This is also a very partial answer.but 0,−1/2,−1 are the only rational roots of the Faulhaber polynomial.
A high school-level proof is possible without using Bernoulli numbers. When a and b are coprime integers (b > 0) satisfying S_m(a/b) = 0, we first show the following two points:
・When m is odd, a/b = -1 or 0
・When m is even, if a/b ≠ -1 or 0, then b is even
Finally, I provided a complete proof for the case where m is even. Hereafter, we assume $m \geq 1$.
Theorem 1 (1.1).
The identity $S_m(-x) = (-1)^{m+1} S_m(x-1) \quad \cdots(1.1)$ holds.
Proof.
It suffices to show that (1.1) has infinitely many solutions.
Since $S_m(0) = (-1)^{m+1} S_m(-1) = 0$, (1.1) holds when $x = 0$.
Assume (1.1) holds when $x = k$. Then
$$\begin{align*}
S_m(-k-1) &= S_m(-k) - (-k)^m \\
&= (-1)^{m+1} S_m(k-1) + (-1)^{m+1} k^m \\
&= (-1)^{m+1} S_m(k)
\end{align*}$$
Therefore, (1.1) also holds when $x = k+1$. □
Let $k$, $l$, $N$ be integers, and let $k$ be a multiple of $N$ (where $N > 0$).
Theorem 2 (1.2).
The congruence $S_m(k+l) \equiv S_m(k) + S_m(l) \pmod{N} \quad \cdots(1.2)$ holds.
Proof.
The case $l = 0$ is trivial. Let $l \geq 1$.
$$\begin{align*}
S_m(k+l) - S_m(k) &= (k+1)^m + (k+2)^m + \cdots + (k+l)^m \\
&\equiv 1^m + 2^m + \cdots + l^m \pmod{N} \\
&\equiv S_m(l) \pmod{N}
\end{align*}$$
Also, from (1.1),
$$\begin{align*}
S_m(k-l) - S_m(k) &= -(k-l+1)^m - (k-l+2)^m - \cdots - (k-l+l)^m \\
&\equiv (-1)^{m+1} \{(l-1)^m + (l-2)^m + \cdots + 0^m\} \pmod{N} \\
&\equiv (-1)^{m+1} S_m(l-1) \pmod{N} \\
&\equiv S_m(-l) \pmod{N}
\end{align*}$$
□
Theorem 3 (1.3).
The congruence $S_m(kl) \equiv l \cdot S_m(k) \pmod{N} \quad \cdots(1.3)$ holds.
Proof.
First, we show the case when $k = N$.
The case $l = 0$ is trivial. When $l \geq 1$, from (1.2),
$$\begin{align*}
S_m(Nl) &= S_m(0) + \sum_{j=0}^{l-1} \{S_m(N(j+1)) - S_m(Nj)\} \\
&\equiv \sum_{j=0}^{l-1} S_m(N) \pmod{N} \\
&\equiv l \cdot S_m(N) \pmod{N}
\end{align*}$$
From this and (1.1),
$$\begin{align*}
S_m(N(-l)) - (-l) \cdot S_m(N) &= (-1)^{m+1} S_m(Nl-1) + l \cdot S_m(N) \\
&\equiv (-1)^{m+1} S_m(Nl) + l \cdot S_m(N) \pmod{N} \\
&\equiv \{(-1)^{m+1} + 1\} \cdot l \cdot S_m(N) \pmod{N}
\end{align*}$$
From this, when $m$ is even, (1.3) holds.
Let $m$ be odd. Then, if $N$ is odd,
$$\begin{align*}
S_m(N) &= 0^m + 1^m + 2^m + \cdots + \left(\frac{N-1}{2}\right)^m + \left(N - \frac{N-1}{2}\right)^m + \cdots + (N-1)^m + (N-0)^m \\
&\equiv 0^m + 1^m + 2^m + \cdots + \left(\frac{N-1}{2}\right)^m - \left(\frac{N-1}{2}\right)^m - \cdots - 1^m - 0^m \pmod{N} \\
&\equiv 0 \pmod{N}
\end{align*}$$
Therefore, $S_m(N) \equiv 0 \pmod{N} \quad \cdots(1.3.1)$, so when $N$ is odd, (1.3) holds.
When $N$ is even, the center term remains and $S_m(N) \equiv (N/2)^m \pmod{N}$, so
$$\begin{align*}
S_m(N(-l)) - (-l) \cdot S_m(N) &= \{(-1)^{m+1} + 1\} \cdot l \cdot S_m(N) \\
&\equiv 2l \cdot (N/2)^m \pmod{N} \\
&\equiv l \cdot N \cdot (N/2)^{m-1} \pmod{N} \\
&\equiv 0 \pmod{N}
\end{align*}$$
Therefore, (1.3) also holds when $N$ is even.
