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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.

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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.2012-01-30
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    @lhf: I don't see how knowing that $\pi$ is irrational helps. There are lots of irrational algebraic numbers.2012-01-30
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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 $u^2\in F$.2012-01-30
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    The question you ask in your third sentence is stronger than the question you ask in your first sentence.2012-01-30
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    @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).2012-01-30
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    Maybe I'm missing something, but doesn't squaring the circle construct $\sqrt{\pi}$?2012-01-30
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    @PeterTaylor, sure. But $\sqrt{\pi}$ is constructible iff $\pi$ constructible.2012-01-30
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    Sure, but you were asking for a direct proof. That seems fairly indirect.2012-01-30
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    @PeterTaylor, yes, sure, I'll settle for any direct proof that $\sqrt{\pi}$ is not constructible.2012-01-30
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    I'm not quite sure the "number theory" tag fits best here.2012-01-30
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    @DougSpoonwood, I've added a couple of tags but none really captures the question though.2012-01-30
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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 $S_{2^n}$.2012-01-30
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    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?2012-02-24
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    I wonder whether Vieta's product series could be used as the foundation of such a proof?2012-05-19
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    @MarkBennet, perhaps! How would you proceed?2012-05-19
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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)).2012-07-03
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    @Dave, your memory is not playing tricks on you. Your previous comment was on a now-deleted answer.2012-07-03
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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....2012-07-04
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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.2012-07-10
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    @Dave, I wouldn't call the Selberg/Erdos proof simpler than the proofs that use complex analysis.2012-07-10

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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]

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    BTW, it's a wonderful book!2012-10-08
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    I see that the book was written $8$ years ago, maybe the information is not up to date, though I doubt that it's not [+1]2012-10-09
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    @Belgi, I think that if things had changed since, we'd have heard!2012-10-09