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Prove that $x + \frac{1}{x}$ is rational if $x^4 + \frac{1}{x^4}$ and $x^5 + \frac{1}{x^5}$ is rational. ($x \in \mathbb{R}, x \neq 0$)

(Please note the direction of the implication! It is the reverse from what you may assume.)

I have made many different attempts to this problem but always hit a brick wall during the proof. I'd be glad for any help. Thank you!

The duplicate referenced is a similar question, but what I want to prove here is a given condition there. So I think they are not duplicates and the question should be reopened.

Akiva Weinberger
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math_inquiry
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    Couldn't find a question like this. Can you provide the url for the original? – math_inquiry Jan 24 '24 at 22:05
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    Comments and the duplicate closure assume one problem direction, but the body and title both make clear the other is sought. Proving that if $x^4+1/x^4,,x^5+1/x^5$ are rational so is $x+1/x$ is much less trivial. Since the questions are different, I've voted to reopen. – J.G. Jan 24 '24 at 23:35
  • I'm sorry but i fail to see how that method could be applied here. If i use the identity you have given i'm still left with factors that i have no information about. Tried some kind of "descending deduction" (if it holds for n than it holds for n-1) but with no clear result. More guidance is needed here. The solutions given above are not correct for this problem i believe. – math_inquiry Jan 25 '24 at 05:39
  • @BillDubuque : I also do not see how to apply your argument from the alleged duplicate backwards. The argument there relies on knowing that $ y + z $ is rational, which is what we're trying to prove here. If we had that, then we could go from $ y ^ { n + 1 } + z ^ { n + 1 } \in \mathbb Q $ and $ y ^ n + z ^ n \in \mathbb Q $ to $ y ^ { n - 1 } + z ^ { n - 1 } \in \mathbb Q $, and so on backwards. But we don't have that. Is there a way to reformulate the identity to use, say, $ y ^ 4 + z ^ 4 $ instead of $ y + z $? – Toby Bartels Jan 25 '24 at 06:19
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    @Toby Oops, I mixed up the dupe targets while working on this and related questions. Now I don't recall the target I had for this question, and I don't have time now to search again so I've voted to reopen. Thanks for catching that. – Bill Dubuque Jan 25 '24 at 07:22
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    As of 10 minutes ago, the person who marked this as a duplicate has acknowledged that it's not a duplicate (at least not of the question it was marked as). Yet only 8 minutes ago, ‘the community’ decided not to reopen it because it is a duplicate (at least, that's what saying that the closure reasons are unresolved seems to mean that that's the closure reason). Something is broken in this process. – Toby Bartels Jan 25 '24 at 07:34
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    This could be a duplicate target : prove that if $x^n+\frac 1{x^n}$ and $x^{n+1}+\frac 1{x^{n+1}}$ are rational and $x$ is positive, then so is $x+\frac 1x$. Some slightly brute-force procedure can be found on AoPS here for the particular case $n=4$. – Sarvesh Ravichandran Iyer Jan 25 '24 at 07:51
  • @TobyBartels Ideally the review would have gone as you hoped, but people make mistakes. Luckily the site is self-correcting in such matters. Bringing a dubious review outcome to the notice of the crowd in C.U.R.E.D., together with an explanation of what went wrong, usually speeds up the process. In other words, I wouldn't say that the system is broken. Rather that we should not expect to get all the decisions "right" the first time (even in the cases where there is an objective "right" outcome). – Jyrki Lahtonen Jan 26 '24 at 09:40
  • @JyrkiLahtonen : I was not aware of CURED; that seems like a useful chatroom. (As it was, I flagged a moderator.) – Toby Bartels Jan 29 '24 at 09:50

1 Answers1

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After a lot of algebra, I have determined the following.