For the general case of $k$, $k$ can be expressed as $k = Nk'$ for some integer $k'$, so
$$S_m(kl) \equiv S_m((Nk')l) \equiv k' \cdot l \cdot S_m(N) \equiv l \cdot S_m(Nk') \equiv l \cdot S_m(k) \pmod{N}$$
Therefore, (1.3) holds. □
Hereafter, (1.1), (1.2), (1.3) are used without notice.
Let integers $a$, $b$, $c$ ($b > 0$) satisfy $S_m(a/b) = 0$, and let $a$ and $b$ be coprime.
Also, let integer $N$ be a divisor of $a - bc$.
Lemma 1 (2.1).
There exist integers $s$, $t$ such that $\displaystyle\frac{a}{b} = \frac{Ns - c}{N^2t - 1} \quad \cdots(2.1)$.
Proof.
Since $a$ and $b$ are coprime, $a - bc$ and $b$ are coprime. Therefore, $aN$ and $b$ are coprime.
Furthermore, since $(a - bc)/N$ is an integer, there exist integers $s$, $t$ satisfying the Diophantine equation $aNt - bs = (a - bc)/N \quad \cdots(2.1.1)$.
Transforming (2.1.1) yields (2.1). □
Lemma 2 (2.2).
The following identity holds (where $y \neq 0$):
$$y^{2m+1} S_m(x/y) = y^m \{S_m(x+y) - S_m(y)\} - \sum_{j=0}^{m-1} \binom{m}{j} y^{m-1-j} S_{m-j}(y) \{y^{2j+1} S_j(x/y)\} \quad \cdots(2.2)$$
Proof.
When $n$, $y > 0$ are integers,
$$\begin{align*}
S_m((n+1)y) - S_m(y) &= \sum_{l=1}^{n} \{S_m((l+1)y) - S_m(ly)\} \\
&= \sum_{l=1}^{n} \sum_{k=1}^{y} (ly + k)^m \\
&= \sum_{l=1}^{n} \sum_{k=1}^{y} \sum_{j=0}^{m} \binom{m}{j} l^j y^j k^{m-j} \\
&= \sum_{l=1}^{n} \sum_{j=0}^{m} \binom{m}{j} l^j y^j S_{m-j}(y) \\
&= \sum_{j=0}^{m} \binom{m}{j} y^j S_{m-j}(y) S_j(n)
\end{align*}$$
Since there are infinitely many $n$, $y$ satisfying this equality, it is an identity.
Therefore, when $y \neq 0$, we can substitute $n = x/y$:
$$\begin{align*}
S_m(x+y) - S_m(y) &= \sum_{j=0}^{m} \binom{m}{j} y^j S_{m-j}(y) S_j(x/y) \\
&= y^{m+1} S_m(x/y) + \sum_{j=0}^{m-1} \binom{m}{j} y^j S_{m-j}(y) S_j(x/y) \\
&= y^{m+1} S_m(x/y) + \sum_{j=0}^{m-1} \binom{m}{j} y^{-1-j} S_{m-j}(y) \{y^{2j+1} S_j(x/y)\}
\end{align*}$$
Multiplying both sides by $y^m$ yields (2.2). □
Theorem 4 (2.3).
The congruence $S_m(a - bc) \equiv -b \cdot S_m(c) \pmod{N} \quad \cdots(2.3)$ holds.
Proof.
For the integers $s$, $t$ in (2.1), let $x = Ns - c$, $y = N^2t - 1$.
Then, since $x/y = a/b$, $y^{2 \cdot 0 + 1} S_0(x/y) = x = Ns - c$ is an integer.
Also, from (2.2), when $y^{2j+1} S_j(x/y)$ is an integer for $j = 0, 1, \ldots, m-1$, $y^{2m+1} S_m(x/y)$ is also an integer.
Therefore, both the left side and the terms inside $\{\,\}$ on the right side of (2.2) can be integers.
