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  "body": "\n## Proof\n\nLet $n \\ge 3$, let $\\zeta = e^{2\\pi i/n}$, and let $\\mathcal{C}$ be the set of the [[constructible numbers|0FMN2MN7TM]]. By the [[field theorem|0PN9WMRWT7]], $\\mathcal{C}$ is a subfield of $\\C$ closed under square roots. By [[Euler's formula|04R3CY41VJ]] and [[de Moivre's formula|03MTBVG3VE]], the powers $\\zeta^k = e^{2\\pi i k/n}$, $0 \\le k \u003c n$, are the vertices of a regular $n$-gon with centre $0$. If $\\zeta \\in \\mathcal{C}$, they all lie in the field $\\mathcal{C}$, so this $n$-gon is constructible. Conversely, let a regular $n$-gon with centre $c$ be constructed, and let $v$ and $w$ be adjacent vertices. Its vertices are $c + (v - c)\\zeta^k$ for $0 \\le k \u003c n$. Their sum is $nc$, because [[$\\sum_{k=0}^{n-1} \\zeta^k = 0$|02PETK9PQ4]]. So $c \\in \\mathcal{C}$. Also $(w - c)/(v - c)$ is $\\zeta$ or $\\zeta^{-1}$, so $\\zeta \\in \\mathcal{C}$. So we must show: $\\zeta \\in \\mathcal{C}$ if and only if $n$ has the stated form. Let $\\varphi$ be [[Euler's totient function|0VV04M41AN]].\n\nThe degree of $\\zeta$.\n\n1. The [[cyclotomic polynomial|0W0KVBM54S]] $\\Phi_n$ is monic, has the root $\\zeta$ (the factor with $k = 1$), and has degree $\\varphi(n)$. Indeed, for $n \\ge 2$ neither $k = n$ nor $0$ is coprime to $n$, so both ranges give the same count. It is [[irreducible in $\\Q[x]$|0SK1VHZ5R2]].\n2. So the [[minimal polynomial|0GK1WB9R98]] of $\\zeta$, which [[divides|06K7SARZKR]] $\\Phi_n$, equals $\\Phi_n$. By the [[degree of a simple extension|0N5QW9K7MH]], $[\\Q(\\zeta) : \\Q] = \\varphi(n)$.\n\nNecessity.\n\n3. Suppose $\\zeta \\in \\mathcal{C}$. Then $\\varphi(n)$ is a [[power of $2$|0SSSM1D475]], say $2^r$.\n4. Write $n = 2^k p_1^{a_1} \\cdots p_t^{a_t}$ with distinct odd primes $p_j$ and $a_j \\ge 1$. By [[multiplicativity|0FSJA9M43W]] and the [[totient of a prime power|0AVY8HKBGD]], $\\varphi(n) = \\varphi(2^k) \\prod_j p_j^{a_j - 1} (p_j - 1)$.\n5. Each factor $p_j^{a_j - 1}(p_j - 1)$ divides $2^r$, so it is a power of $2$. Since $p_j$ is odd, $a_j = 1$, and $p_j - 1 = 2^{m_j}$ with $m_j \\ge 1$.\n6. Suppose that $m = m_j$ has an odd divisor $b \u003e 1$, say $m = cb$. The polynomial $y^b + 1$ has the root $-1$, as $b$ is odd. By the [[factor theorem|0K3MWA7DMC]], $y + 1$ divides $y^b + 1$ in $\\Z[y]$.\n7. Putting $y = 2^c$ shows that $2^c + 1$ divides $2^m + 1 = p_j$. Since $c \u003c m$, we have $1 \u003c 2^c + 1 \u003c p_j$, which contradicts the primality of $p_j$.\n8. So $m_j = 2^{s_j}$ for some $s_j \\ge 0$, and $p_j = 2^{2^{s_j}} + 1$ is a [[Fermat prime|0N9E4G2P97]]. Hence $n = 2^k p_1 \\cdots p_t$ with distinct Fermat primes $p_j$.