If $p=x^4+\dfrac1{x^4}$ and $q=x^5+\dfrac1{x^5}$, then $$\boxed{x+\frac1x=\frac{p^2q-pq-q}{p^3-2p-q^2+1}}$$

Akiva Weinberger
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  • The answer given here is cleaner and more general, but it does not give $x+x^{-1}$ as an explicit rational function of $x^n+x^{-n}$ and $x^{n+1}+x^{-n-1}$ as I have done: https://math.stackexchange.com/a/2829878/166353 – Akiva Weinberger Jan 25 '24 at 09:21
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    As a bonus, this same exact formula also can tell us $2\cos(x)$ in terms of $2\cos(4x)$ and $2\cos(5x)$, as well as telling us $2\cosh(x)$ in terms of $2\cosh(4x)$ and $2\cosh(5x)$. – Akiva Weinberger Jan 25 '24 at 09:29
  • There is the subtlety of verifying that the denominator is never zero. – Akiva Weinberger Jan 25 '24 at 09:54
  • It would be interesting to see the derivation! I think the existence of such a formula follows from a bit of Galois theory and possibly Lüroth's theorem. – Jyrki Lahtonen Jan 25 '24 at 10:21
  • Thank you! Both the linked question and the answer is much appreciated. – math_inquiry Jan 25 '24 at 12:43
  • If the denominator is zero then the numerator must be, too. The numerator factors, so it's easier to check. – Akiva Weinberger Jan 25 '24 at 22:37
  • One should not post unjustified claims as an answer, nor @math_inquiry should one accept them as answer. Please give a proof. This site is for mathematics, not magic. – Bill Dubuque Jan 25 '24 at 23:10
  • @BillDubuque Checking that this is true is trivial with a computer algebra program (or with a graphic calculator, like Desmos). I will amend the answer later with how I derived it. – Akiva Weinberger Jan 26 '24 at 03:24
  • @Akiva Of course I know that, but my critique still stands. – Bill Dubuque Jan 26 '24 at 03:44
  • FWIW I upvoted this as I'm a bit curious about the process. If I find the time to think about this, I will try and describe some related theory (in the linked thread). I want to add the somewhat obvious comment that the way of writing $x+1/x$ as a rational function of $p$ and $q$ is not unique. This is because $\Bbb{Q}(x)/\Bbb{Q}$ is a transcendence degree one extension. Therefore any two elements of $\Bbb{Q}(x)$, transcendental over $\Bbb{Q}$, are algebraically dependent over $\Bbb{Q}$. So there exists non-trivial bivariate polynomials $F(u,v)\in\Bbb{Q}[u,v]$ such that $F(p,q)=0$. – Jyrki Lahtonen Jan 26 '24 at 09:47
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    @JyrkiLahtonen The simplest example is $T_5(p)-T_4(q)$. I don't know if this generates all others. – Akiva Weinberger Jan 26 '24 at 16:19
  • @JyrkiLahtonen Sorry, I had gotten confused. I meant $C_5(p)-C_4(q)$ where $C_n$ is the polynomial such that $C_n(x+x^{-1})=x^n+x^{-n}$. Indeed $f_n(x)=2T_n(\frac12x)$, where $T_n$ is the $n$th Chebyshev polynomial, hence the confusion. (See: https://en.wikipedia.org/wiki/Chebyshev_polynomials#Families_of_polynomials_related_to_Chebyshev_polynomials, "Families of polynomials related to Chebyshev polynomials") – Akiva Weinberger Jan 28 '24 at 02:39
  • No problem @AkivaWeinberger. Your comment was clear enough. I have seen those called Dickson polynomials for they are called that in the context of finite fields. The connection to Chebyshev polynomials is clear also. – Jyrki Lahtonen Jan 28 '24 at 04:29
  • So i have a more elementary answer: let the 4th degree sum be a, and the 5th degree van b. Then the 8th (c) and 10th (d) degree sums are rational as well (squaring and substracting 2). Look at the product of c and the second degree sum. The result is d + a, both of them rational therefore the 2nd degree sum is rational as well. Use the identity y^n + z^n on b. Because of the previous thoughts y+z = x + 1/x must be rational. – math_inquiry Feb 06 '24 at 08:41
  • @math_inquiry When you multiply $x^8+1/x^8$ and $x^2+1/x^2$ won't you get the sum of the tenth powers as well as the sixth powers? It's not clear to me how the logic proceeds. – Jyrki Lahtonen Feb 08 '24 at 03:47