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I was wondering if there exists non-polynomial functions $f : \mathbb{Z} \longrightarrow \mathbb{Z}$ such that $f(k) \equiv f(k+n) \hspace{1mm}(\text{mod } n)$ for every integer $k$ and integer $n \geq 2$.

My reason for asking this was because polynomials in $\mathbb{Z}[x]$ or integer valued polynomials in $\mathbb{Q}[x]$ obviously satisfy above. However, I do not seem to know of any other function that does the same. I was wondering if anyone could provide some examples and/or whether such things have been studied before.

HumbleStudent
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    The integer valued function $f(x)=[x(x+1)]/2$ does not have your congruence property for the case $n=2.$ – coffeemath Nov 23 '22 at 02:31
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    Closely related question (which only imposes your condition for prime $n$): https://math.stackexchange.com/questions/2046098/are-certain-integer-functions-well-defined-modulo-different-primes-necessarily-p/2046134. Robert Israel's answer works for your question, as does a slight tweak of Eric Wofsey's. – Ravi Fernando Nov 23 '22 at 03:07
  • It is known that any integer valued polynomial with rational coefficients is a linear combination of binomial functions $g(n)=\binom{n}{k}$ with integer coefficients. The latter basic functions $g(n)$ can be studied to give their least periods mod $m$ for various $m.$ For example $\binom{n}{2}$ has period $4$ for the modulus $2.$ These might be a good place to start when looking for congruence properties. – coffeemath Nov 23 '22 at 03:10

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There are all kinds of bizarre functions which satisfy your requirement but are not polynomials with rational coefficients. In fact, the set $S$ of all such functions is uncountable.

Fix a bijection $a : \mathbb{N} \to \mathbb{Z}$ and consider some function $g: \mathbb{N} \to S$. We will build a function $f : \mathbb{Z} \to \mathbb{Z}$ which satisfies your property but is not in the range of $g$.

Suppose we have already defined $f(a_1), \ldots, f(a_{n - 1})$. These choices determine $f(a_n)$ mod $lcm(\{a_n - a_i \mid i < a\})$ using a routine Chinese remainder theorem argument. This leaves us with infinitely many choices for $f(a_n)$, so we pick one such that $f(a_n) \neq g_n(a_n)$ (perhaps the smallest eligible natural number, for concreteness). The resulting $f$ is clearly not equal to any $g_n$, since $f(a_n) \neq g_n(a_n)$ by construction.

This confirms that $S$ is indeed uncountable.

Mark Saving
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