Polynomial whose roots are not integers but almost so

Question

Let $\varepsilon \in (0,\frac{1}{2})$. Say that a real number $x$ is an $\varepsilon$-pseudointeger if it is not an integer but at distance at most $\varepsilon$ from some integer (thus $|x-i|\leq \varepsilon, x\neq i$ for some integer $i$).

Is it true that for any $d\geq 2$ and $\varepsilon$, there is a monic polynomial of degree $d$ with integer coefficients, all of whose roots are simple, real and $\varepsilon$-pseudointegers ?

When $d=2$, for example, $X^2-(n^2+1)$ where $n=\lceil \frac{1}{2\varepsilon} \rceil$ is an answer.

I'm looking for an explicit solution rather than a non-constructive proof.

Answer 1

Yes, it seems to be true.

If $d = 2k$, then example can be constructed using idea, similar to one, used in your example, meaning that polynomial will look like $\prod_{i = 1}^{k}(X^2 - i^2(n^2 + 1))$. It's clearly monic, has $d$ distinct real roots and by choosing $n = \lceil\frac{d}{2ε}\rceil$ we can account for increase in initial difference $(\sqrt{n^2 + 1} - n)$ by a factor of $d$.

If $d$ is odd (particularly $3 + 2k$), example will be a little harder to find. Still, let our needed polynomial be $(P(X) + 1)*\prod_{i = 1}^{k}(X^2 - i^2(n^2 + 1))$ where $P(X)$ is of the form $X(X^2 - p^2)$. $P(.)$ has 3 integer roots $\{-p, 0, p\}$ but what is interesting is values of $P'(.)$ corresponding to those roots: $2p^2, -p^2$ and $2p^2$ respectively. This observation gives an idea that by choosing appropriate large enough $p$ we will not change roots much when taking $P(.) + 1$ instead of $P(.)$.

Now to a question of actually finding $p$ needed depending on $ε$: it seems obvious that it exists and I might try to present particulars, but for now it seems that they would be relatively lengthy and not very exciting per se.

Answer 2

I interpret "all roots are simple, real and $\epsilon$-pseudointegers" as "all roots are simple, real but not integer, and $\epsilon$-pseudointegers"; otherwise $ (X-1)(X-2) \cdots (X-d) $ is a trivial solution.

The answer is yes if $d$ is even.

Your idea of obtaining solutions of the form $ \sqrt{n^2 +1} $ is very good and can be extended quite easily. Indeed, all the numbers of the form $ \sqrt{N^2 +1} $ are closer to their respective integer $N$ than $ \sqrt{n^2 +1} $ is to $n$, if $N > n$. Rigorously,

$$ N>n \implies \sqrt{N^2 +1} - N < \sqrt{n^2 + 1} - n $$

a fact that is not hard to prove.

Therefore

$$ (X^2 - (N_1^2 + 1)) \cdot (X^2 - (N_2^2 + 1)) \cdots (X^2 - (N_d^2 + 1)) $$

is a polynomial of degree $2d$, with $N_1, N_2, \dots, N_d$ distinct integers, and $N_i \geq \lceil {\frac{1}{2 \epsilon}} \rceil, \forall i \in \{ 1,2, \dots, k \} $, whose solutions are distinct, real but not integers, and $\epsilon$-pseudointegers.