Is there a polynomial $f$ with integer coefficients such that $f(1)=1$, $f(2)=2$, $f(3)=3$, $f(4)=4$, and $f(5)=17$?

Question

I was wondering if there exists a polynomial $f(x)$ with integer coefficients that satisfies $f(1)=1$, $f(2)=2$, $f(3)=3$, $f(4)=4$, and $f(5)=17$.

I thought there such polynomial would exist because $m-n$ divides $f(m)-f(n)$ for every pair of integers $m$ and $n$. However, I failed to find real evidence that such polynomial exists.

(By the way, if we are finding polynomial with rational number coefficients , there is an answer. See Lagrange polynomial)

Answer 1

Any polynomial $f(x)\in\mathbb{C}[x]$ satisfying $f(1)=1$, $f(2)=2$, $f(3)=3$, $f(4)=4$, and $f(5)=17$ is of the form $$\begin{align}f(x)&=x+\frac12\,(x-1)(x-2)(x-3)(x-4)\\ &\phantom{aaaaa}+(x-1)(x-2)(x-3)(x-4)(x-5)\,P(x)\end{align}$$ for some $P(x)\in\mathbb{C}[x]$. Thus, $f(x)\in\mathbb{Z}[x]$ if and only if $$Q(x):=(x-5)\,P(x)+\frac12\in\mathbb{Z}[x]\,.$$ However, if $Q(x)\in\mathbb{Z}[x]$, then $Q(5)\in\mathbb{Z}$, but $$Q(5)=\frac12\notin\mathbb{Z}\,,$$ which is a contradiciton. Therefore, $f(x)$ cannot have integral coefficients.

The condition that $m-n$ must divide $f(m)-f(n)$ for all $m,n\in\mathbb{Z}$ is a necessary condition for $f(x)$ to have integral coefficients. However, as the paragraph above illustrates, this is not a sufficient condition. From the link, given by Robert Z, we have the following result.

Proposition. Let $n\in\mathbb{Z}_{>0}$ and $x_1,x_2,\ldots,x_n\in\mathbb{Z}$ be such that $x_1,x_2,\ldots,x_n$ are pairwise distinct. For $y_1,y_2,\ldots,y_n\in\mathbb{Z}$, there exists a polynomial $f(t)\in\mathbb{Z}[t]$ for which $f\left(x_i\right)=y_i$ for each $i=1,2,\ldots,n$ if and only if the Lagrange interpolation polynomial $L(t)$ of the points $\left(x_1,y_1\right)$, $\left(x_2,y_2\right)$, $\ldots$, $\left(x_n,y_n\right)$ has integral coefficients.