TITLE: Suppose $a \in \mathbb{R}$, and $\exists n \in \mathbb{N}$, that $a^n \in \mathbb{Q}$, and $(a + 1)^n \in \mathbb{Q}$
QUESTION [10 upvotes]: 1) 
Suppose $a \in \mathbb{R}$, and $\exists n \in \mathbb{N}$, that $a^n \in \mathbb{Q}$, and $(a + 1)^n \in \mathbb{Q}$.
Prove: Is it true that $a \in \mathbb{Q}$?
2) Suppose $a \in \mathbb{C}$, and $\exists n \in \mathbb{N}$, that $a^n \in \mathbb{Q}$, and $(a + 1)^n \in \mathbb{Q}$.
Prove: Is it true that $a \in \mathbb{Q}$?
Somebody explain please..
Thank you..

REPLY [8 votes]: Since the other answer has a major gap, here is an approach using standard results.
HINT $\ $ For the nontrivial problem (1): $\:$  if $\rm\ f({\it a}) = 0 = g({\it a})\:$ then $\:a\:$ is also a root of $\rm\:gcd(f,g)\ =\ h\ f + k\ g\ $ by Bezout. In particular, if $\rm\:f\:$ is irreducible then $\rm\ f\ |\ g\ $ in $\rm\:\mathbb Q[x]\:.\:$ 
Thus, in your case, if $\:a\:$ is a root of the irreducible $\rm\ f(x) = x^n - q,\ q\in \mathbb Q\:$ then your hypothesis implies that $\:a\:$ is also a root of $\rm\:g(x) = (x+1)^n - r,\ r\in \mathbb Q\:,\:$ so $\rm\ f\ |\ g\ \Rightarrow\ f = g\:,\: $ hence $\ \cdots$
It remains to determine when such binomials are irreducible. Here there are classic results, e.g.
THEOREM $\ $ Suppose $\rm\:c\in F\:$ a field,  and $\rm\:0 < n\in\mathbb Z\:.$
$\rm\quad x^n - c\ $ is irreducible over $\rm\:F \iff c \not\in F^p\:$ for all primes $\rm\:p\: |\: n\:$ and $\rm\ c\not\in -4\:F^4\:$ when $\rm\: 4\ |\ n\:. $
A proof is in many Field Theory textbooks, e.g. Karpilovsky, Topics in Field Theory, Theorem 8.1.6.