Furthermore, for $S_m(y)$, $S_{m-j}(y)$ on the right side, letting $j' = 0, 1, \ldots, m$,
$$S_{j'}(y) = S_{j'}(N^2t - 1) \equiv Nt \cdot S_{j'}(N) + S_{j'}(-1) \equiv 0 \pmod{N}$$
Therefore, taking mod $N$ on both sides of (2.2),
$$\begin{align*}
(N^2t - 1)^{2m+1} S_m(a/b) &\equiv (N^2t - 1)^m S_m(Ns - c + N^2t - 1) \pmod{N} \\
&\equiv (-1)^m S_m(N^2t + Ns - c - 1) \pmod{N} \\
&\equiv -S_m(-N^2t - Ns + c) \pmod{N}
\end{align*}$$
Since $S_m(a/b) = 0$, we have $S_m(-N^2t - Ns + c) \equiv 0 \pmod{N}$. Also,
$$\begin{align*}
S_m(-N^2t - Ns + c) &\equiv -Nt \cdot S_m(N) - s \cdot S_m(N) + S_m(c) \pmod{N} \\
&\equiv -s \cdot S_m(N) + S_m(c) \pmod{N}
\end{align*}$$
Therefore, $-bs \cdot S_m(N) \equiv -b \cdot S_m(c) \pmod{N}$. Furthermore,
$$\begin{align*}
S_m(a - bc) &= S_m((aNt - bs)N) \\
&\equiv aNt \cdot S_m(N) - bs \cdot S_m(N) \pmod{N} \\
&\equiv -bs \cdot S_m(N) \pmod{N}
\end{align*}$$
From the above, $S_m(a - bc) \equiv -b \cdot S_m(c) \pmod{N}$. □
Theorem 5 (2.4).
When $m$ is odd, $S_m(x)$ has no linear factors with rational coefficients other than $x$ and $x+1$.
Proof.
When $m$ is odd, it suffices to show that $a/b = 0, -1$.
Substituting $c = 1, -2$ into (2.3),
• When $c = 1$: $S_m(a - b) \equiv -b \pmod{N}$
• When $c = -2$: $S_m(a + 2b) \equiv -b \pmod{N}$
Since $-b$ and the divisor $N$ of $a - bc$ are coprime, letting $d \in \{a - b, a + 2b\}$, $d$ and $S_m(d)$ are coprime.
Here, when $d$ can be expressed as $d = d' p^2$ for some integer $d'$ and prime $p$,
$$S_m(d) = S_m(d' p^2) \equiv d' p \cdot S_m(p) \equiv 0 \pmod{p}$$
so $d$ and $S_m(d)$ are not coprime.
Also, when $d$ can be expressed as $d = d' p$ for some integer $d'$ and odd prime $p$, from (1.3.1),
$$S_m(d) = S_m(d' p) \equiv d' \cdot S_m(p) \equiv 0 \pmod{p}$$
so $d$ and $S_m(d)$ are not coprime.
Therefore, $d \in \{\pm 1, \pm 2\}$, and the $a$, $b$ satisfying this are only $(a, b) = (-1, 1), (0, 1)$.
In this case, $a/b = 0, -1$, which indeed satisfy $S_m(a/b) = 0$. □
Theorem 6 (2.5).
When $m$ is even, if $a/b \neq 0, -1$, then $b$ is even.
Proof.
Substituting $c = 0, -1$ into (2.3),
• When $c = 0$: $S_m(a) \equiv 0 \pmod{N}$
• When $c = -1$: $S_m(a + b) \equiv 0 \pmod{N}$
Assume $b$ is odd. Then one of $a$, $a+b$ is even.
Let the even one be $d = 2^j d'$ ($j \geq 1$, $d'$ is odd). Then, since $a/b \neq 0, -1$, we have $d, d' \neq 0$.
Since $S_m(d)$ is a multiple of $N$, $S_m(2^j d') \equiv 0 \pmod{2^j}$.
Also, since $S_m(2^j d') \equiv d' S_m(2^j) \pmod{2^j}$, we have $S_m(2^j) \equiv 0 \pmod{2^j}$.
Here, $S_m(2^1) \equiv 1^m + 2^m \equiv 1 \pmod{2^1}$. Also, when $j \geq 2$, assume $S_m(2^{j-1}) \equiv 2^{j-2} \pmod{2^{j-1}}$.
Since $m$ is even, for integer $k$, $(k + 2^{j-1})^m \equiv k^m + m \cdot 2^{j-1} k^{m-1} \equiv k^m \pmod{2^j}$.
Therefore,
$$\begin{align*}
S_m(2^j) &= \sum_{k=1}^{2^{j-1}} k^m + \sum_{k=1}^{2^{j-1}} (k + 2^{j-1})^m \\
&\equiv \sum_{k=1}^{2^{j-1}} k^m + \sum_{k=1}^{2^{j-1}} k^m \\
&\equiv 2 S_m(2^{j-1}) \equiv 2^{j-1} \pmod{2^j}
\end{align*}$$
Therefore, $S_m(2^j) \not\equiv 0 \pmod{2^j}$, which is a contradiction. □
Lemma 3 (3.1).
