Shw that $ \ (1+X)^a \in \mathbb{Q}$ both in $\mathbb{R}$ and $ \mathbb{Q}_p$ under some condition

Solution 1:

Not a complete answer, just a collection of hints and remarks.

The series is

$$\displaystyle (1+X)^a = \sum_{k=0}^\infty \binom{a}{k}X^k .$$

Real convergence: If $a\in \Bbb Z_{\ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum, hence converges for arbitrary $X$. If $a\notin \Bbb Z_{\ge 0}$, one has to use estimates of the usual absolute value of the binomial coefficients. I know next to nothing about this and just googled a bit. According to http://emis.math.tifr.res.in/journals/JIPAM/images/061_06_JIPAM/061_06.pdf, as soon as $a>-1$, certainly $\vert X\vert < 1$ is sufficient (and I have the feeling that this bound is reasonable, if not necessary, in general). the series converges for $\vert X\vert < 1$ and diverges for $\vert X\vert> 1$, as shown with an argument from complex analysis by reuns in a comment.

Real rationality: Answered in Wojowu's comment on How to find all $x \in \mathbb{Q}$ and $r \in \mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?.

$p$-adic convergence: Again if $a\in \Bbb Z_{\ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum. For general $a\notin \Bbb Z_{\ge 0}$, now one needs estimates for the $p$-adic absolute value of the binomial coefficients. A good thing is that because we are in an ultrametric, we only need to check whether

$$\lim_{k\to \infty}\vert \binom{a}{k}X^k\vert_p = \lim_{k\to \infty}\vert \binom{a}{k} \vert_p \cdot \vert X\vert_p^k \stackrel{?}=0.$$

Again I have to leave this open in general; in analogy to the real case however, one definitely has (cf. http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/binomialcoeffpadic.pdf, Definition of $p$-adic $(1+x)^\alpha$ via binomial series and log/exp): If $\vert a\vert_p \le 1$, then all $\vert \binom{a}{k}\vert_p \le 1$, meaning that the series certainly converges for those $X$ with $\vert X\vert_p < 1$.

$p$-adic rationality: Here is a subtlety. E.g. look at

$$\displaystyle (1+(-\frac78))^\frac13 = \sum_{k=0}^\infty \binom{1/3}{k}(-7/8)^k .$$

In $\Bbb R$, the series converges to $1/2$, which is indeed a cube root (more precisely: the unique positive real cube root) of $1/8$. The series also does converge in $\Bbb Q_7$, and also to a (!) cube root of $1/8$, but not to $1/2 \in \Bbb Q$, rather to the unique one which is $\equiv 1$ mod $7$, and that is $\zeta \cdot 1/2 \notin \Bbb Q$, where $\zeta \in \Bbb Z_7$ is the primitive third root of unity which is $\equiv 2$ mod $7$. (Note that $\Bbb Z_7$ contains exactly the sixth roots of unity; the primitive sixth ones are $\equiv 3$ resp. $\equiv 5$ mod $7$, the primitive third ones are $\equiv 2$ resp. $\equiv 4$ mod $7$, well and there are $\pm 1$).

So in general, even in the case $\vert a\vert_p \le 1$ and $\vert X\vert_p < 1$ and the real rationality criterion is satisfied, it is not necessarily true that the limit of the $p$-adic series is rational. Rather, it is = (the rational we get from the real consideration) times (some root of unity), so that this product is $\equiv 1$ mod $p$. Whether that can be fulfilled by the only rational roots of unity, namely $\pm 1$, depends on $X, a$ and $p$ again (in your example in the OP, it can, in my above example, it cannot).

To answer a question in the comments: Sure the values of $f(X)$ evaluated in the two different ways can be identical, e.g. in the trivial case that $a \in \Bbb N$. Or also, I think, $(X,a,p)=(-63/64, 1/4,3)$. What one needs is that the real rationality criterion is satisfied, so that we have a rational real value of $(1+X)^a$, and the numerator of $(1+X)^a -1$ must be divisible by $p$.