Faulhaber polynomial of order $p \in \Bbb{N}$ is defined as the unique polynomial of degree $p+1$ satisfying

$$ S_{p}(n) = \sum_{k=1}^{n} k^p $$

for $n = 1, 2, 3, \cdots$. For example,

\begin{align*} S_0(x) &= x, \\ S_1(x) &= \frac{x(x+1)}{2}, \\ S_2(x) &= \frac{x(x+1)(2x+1)}{6}, \\ S_3(x) &= \frac{x^2 (x+1)^2}{4}.\\ S_{4}(x)&=\dfrac{x(x+1)(2x+1)(3x^2+3x-1)}{30}\\ S_{5}(x)&=\dfrac{x^2(x+1)^2(2x^2+2x-1)}{12}\\ S_{6}(x)&=\dfrac{x(x+1)(2x+1)(3x^4+6x^3-3x+1)}{42}\\ S_{7}(x)&=\dfrac{x^2(x+1)^2(3x^4+6x^3-x^2-4x+2)}{90}\\ \end{align*} if define $n\in R$,so I conjecture $$S_{k}(x)=0$$ have only rational roots $0,-\frac{1}{2},-1$

In addition, further discovery

when $2|k$,

$f_{k}(x)=0$ have only rational root $0,-\frac{1}{2},-1$

when $2\nmid k,\ge 3$,

$f_{k}(x)=0$ have only rational root $0,-1$(Double Root)

This is Old Results? Thanks

Faulhaber polynomial of order $p \in \Bbb{N}$ is defined as the unique polynomial of degree $p+1$ satisfying

$$ S_{p}(n) = \sum_{k=1}^{n} k^p $$

for $n = 1, 2, 3, \cdots$. For example,

\begin{align*} S_0(x) &= x, \\ S_1(x) &= \frac{x(x+1)}{2}, \\ S_2(x) &= \frac{x(x+1)(2x+1)}{6}, \\ S_3(x) &= \frac{x^2 (x+1)^2}{4}.\\ S_{4}(x)&=\dfrac{x(x+1)(2x+1)(3x^2+3x-1)}{30}\\ S_{5}(x)&=\dfrac{x^2(x+1)^2(2x^2+2x-1)}{12}\\ S_{6}(x)&=\dfrac{x(x+1)(2x+1)(3x^4+6x^3-3x+1)}{42}\\ S_{7}(x)&=\dfrac{x^2(x+1)^2(3x^4+6x^3-x^2-4x+2)}{24}\\ \end{align*} if define $n\in R$,so I conjecture $$S_{k}(x)=0$$ have only rational roots $0,-\frac{1}{2},-1$

In addition, further discovery

when $2|k$,

$f_{k}(x)=0$ have only rational root $0,-\frac{1}{2},-1$

when $2\nmid k,\ge 3$,

$f_{k}(x)=0$ have only rational root $0,-1$(Double Root)

This is Old Results? Thanks

Faulhaber polynomial of order $p \in \Bbb{N}$ is defined as the unique polynomial of degree $p+1$ satisfying

$$ S_{p}(n) = \sum_{k=1}^{n} k^p $$

for $n = 1, 2, 3, \cdots$. For example,

\begin{align*} S_0(x) &= x, \\ S_1(x) &= \frac{x(x+1)}{2}, \\ S_2(x) &= \frac{x(x+1)(2x+1)}{6}, \\ S_3(x) &= \frac{x^2 (x+1)^2}{4}.\\ S_{4}(x)&=\dfrac{x(x+1)(2x+1)(3x^2+3x-1)}{30}\\ S_{5}(x)&=\dfrac{x^2(x+1)^2(2x^2+2x-1)}{12}\\ S_{6}(x)&=\dfrac{x(x+1)(2x+1)(3x^4+6x^3-3x+1)}{42}\\ S_{7}(x)&=\dfrac{x^2(x+1)^2(3x^4+6x^3-x^2-4x+2)}{90}\\ \end{align*} if define $n\in R$,so I conjecture $$S_{k}(x)=0$$ have only rational roots $0,-\frac{1}{2},-1$

This is Old Results? Thanks

Faulhaber polynomial of order $p \in \Bbb{N}$ is defined as the unique polynomial of degree $p+1$ satisfying

$$ S_{p}(n) = \sum_{k=1}^{n} k^p $$

for $n = 1, 2, 3, \cdots$. For example,

\begin{align*} S_0(x) &= x, \\ S_1(x) &= \frac{x(x+1)}{2}, \\ S_2(x) &= \frac{x(x+1)(2x+1)}{6}, \\ S_3(x) &= \frac{x^2 (x+1)^2}{4}.\\ S_{4}(x)&=\dfrac{x(x+1)(2x+1)(3x^2+3x-1)}{30}\\ S_{5}(x)&=\dfrac{x^2(x+1)^2(2x^2+2x-1)}{12}\\ S_{6}(x)&=\dfrac{x(x+1)(2x+1)(3x^4+6x^3-3x+1)}{42}\\ S_{7}(x)&=\dfrac{x^2(x+1)^2(3x^4+6x^3-x^2-4x+2)}{90}\\ \end{align*} if define $n\in R$,so I conjecture $$S_{k}(x)=0$$ have only rational roots $0,-\frac{1}{2},-1$

In addition, further discovery

when $2|k$,

$f_{k}(x)=0$ have only rational root $0,-\frac{1}{2},-1$

when $2\nmid k,\ge 3$,

$f_{k}(x)=0$ have only rational root $0,-1$(Double Root)

This is Old Results? Thanks