Given $a\_1,a\_2,...,a\_m$ positive integer. Denominator $d$ is smallest positive integer for $b\_l$ integer coefficient.


$$\sum\_{k=1}^m\binom{n}k a\_k= \frac{1}d\sum\_{l=1}^mb\_ln^l$$


Now consider $n=dt+r$ where $d>r\ge 0$.


can it be shown that, above equation transform as


$$\frac{1}d\sum\_{l=1}^mb\_l(dt+r)^l=\sum\_{u=0}^{m-1}(x\_ut+y\_u)(dt+r)^u$$


With $x\_u$ and $y\_u$ integers.


**Example**


Let $a\_1=1,a\_2=1,a\_3=3$ 


then $\sum\_{k=1}^3\binom{n}k a\_k=(n^3-2n^2+3n)/2$ here $d=2$


Case$(1)$ $n=2t$, 


$\frac{n^3-2n^2+3n}2=(t-1)n^2+n+t$


Case$(2)$ $n=2t+1$, 


$\frac{n^3-2n^2+3n}2=(t-1)n^2+(t+2)n$


Thank you.



Problem related to transforming polynomial
Denote $\sum b\_l(x+r)^l=\sum c\_l x^l$. The numbers $c\_l$ are still integers, $c\_0=\sum b\_l r^l$ is divisible by $d$, and we have, denoting $dt+r=z$,
$$
\frac1d\sum b\_l(dt+r)^l=\frac1d\sum c\_l (dt)^l=\frac{c\_0}d+\sum\_{l>0} c\_l t\cdot (dt)^{l-1}=\\
\frac{c\_0}d+\sum\_{l>0} c\_l t\cdot (z-r)^{l-1}=
\frac{c\_0}d+\sum\_{l>0,0\leqslant j\leqslant l-1} c\_l {l-1\choose j}(-r)^{l-1-j}z^jt,
$$
that has desired form if you group all terms with $z^j$ together. 


