https://aops.com/community/p11413318 kaede wrote: Lemma A For any polynomial $ f( x) \in \mathbb{C}[ x]$ and for any complex number $ a$, $ \mathrm{deg}( f( x+a) -f( x)) =\mathrm{deg}( f) -1$ and the leading coefficient of $ f( x+a) -f( x)$ is $ aL\cdot \mathrm{deg}( f)$ where $ L$ is the leading coefficient of $ f( x)$. Lemma B Let $ M\subset \mathbb{Q}[ x]$ be free $ \mathbb{Z}$-module with basis $ \{x( x-1) ...( x-( k-1)) /k!\mid \ k\in \mathbb{Z}_{\geq 0}\}$ $ $ For any polynomial $ f( x) \in \mathbb{C}[ x]$, if $ f( x)$ is integer-valued polynomial, then $ f( x) \in M$ Solution Let $ f( x) =\sum ^{d}_{i=0} c_{i} x^{i} \in \mathbb{C}[ x]$ where $ d=\mathrm{deg}( f)$ By lemma A, we can find $ a_{0}$,$ a_{1}$,..,$ a_{2^{d} -1} \in \{-1$,$ 1\}$ such that $ \sum ^{2^{d} -1}_{i=0} a_{i} \cdot f( x+i) =c_{d} \cdot d!\cdot 2^{d( d-1) /2}$ ...($ \spadesuit $) By lemma B, we can say $ |c_{d} |\cdot d!\in \mathbb{Z}^{+}$. Let $ m=|c_{d} |\cdot d!\cdot 2^{d( d-1) /2}$. From $ ( \spadesuit )$, it is sufficient to find at least one integer that gives the same residue modulo $ m$ as $ k$ which can be expressed in the required form. We may assume $ m\geq 2$. For any prime $p$, there exist $ h\in \mathbb{Z}$ for which $ \gcd( p,f( h)) =1$. So there exist $ r\in \mathbb{Z}^{+}$ for which $ \gcd( m,f( r)) =1$ by CRT. Let $ v=k-\sum ^{r-1}_{j=1} f( j)$ We can take $t \in \mathbb{Z}^{+}$ for which $ tf( r)\equiv v \pmod{m}$. Then we have $ \sum ^{t-1}_{j=0} f\left( r+j\left( 2^{d} +1\right)\right) \equiv tf( r) \equiv v\pmod m$. Finally $ \sum ^{r-1}_{j=1} f( j) +\sum ^{t-1}_{j=0}\left( f\left( r+j\left( 2^{d} +1\right)\right) +\sum ^{2^{d}}_{i=1} a_{i-1} \cdot f\left( r+i+j\left( 2^{d} +1\right)\right)\right) \equiv k\pmod m$ $ \blacksquare $