[Faulhaber polynomial](http://en.wikipedia.org/wiki/Faulhaber%27s_formula) 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