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Is there a direct proof that $\pi$ is not constructible, that is, that squaring the circle cannot be done by rule and compass?

Of course, $\pi$ is not constructible because it is transcendental and so is not a root of any polynomial with rational coefficients. But is there a simple direct proof that $\pi$ is not a root of polynomial of degree $2^n$ with rational coefficients?

The kind of proof I seek is one by induction on the height of a tower of quadratic extensions, one that ultimately relies on a proof that $\pi$ is not rational. Does any one know of a proof along these lines or any other direct proof?

I just want a direct proof that $\pi$ is not constructible without appealing to transcendence.

Harish Chandra Rajpoot
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lhf
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    So far as I know, there is no proof that $\pi$ doesn't have degree $2^n$ that doesn't also prove $\pi$ is transcendental. I leave this as a comment, rather than an answer, because I don't know how to substantiate it. – Gerry Myerson Jan 30 '12 at 00:20
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    @lhf: I don't see how knowing that $\pi$ is irrational helps. There are lots of irrational algebraic numbers. – Zhen Lin Jan 30 '12 at 00:30
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    @Zhen Lin: What lhf means is, I think, consider a tower of quadratic extensions, starting from Q. Note that $\pi\not\in \mathbb Q$ as an induction basis and then, by some magic induction argument, if $\pi\not\in F\implies \pi\not\in F(u)$, where $u^2\in F$. – Myself Jan 30 '12 at 00:49
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    The question you ask in your third sentence is stronger than the question you ask in your first sentence. – Qiaochu Yuan Jan 30 '12 at 02:55
  • @QiaochuYuan: You are right but you could make your point more clearly: what OP seems to ask for is a proof that $\pi$ is not algebraic with a _minimal polynomial_ whose degree is a power of two, whereas being the root of a polynomial of degree $2^n$ is just equivalent to being algebraic (as one can always throw in dummy factors to complete the degree of any polynomial to a power of two). – Marc van Leeuwen Jan 30 '12 at 08:40
  • Maybe I'm missing something, but doesn't squaring the circle construct $\sqrt{\pi}$? – Peter Taylor Jan 30 '12 at 12:30
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    @PeterTaylor, sure. But $\sqrt{\pi}$ is constructible iff $\pi$ constructible. – lhf Jan 30 '12 at 12:37
  • Sure, but you were asking for a direct proof. That seems fairly indirect. – Peter Taylor Jan 30 '12 at 13:04
  • @PeterTaylor, yes, sure, I'll settle for any direct proof that $\sqrt{\pi}$ is not constructible. – lhf Jan 30 '12 at 13:10
  • I'm not quite sure the "number theory" tag fits best here. – Doug Spoonwood Jan 30 '12 at 13:25
  • @DougSpoonwood, I've added a couple of tags but none really captures the question though. – lhf Jan 30 '12 at 15:06
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    @Marc: that's still not the question being asked in the first sentence. Most irreducible polynomials of degree $2^n$ do not have constructible numbers as their roots; in fact, generically they have Galois group $S_{2^n}$. – Qiaochu Yuan Jan 30 '12 at 22:52
  • If there were such a direct proof wouldn't we have seen far fewer cranks trying to square the circle in the last century or so? – Brian B Feb 24 '12 at 22:06
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    I wonder whether Vieta's product series could be used as the foundation of such a proof? – Mark Bennet May 19 '12 at 15:23
  • @MarkBennet, perhaps! How would you proceed? – lhf May 19 '12 at 18:09
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    Strange ... I could have sworn that I previously made the following comment in this thread ... Anyway, for what it's worth, I've been wondering the same thing for many years, and even asked about it once in sci.math (see my 21 July 2009 [Math Forum archived post](http://mathforum.org/kb/message.jspa?messageID=6791444)). – Dave L. Renfro Jul 03 '12 at 19:42
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    @Dave, your memory is not playing tricks on you. Your previous comment was on a now-deleted answer. –  Jul 03 '12 at 20:57
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    Ha. Dave Renfro's sci.math post was a response to one of mine. I guess when the conversation turns to $\pi$, things naturally keep coming 'round.... – Gerry Myerson Jul 04 '12 at 04:00
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    @Rahul Narain, Gerry Myerson: I ended that sci.math post with: *It seems to me that if no such simpler proof is known, then anyone finding such a proof will be making a fairly major contribution. Not quite as major as finding an elementary proof of Fermat's Last Theorem, but still a major contribution.* For more clarity on what I was thinking of, I should have cited as an example the Selberg/Erdős proof of the Prime Number Theorem that doesn't make use of complex analysis methods. In fact, I was thinking of the Selberg/Erdős result when I wrote those comments. – Dave L. Renfro Jul 10 '12 at 11:53
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    @Dave, I wouldn't call the Selberg/Erdos proof simpler than the proofs that use complex analysis. – Gerry Myerson Jul 10 '12 at 13:27
  • The Math Forum link to my 21 July 2009 sci.math post (a post mentioned in previous comments), no longer works. However, [this link at google-groups](https://groups.google.com/g/sci.math/c/WJiHE5-NbvU/m/8jIGi-3UYaUJ) still works. – Dave L. Renfro Jun 01 '21 at 18:28

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I've just read in the book The Number $\pi$ by Eymard and Lafon that no such proof is known.

“The proof that it is impossible to square the circle does not involve direct demonstration of the non-constructibility of the number $\pi$. As far as we are aware, it is not known how to do this! One proves that it is not algebraic, which is much more restrictive, then one uses the fact that a constructible number is algebraic.” [§4.2, p. 134]

lhf
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