How to show that this polynomial is a rational polynomial? In a algebra class, the following polynomial was given: $$f(x) = x^{rs}+11x^{rs-1}+x^{rs-2}+2016x^{rs-3}+rx+s,$$ where $r$ and $s$ are distinct primes (there were several problems stated, but the one I'm trying to solve is for $r=3$, $s=17$). Furthermore, for every $z \in \mathbb{Z}$ the polynomial $g_z(x) = \prod_{i=1}^{rs}(x-\lambda_i^z)$ is defined. Here the $\lambda_i$ are the zeros of $f(x)$ and these zeros do not have to be distinct. It is asked to show that for every $z$, the polynomial $g_z(x) \in \mathbb{Q}[x]$. I've tried to prove this by induction on $z$, as for $z=0$ and $z=1$ the statement is true, but I'm having a hard time continuing the prove of this statement. Is there a better way to solve this?
 A: Suppose first that $z\geq0$. Clearly, when expanding $g_z=\prod_{i=1}^{rs}(x-\lambda_i^z)$, the coefficients of every $x^k$ will be symmetric rational (even integer) polynomials in the $\lambda_j$'s. By the Fundamental Theorem of Symmetric Polynomials, they are rational (even integer) polynomials in $e_1,\ldots,e_n$, but these are precisely (up to sign) the coefficients of $f$ (see e.g. here). So the coefficients of $g_z$ are rational (even integer).
If $z<0$ we can apply the same on $g_z\cdot\prod_{i=1}^{rs}\lambda_j^{|z|}$: its coefficients are again integer polynomials in the $\lambda_j$'s, so $g_z\cdot\prod_{i=1}^{rs}\lambda_j^{|z|}\in\Bbb Z[x]$ as before, and because $\prod_{i=1}^{rs}\lambda_j=\pm\,s$ we get $g_z\in\frac1{s^{|z|}}\Bbb Z[x]\subset\Bbb Q[x]$.
Note that this holds for any polynomial: if $R$ is an integral domain, $f\in R[x]$ and $g_z$ as in the question, then $g_z\in R[x]$ for $z>0$ and $g_z\in\frac1{g(0)^{|z|}}R[x]$ for $z<0$ (w.l.o.g. we may assume that $g(0)\neq0$).
