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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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    Please don't ask questions as if you are giving us homework. Show us what you already tried and where you got stuck!2011-12-20
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    Do you see why it's 'obvious' that in (1) we can't have irrational a? For (2) things get more complicated - however, what can you tell us about roots of unity?2011-12-20
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    @Adam If he could see that (1) is obvious, why would he post that part of the question?2011-12-20
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    (1) if a is irrational number, then there is no such n. Now think about it ..2011-12-20
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    I do not see the end of the proof.. help, some hint please..2011-12-20
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    I imagine a = t^(1/k) $t \in \mathbb{Q}$, $k \in \mathbb{N}$, as $a^n \in \mathbb{Q}$, that $n = mk$. Now t = (p/q)^(1/m). I do not understand how to convert $(a + 1)^n$..2011-12-20
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    @Adam: It's not obvious to me!2011-12-20
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    Ane me... please explain ..2011-12-20
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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,2011-12-20
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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,2011-12-20
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    Geometry and Algebra (Number theory, cryptography, ..., fields, rings, groups).2011-12-20
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    $2^m(3 - sqrt(5))^m$2011-12-20
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    @MattE pow is always even? We have to prove that for any rational - there is no n. This can be done without the binomial theorem?2011-12-20
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    Dear brainail, The only solution I know to this problem uses some Galois theory. This may be overkill; I don't know if e.g. @Adam has a simpler answer in mind. Did your course cover Galois theory? Regards,2011-12-20
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    @MattE No. If your proof is not very confusing, so I understand.... Do not we need to show that if a - is irrational, then (1 + a)^n is always irrational, and all?2011-12-20
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    @Adam please explain your proof.2011-12-20
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    Daer Brainail, If you haven't studied Galois theory, then there is no point to me explaining my proof. Also, it is not a question of showing that if $a$ is an irrational real number then $(1+a)^n$ is always irrational. E.g. $-1+\sqrt{5}$ is irrational, but $(1 + (-1+\sqrt{5}))^2$ is rational. You have to use that fact that the original $a$ is the $n$th root of some rational number (so a very special kind of irrational!). Regards,2011-12-20
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    @MattE Yep, you are right.2011-12-20
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    Clearly the 2nd equation is false (3 rd roots of unity!). The first is trickier - pick n minimal then look to the expression (a+1)^n - a^n to see this must imply a + na^2 + ... + (n,k)a^k + ... + na^(n-1) is rational - here (n,k) is n choose k. So we take the a out (assuming it's irrational) to see na + ... + (n,k)a^(k-1) + ... + na^(n-2) is irrational (as rational + rational = rational and irrational times rational is irrational). Continue in such a fashion to see this means that a is rational, a contradiction.2011-12-20
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    @Adam: Dear Adam, Where in your sketched argument do you use the distinction between $\mathbb R$ and $\mathbb C$? Regards,2011-12-20
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    I disproved the second with one of the third roots of 1. I don't understand what you;re saying however - rational and irrational are adjectives used to describe something that's in Q or R\Q. Would you call i rational, irrational ? Is that what you mean? I used this step repeatedly; 1 + p(a) is irrational implies p(a) irrational [where p(a) is a polynomial of degree greater than 1 in the ineterminate a] to give a culling of powers - as then we can factor out a in p(a) and repeat the argument].2011-12-20
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    @Adam a + na^2 + ... + (n,k)a^k + ... + na^(n-1), you used the Binomial theorem?2011-12-20
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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,2011-12-20
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    @Matt E: If it wouldn't take too long, I wouldn't mind seeing the proof you've mentioned using Galois theory.2011-12-21
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    Adam's proof is correct?2011-12-21
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    Dear brainail, I don't understand it as written, because I don't see where it is using the distinction between $\mathbb R$ and $\mathbb C$. Maybe I am missing something? Regards,2011-12-21
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    Adam's proof simply doesn't work, as far as I can see.2011-12-21
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    @Jason: Dear Jason, I don't think there is much point; it was somewhat in the spirit of Andre's proof below, but much more convoluted. Regards,2011-12-21

2 Answers 2

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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

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 $

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..2011-12-20
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    @brainail: You could try $a = \sqrt{-3}$ as well. Regards,2011-12-20
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    Yep, and n = 6.2011-12-20
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    Dear Andre, I agree with your argument, although I think it would be good to make it explicit that $p$ and $q$ can be arranged to be positive (to avoid the problem of $i = 1^{1/4}$ and $1+i = (-4)^{1/4}$, both of which are of the same degree --- but $2$ rather than $4$!). Regards,2011-12-21
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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,2011-12-21
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    @AndréNicolas Of course the first part of a much more complicated. I was wrong.2011-12-21
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    @AndréNicolas Very nice. I will try to understand today. Many thanks to all of you that spend time on this issue. Is it the final proof?2011-12-21
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    @AndréNicolas Only say that at first $a \in \mathbb{R}$.. What does show your proof is written above?2011-12-21
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    The proof in the post is only valid for real $a$. I would not be surprised if for non-real $a$, the condition forces the *norm* to be rational.2011-12-21
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    @AndréNicolas Ok, I understood. I will try to understand written above, and wait for your final proof.2011-12-21
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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.2011-12-21
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    @AndréNicolas I apologize for my stupidity. Could you explain some things: 1) Bring these rationals to a common denominator, which can be taken to be a perfect n-th power $r^n$, and $r$ is integer. How can I do it? 2) Note that w is a real irrational.. Explain more please. I'm sorry again.2011-12-21
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    @Bill: Dear Andre and Bill, Indeed, the key fact being used here is that if $p$ is a positive integer and at least one of the exponents in the prime factorization of $p$ is prime to $n$, then $p^{1/n}$ has degree $n$ over $\mathbb Q$. Regards,2011-12-21
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    @MattE I am aware of various ways to prove this. IMO the above is *far* from a complete proof. Essential points have been glossed over without any mention. Since clearly you have already thought much about this, I recommend that you post your approach in an answer.2011-12-21
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    @MattE Can you explain this more ""minimality of n" implies that q^{1/n} has degree n"", or give a source where it is described.2011-12-22
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    @AndréNicolas Can you explain this more ""minimality of n" implies that q^{1/n} has degree n"", or give a source where it is described. Thank you.2011-12-22
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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.