= Solution
Begin with two points at distance one. A ruler draws the line through two known points; a compass draws a circle with a known centre and a constructible radius. New points are intersections of such lines and circles. A <real number> is constructible when it occurs as a coordinate of a point obtained by finitely many such operations; a complex point is constructible when both coordinates are.
If current coordinates lie in a <field> $K\subseteq\mathbb R$, a line-line intersection solves <linear equations> over $K$, while line-circle or circle-circle intersections reduce to a <quadratic equation> over $K$. Thus each step requires at most adjoining a <square root>. Consequently every <constructible number> lies in a tower
$$
\mathbb Q=K_0\subseteq K_1\subseteq\cdots\subseteq K_m,\qquad [K_{i+1}:K_i]\le2.
$$
Conversely <field> sums and differences are obtained by laying off segments, products and quotients by similar triangles, and a positive <square root> by the geometric-mean construction in a semicircle. Hence every <real number> in such a tower is constructible. This proves the quadratic-tower criterion, not merely its necessity.
In particular a constructible <algebraic number> has degree a power of two. Degree alone is not sufficient in general; the exact criterion is containment in a quadratic tower. Equivalently the <Galois group> of its <normal closure> is a finite two-group: a tower's <normal closure> remains a two-extension, and a two-group admits a chain of subgroups with successive index two, yielding the converse by <Galois correspondence>.
The standard impossibility results follow. Doubling a unit-volume cube requires $\sqrt[3]2$, whose <polynomial> $X^3-2$ is irreducible by Eisenstein and has degree three. Trisecting an arbitrary angle would in particular trisect $60$ degrees, requiring $2\cos20^\circ$, a root of $X^3-3X-1$. This cubic has no rational root and is irreducible, so this construction is impossible, although some particular angles can be trisected. Squaring the <unit circle> requires side length $\sqrt\pi$, impossible because $\pi$ is transcendental whereas <constructible numbers> are algebraic.
For a regular $n$-gon, constructibility is equivalent to that of $e^{2\pi i/n}$. The cyclotomic extension is Galois of degree $\varphi(n)$, so it is constructible exactly when this degree is a power of two. Writing the <prime factorization> of $n$ shows the equivalent condition
$$
\boxed{n=2^a p_1\cdots p_s,\quad p_i\text{ distinct Fermat primes}.}
$$
Indeed <odd prime> exponents must be one, and each $p_i-1$ must be a power of two. If $2^m+1$ is <prime>, $m$ must itself be a power of two by factoring when $m$ has an odd divisor. Conversely this factorization makes the cyclotomic <Galois group> a two-group and provides a quadratic tower. This gives both positive constructions, such as the regular seventeen-gon, and obstructions, such as the regular seven-gon.
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