Prove the product of a polynomial function of the roots of another polynomial is an integer. I noticed this while solving another problem on this site.
Let $P(x)$ be a polynomial in $x$ with integer coefficients, and let the roots of $P(x)=0$ be $r_1, r_2 \ldots ,r_n$, where multiple $r_i$ might be equal if there are roots with multiplicity higher than one.  Let Q(x) be some other polynomial in $x$, also with integer coefficients.
Prove that 
$$
\prod_i Q(r_i) \in \Bbb{Z}
$$ 
For example, if $P(x) = x^5+2x^2+1$ and $Q(x) = x^2-2$ then $\prod Q(r_i) = -7$.
I am pretty sure it is true, because you can express each term in the product of those polynomials in a form like
$$
\sum_{i<j<\ldots <n} r_i^{p_1} r_j^{p_2} \ldots
$$
and laboriously express those sums as sums of  products of combinations of the roots that match expressions determined by the (integer) coefficients of $P(x)$.  But making that constructive proof anything more than hand-waving seems  difficult.
I wonder if any ideals in the theory of rings, for instance, can make this proposition easier to prove.
NOTE Afterward 
A counterexample would also nicely resolve the question, showing that the conjecture is false.

I agree with Bill that symmetric polynomials should somehow be the standard solution, but I wanted to point out that Galois theory makes this straightforward.
It is easy to show that $s = \prod_i Q(r_i)$ is an algebraic integer, so it is enough to show that $s\in\mathbb{Q}$.  By the Galois correspondence, this is the same as checking that $\sigma(s)=s$ for every automorphism $\sigma$ of a splitting field $\mathbb{Q}\subset K$ of $P(X)$.
But any such automorphism just permutes the roots of $P$, and therefore permutes the terms of the product $\prod_i Q(r_i)$.
