$\color{brown}{\textbf{The setup and observations}}$

For $d\in{\bf N}$ and $n\in{\bf N}^\star$, denote by $S_d(n)$ the sum of the $d$-powers of integers $1$ to $n$: $$S_d(n)=\sum_{k=1}^nk^d.$$

A simple use of the binomial theorem in developping the quantity $(k+1)^{d+2}-k^{d+2}$ and summing it yields the following recursive relation: $$S_{d+1}(n)=\frac{1}{d+1}\left\{(n+1)^{d+2}-1-\sum_{k=0}^d\binom{d+2}{k}S_k(n)\right\}.$$

This allowed me to implement a $\verb|Python|$ script to compute them inductively. I calculated the first 100 polynomials. Here are some examples:

  • $\small S_8(n)=\displaystyle\frac{n(n + 1)(2n + 1)(5n^6 + 15n^5 + 5n^4 - 15n^3 - n^2 + 9n - 3)}{90}$

  • $\small S_9(n)=\displaystyle\frac{n^2(n + 1)^2(n^2 + n - 1)(2n^4 + 4n^3 - n^2 - 3n + 3)}{20}$

  • $\small S_{10}(n)=\displaystyle\frac{n(n + 1)(2n + 1)(n^2 + n - 1)(3n^6 + 9n^5 + 2n^4 - 11n^3 + 3n^2 + 10n - 5)}{66}$

  • $\small S_{11}(n)=\displaystyle\frac{n^2(n + 1)^2(2n^8 + 8n^7 + 4n^6 - 16n^5 - 5n^4 + 26n^3 - 3n^2 - 20n + 10)}{24}$

Amongst all things, there are a few of interest:

  • Every $S_d(n)$ is a rational polynomial in $n$, where OEIS:A064538 appears.

  • The roots of the polynomials seem to follow some interesting pattern, as described on this thread.

  • For odd values of $d$, $S_d$ is divisible by $\left(\frac{n(n+1)}{2}\right)^2$. For even values of $d$, it is divisible by $\frac{n(n+1)(2n+1)}{6}$.

$\color{brown}{\textbf{Getting into polynomials}}$

For $\delta\in{\bf N}$, denote as $F_\delta(X)$ the Faulhaber polynomial: $$F_\delta(X)=\frac{1}{2^{2\delta+2}(\delta+1)}\sum_{q=0}^\delta\binom{2\delta+2}{2q}(1-q)B_{2q}\left[(8X+1)^{\delta+1-q}-1\right],$$ where $(B_n)_{n\in{\bf N}}$ are the Bernoulli numbers. Then, this polynomial has lowest degree term $X^2$, and Faulhaber showed that:

$$S_{2d+1}(n)=F_d\left(\frac{n(n+1)}{2}\right)\qquad\text{and}\qquad S_{2d}(n)=\frac{n+1/2}{2d+1}F_d'\left(\frac{n(n+1)}{2}\right).$$

The question now comes when one looks at the irreducible decomposition of $S_d$ as a polynomial over ${\bf Q}$. Having the previous properties in mind, set: $$T_d(n)=\begin{cases}\frac{4}{n^2(n+1)^2}S_d(n)&\text{if $d$ is odd,}\\\frac{6}{n(n+1)(2n+1)}S_d(n)&\text{otherwise.}\end{cases}$$

Then my script returned that $T_d$ was always irreducible, except for the values $d=9$ and $d=10$. Just so that we are on the same page, here are the corresponding examples from before:

  • $\small T_8(X)=\displaystyle\frac{5X^6 + 15X^5 + 5X^4 - 15X^3 - X^2 + 9X - 3}{15}$

  • $\small T_9(X)=\displaystyle\frac{(X^2 + X - 1)(2X^4 + 4X^3 - X^2 - 3X + 3)}{5}$

  • $\small T_{10}(X)=\displaystyle\frac{(X^2 + X - 1)(3X^6 + 9X^5 + 2X^4 - 11X^3 + 3X^2 + 10X - 5)}{11}$

  • $\small T_{11}(X)=\displaystyle\frac{2X^8 + 8X^7 + 4X^6 - 16X^5 - 5X^4 + 26X^3 - 3X^2 - 20X + 10}{6}$

$\color{brown}{\textbf{The questions}}$

Are there known results regarding this? One can ask the following questions:

  • Is it true that $T_d$ is always irreducible over ${\bf Q}$ for $d\neq 9,10$?

  • Is there an explanation behind those exceptional values $d=9,10$?

  • Or if the first question turns out to be false, what are those values of $d$ for which $T_d$ is reducible? Do they follow some pattern? (e.g. would it happen that an odd value would be followed by an even value, both sharing the property that $T_d$ is reducible?)

$\color{brown}{\textbf{My (failing) ideas}}$

What I thought about doing first is proving that if $P(X)$ is irreducible over ${\bf Q}$, then $P\left(\frac{X(X+1)}{2}\right)$ is irreducible as well. But it turns out it isn't true, for if simply $P(X)=X$, then the statement doesn't hold. This means that studying the Faulhaber polynomial just won't work, and one really has to work with the $T_d$.

Now, working for instance on the odd case scenarios, we seek to prove irreducibility of $$\small T_{2d+1}(X)=\frac{1}{2^{2d}(d+1)X^2(X+1)^2}\sum_{q=0}^d\binom{2d+2}{2q}(1-q)B_{2q}\left[(4X(X+1)+1)^{d+1-q}-1\right]$$ for all $d\neq4$.

I'm lost. I have no clue of where to even begin, the definition seems to rely on the values of the numerators and denominators of the Benoulli numbers, which I know are some very tough problems from number theory. Thus, applying either the Eisenstein criterion, or reducing modulo some prime number, seems like a dead end.

I have made a related question involving an expression I have derived for the odd case: $$\small T_{2d+1}(X)=\frac{1}{2(d+1)}\sum_{\ell=0}^{2d-2}\left\{(-1)^\ell\sum_{k=\ell+2}^{2d}\binom{k}{\ell+2}\binom{2d+2}{k+2}B_{2d-k}\right\}X^\ell.$$

Now, I have no clue whether this is any useful into proving that it is irreducible. Indeed, assuming the inner sum cannot be simplified further (which was the point of the related question), proving irreducibility of $T_{2d+1}$ therefore highly relies on the numerators and denominators of the Bernoulli numbers, which are known to be hard questions...



You must log in to answer this question.