divisor of $e+1$.


