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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..

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9  
Please don't ask questions as if you are giving us homework. Show us what you already tried and where you got stuck! –  sxd Dec 20 '11 at 15:24
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@Adam If he could see that (1) is obvious, why would he post that part of the question? –  Thomas Andrews Dec 20 '11 at 15:54
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@Adam: If we take $a = -\sqrt{5}$, then $a^{2m}$ is in $\mathbb Q$ for any $m$. Then $(1+a)^{2m} = 2^m(3-2\sqrt{5})^m$, and so (1) includes the claim that $(3-2\sqrt{5})^m$ has a non-zero coefficient of $\sqrt{5}$ for every $m$. This is certainly true, but is it obvious? (And note that (1) contains infinitely assetions of this type, so an ad hoc argument that proves one of them won't prove the general case.) Regards, –  Matt E Dec 20 '11 at 17:41
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@brainail: Dear brainail, In what context did this question arise? If it is homework from a course, what is the subject of the course and what tools have you been given? Regards, –  Matt E Dec 20 '11 at 17:43
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@Adam: Dear Adam, I (and I suspect many others) would describe the elements of $\mathbb C \setminus \mathbb Q$ as irrational (so $i$ is irrational), and certainly the statement that you use holds: if $a$ is a complex number and $p$ a polynomial in $\mathbb Q(x)$ such that $1 + p(a)$ is irrational, then $p(a)$ is irrational. So where does your argument break down if applied to e.g. $a = i$? Regards, –  Matt E Dec 20 '11 at 23:53
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2 Answers

up vote 9 down vote accepted

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.

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$a = i$, $n = 4$ .. Too easy, strangely.. –  jofisher Dec 20 '11 at 16:04
    
@brainail: You could try $a = \sqrt{-3}$ as well. Regards, –  Matt E Dec 20 '11 at 17:42
    
Yep, and n = 6. –  jofisher Dec 20 '11 at 18:01
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Also, you need to rule out the case that (say) $n = 6$, $p$ is a square, and $q$ is a cube, but your degree argument does this, I guess (once $p$ and $q$ are positive). Very nice! Regards, –  Matt E Dec 21 '11 at 6:40
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@AndréNicolas Could you please clarify precisely how the "minimality of $n$" implies that $q^{1/n}$ has degree $n$. This is the crux of the proof so it is essential that this be clear. –  Bill Dubuque Dec 21 '11 at 14:13
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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.

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