For integer $l \geq 2$, there exist integers $a_1, a_2, \ldots, a_{l-1}$ such that
$$S_{2l}(x) = \frac{1}{2l+1} \left\{ \frac{x^l(x+1)^l(2x+1)}{2} - \sum_{j=1}^{l-1} a_j S_{2j}(x) \right\} \quad \cdots(3.1)$$
Proof.
$f(x) = x^l(x+1)^l(2x+1)$ can be expressed as follows for some integers $a'_1, a'_2, \ldots, a'_{2l-1}$:
$$f(x) = 2x^{2l+1} + (2l+1)x^{2l} + \sum_{j=1}^{2l-1} a'_j x^j$$
Also, from $f(-x) = (-x)^l(-x+1)^l(-2x+1) = (-1)^{2l+1} x^l(x-1)^l(2x-1) = -f(x-1)$,
$$\frac{f(x) - f(x-1)}{2} = \frac{f(x) + f(-x)}{2} = (2l+1)x^{2l} + \sum_{j=1}^{l-1} a'_{2j} x^{2j}$$
Taking the sum of both sides for $x = 1, 2, \ldots, n$,
$$\frac{f(n)}{2} = (2l+1) S_{2l}(n) + \sum_{j=1}^{l-1} a'_{2j} S_{2j}(n)$$
Replacing $n$ with $x$ and setting $a_j = a'_{2j}$ and rearranging,
$$S_{2l}(x) = \frac{1}{2l+1} \left\{ \frac{x^l(x+1)^l(2x+1)}{2} - \sum_{j=1}^{l-1} a_j S_{2j}(x) \right\}$$
□
Here, for $l \geq 1$, let integer $A_l > 0$ and integer coefficient polynomial $T_l(x) = \sum_{j=0}^{2l-2} C_{l,j} x^j$ (where $C_{l,j}$ are integers) satisfy
$$S_{2l}(x) = \frac{x(x+1)(2x+1) T_l(x)}{A_l}$$
Here, $A_l$ is coprime to the greatest common divisor of each $C_{l,j}$.
Lemma 4 (3.3).
For $l \geq 1$, $C_{l, 2l-2}$ is odd.
Proof.
When a positive integer $N$ is divisible by 2 at most $k$ times, we denote $V_2(N) = k$.
When $l = 1$, $C_{1,0} = 1$ is odd. Also, $A_1 = 6$, so $V_2(A_1) = 1$.
When $l \geq 2$, assume $V_2(A_1), V_2(A_2), \ldots, V_2(A_{l-1}) \leq 1$. From (3.1),
$$S_{2l}(x) = \frac{x(x+1)(2x+1)}{2l+1} \left\{ \frac{x^{l-1}(x+1)^{l-1}}{2} - \sum_{j=1}^{l-1} \frac{a_j T_j(x)}{A_j} \right\} \quad \cdots(3.1.1)$$
From (3.1.1), $A_l$ is a divisor of $(2l+1) \mathrm{lcm}(2, A_1, A_2, \ldots, A_{l-1})$, so
$$V_2(A_l) \leq V_2(2l+1) + V_2(\mathrm{lcm}(2, A_1, A_2, \ldots, A_{l-1})) = 0 + 1 = 1$$
Therefore, $V_2(A_l) \leq 1$.
Also, from (3.1.1), the coefficient of the highest degree term of $S_{2l}(x)$ is $1/(2l+1)$, so
$$\frac{1}{2l+1} = \frac{2C_{l,2l-2}}{A_l} \Longleftrightarrow A_l = 2(2l+1)C_{l,2l-2}$$
From this, $1 \leq V_2(A_l)$, so
$$1 \leq V_2(A_l) = V_2(2(2l+1)C_{l,2l-2}) \leq 1 + 0 + V_2(C_{l,2l-2}) \leq 1 \Longrightarrow V_2(C_{l,2l-2}) = 0$$
Therefore, $C_{l,2l-2}$ is odd. □
Theorem 7 (3.4).
For $l \geq 1$, $S_{2l}(x)$ has no linear factors with rational coefficients other than $x$, $x+1$, and $2x+1$.
Proof.
We show that $a/b = 0, -1, -1/2$.
If $a/b \neq 0, -1, -1/2$, then $a/b$ is a root of $T_l(x)$.
In this case, from Theorem 6 (2.5), $b$ is even, so the coefficient of the highest degree term of $T_l(x)$, $C_{l,2l-2}$, must be even.
However, this contradicts Lemma 4 (3.3). □
