TITLE: Remainders of binomial coefficients.
QUESTION [8 upvotes]: Motivation: It is easy to notice that a polynomial map $f: \mathbb{Z} \to \mathbb{Z}$ does not need to have only integer coefficient. For example, $f(x) = \frac{x(x-1)}{2}$ does have rational coefficients and maps $ \mathbb{Z}$ to $\mathbb{Z}$. This gives an easy reason for an polynomial with integer coefficients not to give all possible remainders modulo some number $m$: for instance, $x^2 - x = 2\frac{x(x-1)}{2}$ is always $0$ modulo $2$, and in a similar spirit $f(x) = x^2$ is always $0$ or $1$ modulo $4$. More generally, one can show that  a polynomial map $f: \mathbb{Z} \to \mathbb{Z}$ is a $\mathbb{Z}$-linear combination of the polynomials $\binom{x}{d}$ for $d \in \mathbb{N}$.
I am looking for a class of polynomials $f: \mathbb{Z} \to \mathbb{Z}$ which have the property that for any $m \in \mathbb N$ and any remainder $r \in \{0,1,\dots,m-1\}$ there exists some $a \in \mathbb N$ such that $f(a) \equiv r \pmod{m}$. The above shows that $f(n) = n^d$ do not have much chance of working, but $f(n) = \binom{n}{d}$ might.
Question: Is it true that for any $d \in \mathbb N$ and for any $m,r \in \mathbb N, r < m$, the map $f(n) = \binom{n}{d}$ has the property that $f(a) \equiv r \pmod{m}$ for some $a \in \mathbb{Z}$ ?

REPLY [3 votes]: This will answer your motivation and in particular the question.
We want to prove that $f(x)=x+a$ are the only possibilities for such a polynomial.
Consider the polynomial $g(x)=f(x+1)-f(x)$. By looking at primes $p$, large enough, so that all denominators in the coefficients of $f$ are prime to $p$, we can assume $f$ has integer coefficients from now on. We want the congruences $f(x)=r\ (\text{ mod }p)$ to have solutions for all $r=0,1,\ldots,p-1$. Then $f(0),f(1),\ldots,f(p-1)$ must give all remainders $0,1,\ldots,p-1$. In particular they all must be different. Therefore $g(0),g(1),\ldots,g(p-1)$ are not zero mod $p$. Otherwise we don't get all the remainders out of $f$.
Lemma: If $g$ is non-constant then we can make $g(n)$ divisible by arbitrary large primes.
Proof: I wrote the proof of this lemma in this answer to another problem. 
Take $g(M)$ for $M$ very large such that $g(M)$ is divisible by a very large prime $p$ with the conditions above (such that this prime is prime with all denominators in the coefficients of $f$). Then $f(x)=r$ will not always be solvable mod this prime $p$.
Therefore $g$ is constant. Hence $f$ is linear. From this we get that the leading coefficient of $f$ must be $1$ (otherwise mod a prime that divides this coefficient we don't get all remainders) and we get the solution $f(x)=x+a$, which does work.