Given a polynomial $P(x)$, find another polynomial $Q(x)$ such that the powers of $x$ in $P(x)Q(x)$ are multiples of an integer $n$ Given a real polynomial $P(x)$, find another real polynomial $Q(x)$ such that the powers of $x$ in $P(x)Q(x)$ are multiples of an integer $n$. Is it always possible to find such a polynomial?
For e.g if $P(x) = 1+3x+2x^2$ I can find $Q(x)=(2 x - 1) (x - 1) (4 x^2 + 1) (x^2 + 1)$ such that $P(x)Q(x) =   16 x^8 - 17 x^4 + 1$  has powers which are multiples of 4. 
I'm bumping the question because I'm not satisfied of the original answer.
 A: Here are three lemmas that underlie one approach to the problem: the first underlies the main idea, and the other two in dealing with the complication of real coefficients instead of complex ones.
Let $\zeta$ be a primitive $n$-th root of unity, and let Let $\bar{R}(x)$ be the polynomial whose coefficients are the complex conjugates of the coefficients of $R(x)$.
Lemma: $R(x) = R(\zeta x)$ if and only if the only monomials in $R(x)$ are of the form $x^{an}$.
Lemma: $R(x)$ is a real polynomial if and only if $R(x) = \bar{R}(x)$
Lemma: If $R(x) = S(\alpha x)$, then $\bar{R}(x) = \bar{S}(\bar{\alpha} x)$.
I think you've already spotted the trick to the problem when $n=2$: take $Q(x) = P(-x)$, so that the product is $P(x) P(-x)$....
A: If $R(x)= x^{n} - a^{n} $ then by factorisation we can see that  $ x^{n} - a^{n} = (x - a)(x^{n - 1} + ax^{n - 2}+ a^2x^{n - 3} + \dots + a^{n-1})$ or by simply observing that $R(a)=0$ implies $x-a$ is a factor of $x^{n} - a^{n}$. 
In the same way we can deduce that $x+a$ is a factor of $x^n -a^n$ when $n$ is even  (if $n =2p$ then  $x^n -a^n = x^{2p} -a^{2p} $  is divisible by $x^2 - a^2$ hence divisible by $x+a$). Also $x+a$ is a factor of $x^n +a^n$ when $n$ is odd.
Let's suppose that $P(x)$ is monic. By the fundamental theorem of algebra $P(x) = \prod^{n}_{i=1} (x_i - a_i)$ . If $a_i$ is real we can find a real polynomial $Q_{a_i} (x)$ such that $Q_{a_i}(x) \times (x-{a_i}) = x^n \pm {a_i}^n$ . In the case $a_i$ isn't real we know that non-real complex roots come in conjugate pairs. 
$(x-a)(x- \bar a) = x^2 -2Re(a)x + |a|^2 $ is a polynomial with real coefficients and so is $(x^n -a^n)(x^n - \bar a^n)$. Hence the quotient  $Q_{a}(x)  Q_{\bar a}(x)$ of  $(x^n -a^n)(x^n - \bar a^n)$ divided by $(x-a)(x- \bar a)$ will also be a polynomial with real coefficients.
Since $ \Pi^{n}_{i=1} (x^n \pm a_i^n)$ is a polynomial such that powers of $x$ are multiples of an $n$, we have shown that such a polynomial $Q(x)$ in the question exists. 
We can also use the fact that $x^n - a_j^n = \Pi^{n}_{i=1} (x- \mathcal{E}_ia_j)$ where $\mathcal{E}_i$ is an $n^{th}$ root of unity.
