Minimizing the number of different values of $\frac{f(c)-f(d)}{c-d}$ for integers $c,d$ and polynomial $f$

Question

I am looking for the following quantity

$\mathcal{E}:=\min \Big( \#\Big\{ \frac{f(c)-f(d)}{c-d}:\text{$f$ is a polynomial s.t. $\text{deg}(f)\ge 2$}, c,d\in A\subseteq \mathbb{Z}~~\text{and}~~|A|\ge 20\Big\}\Big)$

i.e. how many different values of $\frac{f(c)-f(d)}{c-d}$ for integers $c,d$ in the set $A$ that is a subset of the integers.

The assumption of $|A|\ge 20$ is simply made since I want to find a lower bound for this value $\mathcal{E}$ for large $|A|$. It can be seen that $\mathcal{E}\le 2|A|-3$ by the simple example of choosing $f(x)=x^{2}$ and choosing, say the points $\{(x,f(x)):x\in \{-10,-9,...,-1,0,1,...,9,10\}\}$. Are there other constructions that perhaps yield a bound less than $2|A|-3$? How about even lower? I am mainly however looking for a proof for obtaining a nice upper and lower bounds for $\mathcal{E}$. I've shown that clearly $|\mathcal{E}|\le 2|A|-3$, but I cannot seem to find a polynomial $f(x)$ that yields even a lower bound.

My idea currently is the following: we write $f(c)=\sum_{j=0}^{n}a_{j}c^{j}$ and $f(d)=\sum_{j=0}^{n}a_{j}d^{j}$. By taking the difference $f(c)-f(d)$ and dividing by $c-d$ we have the following:

$\frac{f(c)-f(d)}{c-d}=\frac{\sum_{j=0}^{n}a_{j}c^{j}-\sum_{j=0}^{n}a_{j}d^{j}}{c-d}=\frac{a_1(c-d)+...+a_n(c^n-d^n)}{c-d}=\frac{(a_1(c-d)+a_2(c-d)(c+d)+...+a_n(c-d)(c^{n-1}+c^{n-2}d+...+cd^{n-2}+d^{n-1}))}{c-d}=a_{1}+a_{2}(c+d)+...+a_{n}(c^{n-1}+c^{n-2}d+...+cd^{n-2}+d^{n-1})~(*)$

whenever $a_0=0$. I am not sure how to approach two of the following things:

1. how to proceed from the equation $(*)$ to obtain a bound for $\mathcal{E}$ and

2. how to approach the case of $a_0\neq 0$.

For (1) I do have the following thought though: we would need to count how many different values $(*)$ can take for different values of $c$ and $d$, and which $c$ and $d$ could potentially minimize this quantity. I also think that one could potentially use an argument of symmetry for this case in the sense that the points $\{(x,f(x))\}$ should be chosen for $x\in \{-a,-a+1,...,0,1,...,a\}$ for $a\in \mathbb{Z}$ for some $a\ge 10$.

Answer 1

Whatever the function $f$ or the set $A$ we have $$1\le \mathcal{E} \le \frac {|A||A-1|}{2}.$$ Both of these bounds can be attained.

Lower bound See @mathlove's comment.

Given any set of integers $A$, let $f$ be any polynomial whose set of roots contains $A$. Then $\mathcal{E}$=1 and this can clearly not be bettered.

Upper bound

Let $A$ be the set of powers of $2$, $\{1,2,4,16,...\}$, up to any desired number and let $f(x)=x^2$. Then $\mathcal{E}=\frac {|A||A-1|}{2}$and this can clearly not be bettered.