On a weak polynomial version of Erdős conjecture

Question

This question is motivated by Erdős conjecture on arithmetic progressions. It is a weaker conjecture than Erdős' one.

Erdős conjecture on arithmetic progressions states, "If $A$ is a large set, then $A$ contains (nontrivial) arithmetic progressions of every length."

The following conjecture states that, "If $A$ is a large set, then $A$ contains (nontrivial) polynomial progressions of integer coefficients of every length."

Conjecture: If $A\subset \mathbb{N}$ is a large set in the sense that

$$ \sum_{n\in A} \frac{1}{n} = \infty,$$

then for every $k\in\mathbb{N},$ there exists a polynomial of integer coefficients $(c_i)_{i=1}^{n}\ $ with $\ n\geq 1\ $ and $\ c_n\geq 1$,

$$ f_k(x) = c_0 + c_1 x + c_2 x^2 + \ldots + c_n x^n, $$

such that

$$ f_k(i) \in A \quad \forall\ i\in \lbrace{1,\ldots, k\rbrace}. $$

Am I missing something and somehow this is relatively easily proved using elementary number theory, or would you expect this to be almost as difficult as Erdős conjecture itself?

Answer 1

In fact this can be done rather trivially for any infinite set $A$ of integers.

Given $k$, let $$L (x) = \prod_{j=2}^k \frac{x - j}{1-j}$$ a polynomial of degree $k-1$ interpolating $(1,1)$ and $(j,0)$ for $j = 2 \ldots k$. Its coefficients are integers divided by $(k-1)!$. Let $y_1$ and $y_2$ be two members of $A$ that are congruent mod $(k-1)!$. Then the polynomial

$$f_k(x) = y_1 + (y_2 - y_1) L(x)$$

has degree $k-1$, integer coefficients $f_k(1) = y_2$ and $f_k(j) = y_1$ for $j = 2 \ldots k$. If the leading coefficient is negative, interchange $y_1$ and $y_2$ to make it positive.

It's also easy to modify this to get a solution where the values $f_k(j)$ for $j = 1, \ldots, k$ are distinct.

[EDIT] Consider the polyomials $$L_i(x) = \prod_{j \in \{1,\ldots,k\} \backslash \{i\}} \frac{x-j}{i-j}$$ This is the unique polynomial of degree $k-1$ with value $1$ at $i$ and $0$ at the other integers from $1$ to $k$. Note $\sum_{i=1}^k L_i(x) = 1$. Then $$f(x) = \sum_{i=1}^k y_i L_i(x)$$ has $f(i) = y_i$ for $i = 1, \ldots, k$. If all the $y_i$ are congruent modulo the lcm of the denominators, this polynomial has integer coefficients.