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Fix $\epsilon>0$. For a finite set of arbitrary-degree polynomials with integer constant term, $p_1(x), ..., p_m(x)\in \mathbb{R}[x]$ is it possible to find an $n\in \mathbb{N}$ such that $$\max_{i=1,...,m}||p_i(n)||<\epsilon$$ where $||\cdot||$ denotes the distance to the nearest integer?

In the description of the book Small Fractional Parts of Polynomials a problem of this type is listed as an open question for degree greater than 4, but the book is from 1977. Research has continued since then on small fractional parts of polynomials. Has progress improved on this problem, or is it still open?

Updates: Discussion in the comments indicates that this can be proved by Weyl's equidistribution criterion. While this proves the result for $\mathbb{Q}$-linearly independent collections $\{p_1(n),...,p_k(n), 1, n, n^2,...\}$ it does not prove it when the collection is linearly dependent, since equidistribution no longer holds in that case. How can we generalize to when the polynomials may be $\mathbb{Q}$-linearly dependent?

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    $\begingroup$ Not if each polynomial is of the form $P(x) + 1/2$ for some $P \in {\bf Z}[X]$. So some hypothesis is necessary. What is condition imposed in the $p_i$ in your source? $\endgroup$
    – Noam D. Elkies
    Commented Jul 12, 2020 at 0:06
  • $\begingroup$ Ah of course, sorry! It is imposed that the polynomials have integer constant term. I will edit the post. $\endgroup$
    – Joe Previdi
    Commented Jul 12, 2020 at 0:09
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    $\begingroup$ OK, but something still seems to be missing. For example, if $p_1$ never takes integer values on $\mathbb N$, and $p_2 = -p_1$, then $\{ p_1(n) \} + \{ p_2(n) \} = 1$ for all $n \in \mathbb N$, so the fractional parts cannot both be small. $\endgroup$
    – Noam D. Elkies
    Commented Jul 12, 2020 at 1:58
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    $\begingroup$ So it is equivalent to the analogous question about several polynomials of the form $cx^k$, $k>0$, right? $\endgroup$
    – Fedor Petrov
    Commented Jul 12, 2020 at 10:26
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    $\begingroup$ @Fedor Petrov: There's a natural generalization from the circle ${\bf R} / {\bf Z}$ to the torus ${\bf R}^m / {\bf Z}^m$. A sequence of vectors $v_n \in {\bf R}^m / {\bf Z}^m$ is asymptotically equidistributed iff for even nonzero $a \in {\bf Z}^m$ we have $$ \sum_{n=1}^N \exp 2\pi i (a \cdot v_n) = o(N) $$ as $N \to \infty$. This does in the end come to the estimate for a single polynomial. (If $p_1,\ldots,p_m$ are $\bf Q$-linearly dependent then they're equidistributed in some closed subgroup of the torus.) $\endgroup$
    – Noam D. Elkies
    Commented Jul 12, 2020 at 13:45

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The claim is true as stated, and can be arrived at using Weyl's criterion which was pointed out in the comments. As I post this answer, there is no consensus on the rate of convergence of the $N$.

We proceed by induction on $k$, the number of polynomials. For $k=1$, either $p_1$ has only rational coefficients or it has at least one irrational. If it has an irrational one, Weyl's criterion shows that $p_1(n)$ is an equidistributed sequence, and thus has infinitely many $N$ for which $||p_1(N)||<\epsilon$. If $p_1$ has only rational coefficients, we can simply choose $N$ to be the product of every coefficient's denominator to get $||p_1(N)||=0$.

Now suppose the theorem holds for some fixed $k$. Consider a polynomial list $p_0, p_1, ..., p_k$ of length $k+1$. If $\{p_0(x),p_1(x),...,p_k(x), 1, x, x^2,...\}$ is $\mathbb{Q}$-linearly independent, then Weyl's criterion on the $k+1$ torus applies and equidistribution gives the existence of infinitely many $N$ for which $||p_i(N)||<\epsilon$ for all $i=0,...,k$.

If, on the other hand, the aforementioned collection is instead $\mathbb{Q}$-linearly dependent, Weyl's criterion does not apply. However, $$p_0(x)=q_1p_1(x)+\cdots +q_kp_k(x)+r_0+r_1x+\cdots +r_sx^s$$ for $q_i,r_j\in \mathbb{Q}$. If we let $q$ be the product of all denominators of the $r_j$, we have that at any natural number $n$, $$||p_0(qn)||=||q_1p_1(qn)+\cdots +q_kp_k(qn)||.$$

Apply the inductive hypothesis to $q_1p_1(qx),...,q_kp_k(qx)$. Choose $M$ so that $||q_ip_i(qM)||<\epsilon/2k$ for all $i=1,...,k$. At this small scale, the $||\cdot||$ function is subadditive in $k$ arguments, so $$||p_0(qM)||\leq ||q_1p_1(qM)||+\cdots +||q_kp_k(qM)||<k(\epsilon/2k)<\epsilon.$$

Therefore, at $N=qM$, $||p_i(N)||<\epsilon$ for all $i=0,...,k$. We are finished by induction.

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