Consider a function f from a cofinite subset of \mathbb{Q} to \mathbb{Q}. As established here and here f extending to a smooth function on a cofinite subset of \mathbb{R} is not sufficient to ensure that f is a rational function. Furthermore, I believe that the stronger condition that all of f's derivatives also map \mathbb{Q} to \mathbb{Q} (which I will call being '\mathbb{Q}-closed') is not strong enough to ensure that f is a rational function.
The question is are any of these strengthenings involving extensions of f to \mathbb{Q_p} strong enough to ensure that f is rational (or equal to some rational function on its domain). Specifically these are motivated by the fact that a rational function from \mathbb{Q} to \mathbb{Q} extends to smooth functions in \mathbb{R} and \mathbb{Q}_p and the derivatives of a rational function are fixed algebraically and so are the same in \mathbb{R} and \mathbb{Q_p}.
Let K\in\{\mathbb{R},\mathbb{Q_2},\mathbb{Q_3},... \}
- f extends to a continuous function (on some cofinite domain) in every K.
- f extends to a smooth function (on some cofinite domain) in every K.
Write \partial_Kf to mean the derivative of the maximal extension of f to K restricted to \mathbb{Q}.
- \partial^n_Kf exists (on some cofinite domain) and is \mathbb{Q}-closed for arbitrary n and K (note that the derivatives f^{(n)} in different extensions aren't a priori the same when restricted to \mathbb{Q}).
- \partial_{K_1}\partial_{K_2}...\partial_{K_n}f exists (on some cofinite domain) and is Q-closed for arbitrary sequences K_1,K_2,...,K_n.
- Condition 4 with the additional requirement that \partial_{K_1}\partial_{K_2}...\partial_{K_n}f=\partial_{K_1^\prime}\partial_{K_2^\prime}...\partial_{K_n^\prime}f for any two sequences K_1,K_2,...,K_n and K_1^\prime,K_2^\prime,...,K_n^\prime of the same length.
Or any other conditions of this ilk. I'm also curious if conditions 4 and 5 are equivalent.