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Does there exist a function $f$, holomorphic on the whole complex plane $\mathbb{C}$, such that $f\left(\mathbb{Z}\right)=\{0,1,2,3\}$ and

$\forall z\in\mathbb{C}\ (f(z)\in\{0,1,2,3\}\Rightarrow z\in\mathbb{Z})$?

If yes, is it possible to have an explicit construction?

Note that, for example, $h(z)=\frac{1}{6} \left(9-8 \cos \left(\frac{\pi z}{3}\right)-\cos (\pi z)\right)$ is not a valid solution, since, in particular, certain roots of the equation $h(z)=0$ are not integers, but complex numbers.

Also note that this question is answered in positive regarding the function $g$ such that $g\left(\mathbb{Z}\right)=\{0,1,2\}$ and $\forall z\in\mathbb{C}\ (g(z)\in\{0,1,2\}\Rightarrow z\in\mathbb{Z})$. In this case, an example of such function is $g(z)=1-\cos \left(\frac{\pi z}{2}\right)$.

user
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    where does this problem come from? – Conrad Jan 23 '23 at 21:07
  • No, such a function does not exist. A holomorphic function on the complex plane is infinitely differentiable, and thus must have derivatives of all orders. However, since the set of integers is countable, a function that is only defined on the integers cannot be infinitely differentiable. Therefore it cannot be holomorphic on the whole complex plane. –  Jan 24 '23 at 01:08
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    @Dmitry - the function is entire so is analytic on the plane; the condition just talks about specific values of the function at specific points - it is very easy to construct a function like that st it sends the integers surjectively to the given set, the problem is to ensure that no other numbers go there and that is harder; for the smaller set $0,1,2$ a simple solution is actually presented – Conrad Jan 24 '23 at 01:16
  • @Conrad - well, first I've noticed the fact that this is easily solvable if the discrete set is composed of all the $n$-th roots of unity. And then started to check if generalizations are possible. – user Jan 24 '23 at 17:15
  • A candidate would be $\frac32\left(1+\cos\left(\frac{\pi z}3+\alpha\sin \frac{\pi z}3\right)\right)$ with $\alpha=\left(\arccos\left(\frac13\right)-\frac\pi3\right)/\sin\left(\frac\pi3\right)\approx0.21219$, which takes the right values on the integers, but I suspect that it also takes them elsewhere, and I don't know how to easily check for that. Wolfram|Alpha doesn't return any other solutions. – joriki Jan 26 '23 at 11:34
  • @joriki - can you recheck the formula? With $z=2$ I get $f(2)=3/2 (1 + \sin[π/6 - \arccos[1/3]])$, which is not an integer. – user Jan 26 '23 at 12:31
  • @user1950: I'm sorry, I was missing a factor of $2$, it should be $\frac32\left(1+\cos\left(\frac{\pi z}3+\alpha\sin \frac{2\pi z}3\right)\right)$ (with the same $\alpha$). – joriki Jan 26 '23 at 13:15
  • @joriki Your function is $6$-periodic. If there is a $n$-periodic solution $f(z)=g(e^{2i\pi z/n})$ then $g$ is analytic on $\Bbb{C}^*$ so it must be in $\Bbb{C}[q,q^{-1}]$ as otherwise $g(q)$ or $g(q^{-1})$ has an essential singularity at $0$ and we can apply great Picard's theorem – reuns Jan 26 '23 at 13:49
  • @reuns: I'm afraid I don't quite follow. What are $q$ and $\mathbb C[q,q^{-1}]$? (And in case this is an argument that my proposed function takes on the desired values at non-integers, why doesn't it apply to the $4$-periodic function in the question?) – joriki Jan 26 '23 at 14:58
  • @joriki $f$ is entire $n$-periodic iff $g(q) = f(\frac{n\log(q)}{2i\pi })$ is analytic on $\Bbb{C}^*$. Due to great Picard's theorem $g(q)$ must be in the Laurent polynomials ring $\Bbb{C}[q,q^{-1}]$. – reuns Jan 26 '23 at 15:14
  • the problem with @joricki function is that is of infinite order so any $f(z)=a$ should have tons of solutions (many more than integers); for example $f(z)=0$ means $\frac{\pi z}3+\alpha\sin \frac{2\pi z}3=(2k+1)\pi$ and for a fixed large $k$ that has only a few integral solutions since $z/3$ needs to be around $2k+1$ as the other term is bounded on the integers, but (by Picard) it must have infinitely many solutions for all but at most one $k$; a function with the given property kind of have to be of finite order $1$ so the roots of $f(z)=0,1,2,3$ are the right number in any disc of radius $R$ – Conrad Jan 26 '23 at 16:27
  • @Conrad: So any valid solution to the main question must be of the form $\sum_{n=a}^{b} a_n q^n$, where $-\infty<a,b<+\infty$ and $q=e^{\frac{2\pi i z}{k}}$? – user Jan 26 '23 at 16:38
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    why does it need to be periodic? the above applies only to periodic functions – Conrad Jan 26 '23 at 16:42
  • one way to go is to interpolate the values $0,1,2,3$ evenly at integers (first for example the values may be symmetrized and the function made odd say on the integers as that gives better interpolation convergence) and get a function $f(z)=\frac{\sin \pi z}{\pi}(f(0)/z +\sum_{n \ne 0} (-1)^n f(n)/(z-n)+c)$ of order $1$ where $f(n)$ are precisely the choices you make and see if some clever choice of the interpolation values, the function obtained can be show to satisfy the property that only integers are sent there – Conrad Jan 26 '23 at 17:07
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    It follows from Theorem 1 of G.G. Gundersen and C.C. Yang's "On the Preimage Sets of Entire Functions" that the sought function must be periodic. (https://projecteuclid.org/journalArticle/Download?urlId=10.2996%2Fkmj%2F1138036858) – user Jan 29 '23 at 14:48
  • So from @reuns remark it follows that the sought function, if it exists, must be a Laurent polynomial of $e^{\frac{2\pi i z}{n}}$. Looks like the problem is within reach of an expert in Galois theory. – user Jan 29 '23 at 15:13
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    @user1950 - one has to be careful as the cited theorem applies only for multiplicity one values, so in other words if $f^{-1}({0,1,2,3})=\mathbb Z$ and $f'(n) \ne 0$ for any $n \in \mathbb Z$; probably not hard to generalize if the multiplicities are uniformly bounded but unclear what happens if multiplicities at integers can be unbounded – Conrad Jan 29 '23 at 19:51
  • I'd ask for a Laurent polynomial solution in a separate post (ie. $g\in \Bbb{C}[q,q^{-1}]$ such that $g^{-1}({0,1,2,3}) = \mu_n$ for some $n$) – reuns Jan 29 '23 at 20:48
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    I suggest you post this to MathOverflow, adding a link to this MSE post to let readers know you've crossposted. – TheSimpliFire Mar 28 '23 at 22:53
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    Crossposted to MathOverflow: A holomorphic function sending integers (and only integers) to ${0,1,2,3}$ – The Amplitwist Apr 09 '23 at 16:51
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    It is already solved on MO – Alexandre Eremenko Apr 19 '23 at 03:47

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