I can deduce some results about this from the prime number theorem or from results about the primorial function $p\#$ but I'm wondering what the state of the art is:
Given integers $0\le a_1<a_2<\dots <a_c$, what bound can we put on the least modulus $m$ such that for all $i\ne j$, we have $a_i\not\equiv a_j$ (mod $m$)?
I'm not sure about the state of art, but here is a rough estimate in terms of $c$ and $\ell:=a_c-a_1$ in the case $c\ll \ell$.
First, we notice that for an integer $M$ satisfying $$M\# ~>~ \big(\frac{c}{2(c-1)}\ell\big)^{c(c-1)/2} ~\geq~ \prod_{1\leq i<j\leq c} (a_j-a_i),$$ where the second inequality follows from AM-GM, there exists a prime $m\leq M$ that does not divide the r.h.s.
Second, taking $M$ as smallest as possible and estimating primorial $M\#$ as $e^M$, we have $$m\leq M\approx \frac{c}2 + \frac{c^2}2\log\frac{\ell}2.$$ I believe this rough estimate can be made rigorous if needed.
If $f(1),\ldots,f(n)$ are distinct integers, then the least positive integer $m$ such that $f(1),\ldots,f(n)$ are pairwise incongruent modulo $m$ is usually denoted by $D_f(n)$. There are some research papers studying $D_f$ (the discriminator of $f$) for various number-theoretic functions $f$, see the introduction of my 2013 JNT paper On functions taking only prime values. For example, I proved that if $f(k)=2k(k-1)$ for $k=1,2,\ldots,n$, then $D_f(n)$ is the least prime greater than $2n-2$.
If $f(k)$ is the product of the first $k$ primes, in the 2013 paper of mine I conjectured that $D_f(n)$ with $n>1$ is a prime smaller than $n^2$, and this was verified by W.B. Hart for $n=2,\ldots,10^5$.
Quite recently, Q.-H. Yang and L. Zhao (see arXiv:2111.02746) used Kloosterman sums to prove a conjecture of mine which states that the least positive integer $m$ such that $k^3+k\ (k=1,\ldots,n)$ are pairwise incongruent modulo $m^2$ is the least power of three no less than $\sqrt n$.
