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(This question is inspired by a question about remainders of binomial coefficients I asked recently.)

Consider the polynomial maps $f:\ \mathbb{Z} \to \mathbb{Z}$. These include the polynomials with integer coefficients, but also functions like $f(x) = \frac{x(x-1)}{2} = \binom{x}{2}$. One can show that any such polynomial map $\mathbb{Z} \to \mathbb{Z}$ is a $\mathbb{Z}$-linear combination of the polynomials $\binom{x}{d}$ for $d \in \mathbb{N}$. I am interested in polynomial maps which take on all residues modulo all integers. More precisely, a polynomial $f$ is interesting for me if for any integer $m \geq 2$ and for any integer $0 \leq r < m$ there exists an integer $n$ with $f(n) \equiv r \pmod{m}$. One such polynomial is $f(x) = x$.

As explained in the mentioned other questions, polynomials $f(x) = x^d$ don't have the aforementioned property (for $d > 1$). I had a (naive) hope that the polynomials ${x \choose d}$ would do the trick, but they don't work either.

After some more thought, I discovered that degree $2$ polynomials can never work. This is because for $f(x) = ax^2 +bx +c$ the condition $f(n) \equiv r \pmod{p}$ can be rewritten as $$\left(n+\frac{b}{2a}\right)^2 \equiv \frac{1}{a}\left(r-c+\frac{b^2}{4a}\right) \pmod{p}$$ (assuming numerators and denumerators of $a,b,c$ are coprime to $p$). Hence, $f(n)$ does not have the residue $r$ if $p$ is large enough and $\frac{1}{a}\left(r-c+\frac{b^2}{4a}\right)$ is not a quadratic residue modulo $p$.

This leaves me doubtful about polynomials of higher degrees. I can't think of a general argument or an example of an interesting polynomial, though.

Question: Does there exist a polynomial map $f:\ \mathbb{Z} \to \mathbb{Z}$, different than $f(x) = x$, with the aforementioned property that for any integer $m \geq 2$ and for any integer $0 \leq r < m$ there exists an integer $n$ with $f(n) \equiv r \pmod{m}$.

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We may be thinking of different definitions, but I can't see how $\,f(x)=\frac x{x-1}\;$ can be considered a polynomial map... – DonAntonio Jul 29 '13 at 18:35
@DonAntonio: It was supposed to be $x(x-1)/2$. Corrected. Thanks for pointing this out! – Jakub Konieczny Jul 29 '13 at 18:36
@RGB: Are you sure? As far as I understand, for $f(x) = x^2$ what you say is not true. The law of quadratic reciprocity seems to imply that for appropriate choice of $p \pmod{ 4p_1p_2\dots p_k}$ I can ensure that all primes $p_1,p_2,\dots,p_k$ are quadratic residues modulo $p$. Hence, $f(x) \equiv r \pmod{p}$ has solutions for $r$ up to $p_{k+1}$, which can be made larger than $d$. – Jakub Konieczny Jul 29 '13 at 18:55
up vote 4 down vote accepted

We want to prove that $f(x)=x+a$ are the only posibilities.

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 $1$ and we get the solution $f(x)=x+a$, which works.

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