Suppose $a \in \mathbb{R}$, and $\exists n \in \mathbb{N}$, that $a^n \in \mathbb{Q}$, and $(a + 1)^n \in \mathbb{Q}$

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..

Solution 1:

For the "harder" (??) second part of the question, let $a=i$ or let $a=\frac{-1+i\sqrt{3}}{2}$.

Added: The first part of the question is (for me at least) more difficult than the second part. Maybe I am missing something obvious. The solution below uses some algebra, but not Galois Theory, just degrees of extensions.

We prove something that looks stronger but isn't. Let $a$ be a real number. If there exists a positive integer $n$, and a non-zero rational $e$, such that $a^n$ and $(e+a)^n$ are rational, then $a$ is rational. Suppose the result is not correct. Then there is a smallest positive integer $n$, a real irrational $a$, and a non-zero rational $e$ such that $a^n$ and $(e+a)^n$ are rational. It is clear that $n$ must be $\ge 2$.

First we do something completely unnecessary. By assumption $a^n$ and $(e+a)^n$ are rational. Bring these rationals to a common denominator, which can be taken to be a perfect $n$-th power $r^n$. If $n$ is even, then $a^n=\frac{p}{r^n}$ and $(e+a)^n=\frac{q}{r^n}$ for some non-negative integers $p$ and $q$. If $n$ is odd, then, depending on the signs of $a$ and $e+a$, $a^n=\pm\frac{p}{r^n}$ and $(e+a)^n=\pm\frac{q}{r^n}$ for some non-negative integers $p$ and $q$. Then in the even case, $(ar)^n=p$ and $(r+ar)^n=q$, and in the odd case we have the same thing, with $p$ and/or $q$ possibly decorated with minus signs.

Let $w=ar$. Then $w=p^{1/n}$ and $r+w=q^{1/n}$ in the even case, and $w=\pm p^{1/n}$, $r+w=\pm q^{1/n}$ in the odd case. Note that $w$ is a real irrational. Note also the crucial fact that by the minimality of $n$, there is no positive integer $m<n$ such that simultaneously $(p^{1/n})^m$ and $(q^{1/n})^m$ are rational.

Since $\pm p^{1/n}$ and $\pm q^{1/n}$ differ by an integer $r$, they have the same degree. By the minimality of $n$, this degree is $n$. But $p^{1/n}=q^{1/n}+r$. Take the $n$-th power of both sides. We find that $q^{1/n}$ is the root of a polynomial with integer coefficients, of degree $<n$, contradicting the fact that $q^{1/n}$ has degree $n$.

Comment: My first posted "proof" implicitly assumed that $a>0$. Thanks to Matt E for pointing out that modification was needed.

Solution 2:

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.