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In his book, The Annotated Turing, Charles Petzold says (emphasis mine),

Using a straightedge and compass to construct geometrical shapes is equivalent to solving certain forms of algebraic equations.

Does that mean that given any polynomial $p(x)$ in one variable $x$, an algorithm can be written to do the construction of a geometric shape whose equation is $p(x) = 0$ (algebraic equation) using a straightedge and compass?

And what does he mean by certain forms of algebraic equations? Is he referring only to polynomials with real roots?

KReiser
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Kedar Mhaswade
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    Not any polynomial, far from it. This is true for a limited number of polynomials. I believe there should be an explanation later in the book. But anyway, when you intersect between lines, or between a line and a circle, or between two circles, the intersection point is a solution of some algebraic equation. – Mark May 21 '23 at 12:36
  • To elaborate, it's "not any polynomial, far from it" because we can show there are (a lot of) algebraic numbers which aren't constructible – FShrike May 21 '23 at 14:33
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    @FShrike No, a number $\alpha$ is constructible if and only if there is a sequence of field extensions $\mathbb{Q}=F_0\subseteq F_1\subseteq F_2\subseteq...\subseteq F_k$ such that $\alpha\in F_k$ and all extensions are of degree $2$. So $[\mathbb{Q}(\alpha):\mathbb{Q}]$ being a power of $2$ is necessary, but not sufficient. (there are examples of non constructible algebraic numbers of degree $4$) – Mark May 21 '23 at 22:07
  • @Mark Ok, thanks. I had forgotten that point, was probably thinking about Gauss-Wantzel – FShrike May 21 '23 at 22:08
  • Can someone point, as an answer to this question, me to a proof of what is possible (by straightedge and compass) and what's not? – Kedar Mhaswade May 22 '23 at 00:58

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