If $s = 4m - 3$, then $\sigma(p^s) = (1 + p^{2m-1})(1 + p + \ldots + p^{2m-2})$. Is there a similar factorization for $\sigma(q^t)$ when $t=4n$?

Question

Let $p,q$ be (odd) primes, and let $m,n,s,t$ be positive integers.

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.

M. A. Nyblom showed that, if $s = 4m - 3$, then we have the simple factorization $$\sigma(p^s) = (1 + p^{2m-1})(1 + p + \ldots + p^{2m-2}).$$

Here is my inquiry:

Is there a similar factorization for $\sigma(q^t)$ when $t = 4n$?

MY ATTEMPT

Since $t = 4n = 2\cdot{2n}$ is even, then $\sigma(q^t)$ is odd. Since $q$ is odd, this means that there are an odd number ($= 4n + 1$) of odd terms in $$\sigma(q^t) = 1 + q + \ldots + q^{4n-1} + q^{4n},$$ which essentially means that we cannot take the terms successively in pairs (i.e. $(1 + q) \nmid \sigma(q^t)$).

Which brings us to the question: Does $(1 + q + q^2)$ divide $\sigma(q^t)$?

Allow me to consider simpler cases first.

• $t = 4$: $\sigma(q^t) = \sigma(q^4) = 1 + q + q^2 + q^3 + q^4 = (1 + q + q^2) + (q^2 + q^3 + q^4) - q^2 = (q^2 + 1)(1 + q + q^2) - q^2$

Note that $\gcd(q^2,q^2 + 1) = \gcd(q^2, 1 + q + q^2) = 1$ (since $q$ is prime). Hence, $$\gcd\bigg(q^2, (q^2 + 1)(1 + q + q^2)\bigg) = 1,$$ which essentially means that $(1 + q + q^2) \nmid \sigma(q^4)$.

• $t = 8$: $\sigma(q^t) = \sigma(q^8) = 1 + q + q^2 + q^3 + q^4 + q^5 + q^6 + q^7 + q^8 = (1 + q + q^2) + {q^3} (1 + q + q^2) + {q^6} (1 + q + q^2)$

Evidently, we then see that $1 + q + q^2$ divides $\sigma(q^8)$.

• $t=12$: $\sigma(q^t) = \sigma(q^{12}) = 1 + q + q^2 + q^3 + q^4 + q^5 + q^6 + q^7 + q^8 + q^9 + q^{10} + q^{11} + q^{12}$ $= (1 + q + q^2) + {q^3} (1 + q + q^2) + {q^6} (1 + q + q^2) + {q^9} (1 + q + q^2) + q^{12}$

Hence, we "conclude" that $(1 + q + q^2) \nmid \sigma(q^{12})$.

• $t=16$: $\sigma(q^t) = \sigma(q^{16}) = 1 + q + q^2 + q^3 + q^4 + q^5 + q^6 + q^7 + q^8 + q^9 + q^{10} + q^{11} + q^{12} + q^{13} + q^{14} + q^{15} + q^{16}$ $= (1 + q + q^2) + {q^3} (1 + q + q^2) + {q^6} (1 + q + q^2) + {q^9} (1 + q + q^2) + q^{12} (1 + q + q^2) + q^{15} (1 + q)$

Hence, we "conclude" that $(1 + q + q^2) \nmid \sigma(q^{16})$.

And then, I will have to consider whether $(1 + q + q^2 + q^3) \mid \sigma(q^t)$, etc.

Alas, this is where I get stuck.

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I was actually hoping somebody with more (in-depth) knowledge of cyclotomic polynomials could explain what is going on here.

Answer 1

Expanding my comment into an answer:

One interpretation of the question is: "does the polynomial $1 + x + \cdots + x^{4n}$ factor in $\mathbb Z[x]$?" (If it does, then plugging in $x = q$ gives a factorization of the integer you're interested in.) The theory of cyclotomic polynomials shows that this polynomial is irreducible if and only if $4n+1$ is prime. Any divisor $a$ of $4n+1$ gives rise to a divisor $1 + x + \cdots + x^{a-1}$ of the polynomial; in particular, the case $a = 3$ accounts for the results of your numerical investigations.

Let me provide some context about cyclotomic polynomials; since this is valid for all positive integers, I'll write $n$ instead of $4n+1$. The cyclotomic polynomials $\Phi_d(x)$ provide the exact factorization of the polynomials $x^n - 1$ (in $\mathbb Z[x]$, or equivalently by Gauss's lemma in $\mathbb Q[x]$) in the following sense: each $\Phi_d$ is monic and irreducible, and we have $\prod_{d|n} \Phi_d(x) = x^n - 1$. The relevance of this to your question is that we have: $$ x^{n-1} + \cdots + x + 1 = \frac{x^n-1}{x-1} = \frac{x^n-1}{\Phi_1(x)} = \prod_{d|n, d \neq 1} \Phi_d(x). $$ Since the irreducible factors correspond to the divisors $d > 1$ of $n$, this proves my earlier claim that the polynomial is irreducible if and only if $n$ is prime. You can also see the divisors $1 + x + \cdots + x^{a-1}$ in this picture: just observe that if $a$ divides $n$, then $$ \left( \prod_{d|a, d \neq 1} \Phi_d(x) \right) \text{ divides } \left( \prod_{d|n, d \neq 1} \Phi_d(x) \right), $$ since the index set on the left is a subset of the one on the right.

Lastly, although the prime-indexed cyclotomic polynomials $\Phi_p(x) = \frac{x^p - 1}{x-1} = x^{p-1} + \cdots + x + 1$ are all irreducible, it is of course not true that the integer $\Phi_p(q) = q^{p-1} + \cdots + q + 1 = \sigma(q^{p-1})$ will always be prime. We can, however, say two things in this direction. First, it will never have any divisors of the form $q^{a-1} + \cdots + q + 1$, since if $p = ak + b$ with $0 < b < a$, then we can explicitly divide $q^{p-1} + \cdots + q + 1$ by $q^{a-1} + \cdots + q + 1$ with remainder, similarly to your calculations: \begin{align*} q^{p-1} + \cdots + q + 1 & = \left( q^{a-1} + \cdots + q + 1 \right) \cdot q^b \left( q^{a(k-1)} + \cdots + q^a + 1 \right) \\ & \quad + \left( q^{b-1} + \cdots + q + 1 \right), \end{align*} where the remainder $q^{b-1} + \cdots + q + 1$ is positive and less than $q^{a-1} + \cdots + q + 1$. Second, the Bateman-Horn conjecture implies that every cyclotomic polynomial $\Phi_n(x)$ produces infinitely many prime values $\Phi_n(a)$ with $a \in \mathbb Z$. Moreover, if $n$ is not a power of $2$, then applying Bateman-Horn to the pair of polynomials $(x, \Phi_n(x))$ implies that $\Phi_n(x)$ even produces infinitely many prime values at prime inputs. (This isn't true when $n$ is a power of $2$ because $\Phi_{2^k}(x) = x^{2^{k-1}} + 1$ is always even when $x$ is odd.) So if you are interested in $\sigma(q^{4n})$ when $4n+1$ happens to be prime, you should expect that these numbers genuinely will sometimes be prime.