Coefficients of a polynomial function from $\Bbb{N}$ to $\Bbb{N}$ Let $f$ be a polynomial from $\Bbb{N}$ to $\Bbb{N}$.
So let $f(x) = a_0+a_1x+a_2x^2+..a_rx^r$
Then for every natural number $k$, $f(k) \in \Bbb{N}$.
But how does it imply that all the coefficients of $f(x)$ i.e. $a_0, a_1,..a_r$ will belong to $\Bbb{Q}$?
 A: You can retrieve the coefficients by using a Lagrangian interpolator on $r+1$ natural points. They will be rational.
The reasoning also applies to polynomials from $\mathbb Q$ to $\mathbb Q$.
A: Take $r+1$ natural numbers $n_i$ and make the substitution $x=n_i$ which gives a system of linear equations that we can solve to get the $a_k$. This system will have entirely integer coefficients so all the operations of Gaussian elimination will only require rational numbers and we are done.
A: It results that
$\begin{cases}
  a_0+a_1+\ldots+a_i+\ldots+a_r=f(1)\\
  a_0+2a_1+\ldots+2^ia_i+\ldots+2^ra_r=f(2)\\
  a_0+3a_1+\ldots+3^ia_i+\ldots+3^ra_r=f(3)\\
  ......................................\\
  a_0+ra_1+\ldots+r^ia_i+\ldots+r^ra_r=f(r)\\
  a_0+(r+1)a_1+\ldots+(r+1)^ia_i+\ldots+(r+1)^ra_r=f(r+1)
\end{cases}$
Since all the coefficients of the unknowns $a_i$, $0\le i\le r$, and all the constant terms $f(j)$, $1\le\ j\le r+1$, are natural numbers, then
$\Delta=
\begin{vmatrix}
1 & 1 & \cdots & 1 & \cdots & 1\\
1 & 2 & \cdots & 2^i & \cdots & 2^r\\
1 & 3 & \cdots & 3^i & \cdots & 3^r\\
\cdots & \cdots & \cdots & \cdots & \cdots &\cdots\\
1 & r & \cdots & r^i & \cdots & r^r\\
1 & (r+1) & \cdots & (r+1)^i & \cdots & (r+1)^r
\end{vmatrix}$
and
$\Delta a_i=
\begin{vmatrix}
1 & 1 & \cdots & f(1) & \cdots & 1\\
1 & 2 & \cdots & f(2) & \cdots & 2^r\\
1 & 3 & \cdots & f(3) & \cdots & 3^r\\
\cdots & \cdots & \cdots & \cdots & \cdots &\cdots\\
1 & r & \cdots & f(r) & \cdots & r^r\\
1 & (r+1) & \cdots & f(r+1) & \cdots & (r+1)^r
\end{vmatrix}$
are integer numbers and $\Delta\ne0$.
Moreover $$a_i=\frac{\Delta a_i}{\Delta}$$ are all rational numbers.
