Let $d$ be a positive integer. My question is: can we then find a positive integer $r$ (dependent on $d$) with integers $\alpha\_1, \alpha\_2, ..., \alpha\_r$ and $\beta\_1, \beta\_2, ..., \beta\_r$ such that the Polynomials $P(X) = (X - \alpha\_1) \cdots (X - \alpha\_r)$ and $Q(X) = (X - \beta\_1) \cdots (X - \beta\_r)$ satisfy 
$$ \frac{P(X)}{Q(X)} = 1 + O(\frac{1}{X^d})$$
as $X$ tends to $\infty$ but also $P \not= Q$? For example, if $d=1$ we easily find $r=1$, $\alpha\_1 = 1$ and $\beta\_1=0$ is a solution. Similarly, if $d=2$ we have $r = 2$ and $\alpha\_{1/2} = \pm 1$ and $\beta\_1 = \beta\_2 = 0$ since then
$$ \frac{P(X)}{Q(X)} = \frac{X^2 - 1}{X^2} = 1 - \frac{1}{X^2}. $$



Polynomials $P$ with integer roots near to $X^{\mathrm{deg}(P)}$
One can take $r = 1 + \frac{d(d-1)}{2}$. 
Indeed, one can consider the map
$$
(\alpha\_i)\_{i=1}^r \in [|1,N|]^{r} \mapsto (\sum\_{i=1}^r \alpha\_i^k)\_{k=1}^{d-1} \in \prod\_{i=1}^{d-1} [|1,rN^k|].
$$
The source has cardinality $N^r$ while the target has cardinality $r^{d-1} N^{\frac{d(d-1)}{2}}$. If 
$$
(\*) \quad N^r > r! r^{d-1} N^{\frac{d(d-1)}{2}},
$$
then we obtain by the pigeonhole principle a tuple $(s\_k)\_{k=1}^{d-1}$ with $> r!$ preimages. Among these preimages one can find $(\alpha\_i)\_{i=1}^r$ and $(\beta\_i)\_{i=1}^r$ which are not permutations of each other. Thus the polynomials $P(X) = \prod\_i (X-\alpha\_i)$ and $Q(X) = \prod\_i (X-\beta\_i)$ are distinct, but satisfy $P(X) = Q(X) + O(X^{r-d})$.


The condition $(\*)$ is satisfied for large $N$ whenever $r > \frac{d(d-1)}{2}$.


