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 http://www.oddperfectnumber.org/docs/Kronecker%20proof%20of%20irreducibility.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.

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.

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.

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 http://www.oddperfectnumber.org/docs/Kronecker%20proof%20of%20irreducibility.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)(mod e + 1) = a^e + a^{e-1} + ... + a^2 +a + 1.

For a = 1, then Y(mod 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(mod e + 1), then p is composite and not a prime. For a = 2, 3, ..., e, then the following holds:

Y(mod 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 = f1^{e1} f2^{e2} ... fm^{em}. Then

Y = a x sigma(a^{f1-1}) x sigma(a^{f2-1}) x ... x sigma(a^{fm-1}) x sigma((-a)^{f1-1}) x sigma((-a)^{f2-1}) x ... x sigma((-a)^{fm-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(mod 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.

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 http://www.oddperfectnumber.org/docs/Kronecker%20proof%20of%20irreducibility.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.