\n\nSufficiency.\n\n9. Let $n = 2^k p_1 \\cdots p_t$ with distinct Fermat primes $p_j = 2^{2^{s_j}} + 1$. By step 4, $\\varphi(n) = \\varphi(2^k) \\prod_j 2^{2^{s_j}}$, where $\\varphi(2^k) = 2^{k-1}$ for $k \\ge 1$ and $\\varphi(1) = 1$. So $\\varphi(n) = 2^r$ for some $r \\ge 0$.\n10. Let $L = \\Q(\\zeta) \\subseteq \\C$ and $f = x^n - 1$. By the [[description of the roots of unity|057BH3JC9S]], every root of $f$ is $\\zeta^k$ for some $k$ with $0 \\le k \u003c n$.\n11. There are [[exactly $n$ roots of $f$|0E021GF2B8]], so $\\zeta^0, \\dots, \\zeta^{n-1}$ are distinct. Then $f - \\prod_k (x - \\zeta^k)$ has degree less than $n$ and $n$ roots, so it is $0$ by the [[root bound|0KJVQV3E0F]].\n12. So $f$ splits over $L$, and $L = \\Q(\\zeta^0, \\dots, \\zeta^{n-1})$. Thus $L$ is a [[splitting field|02NZZ1J0FH]] of $f$ over $\\Q$.\n13. Also $(-1) \\cdot f + (x/n) \\cdot f' = 1$, so $f$ is [[separable|03J8BS7FMV]]. By the [[characterisation of Galois extensions|P75CXH2Q2B]], $L/\\Q$ is a finite Galois extension.\n14. Let $G = \\operatorname{Gal}(L/\\Q)$, the [[Galois group|0XVR72M51M]]. By the [[order of the Galois group|0BDJBH8B4Z]] and step 2, $\\abs{G} = \\varphi(n) = 2^r$.\n15. For $\\sigma \\in G$, $\\sigma(\\zeta)^n = 1$, so $\\sigma(\\zeta) = \\zeta^{a_\\sigma}$ for some integer $a_\\sigma$ by step 10. Since [[every element of $L$ is $h(\\zeta)$ with $h \\in \\Q[x]$|0ZN39QDTP9]], $\\sigma$ is determined by $\\sigma(\\zeta)$.\n16. For $\\sigma, \\tau \\in G$, we get $\\sigma\\tau(\\zeta) = \\zeta^{a_\\sigma a_\\tau} = \\tau\\sigma(\\zeta)$. So $\\sigma\\tau = \\tau\\sigma$, and $G$ is [[abelian|071DDFVG4J]].\n17. Claim: for every $r \\ge 0$, every subfield $E \\subseteq \\C$ such that $E/\\Q$ is a finite Galois extension with abelian Galois group of order $2^r$ satisfies $E \\subseteq \\mathcal{C}$. We prove it by induction on $r$.\n18. If $r = 0$, then $[E : \\Q] = 1$ by the order of the Galois group. So $E = \\Q$, and $\\Q \\subseteq \\mathcal{C}$ because the subfield $\\mathcal{C}$ contains $1$.\n19. Let $r \\ge 1$ and $G_E = \\operatorname{Gal}(E/\\Q)$. By [[Cauchy's theorem|0JM7PA288B]], some $h \\in G_E$ has order $2$. As $G_E$ is abelian, $H = \\set{1, h}$ is a [[normal subgroup|0W77CVBCHQ]].\n20. Let $E' = E^H$ be the [[fixed field|0KMJ6919WS]]. By the [[fundamental theorem of Galois theory|0T1Q8D0FW8]], $[E : E'] = 2$, $E'/\\Q$ is normal, and $\\operatorname{Gal}(E'/\\Q) \\cong G_E/H$, which is abelian of order $2^{r-1}$.\n21. Every element of $E'$ lies in $E$, so it is separable over $\\Q$. So $E'/\\Q$ is a finite Galois extension, and $E' \\subseteq \\mathcal{C}$ by the induction hypothesis.\n22. Choose $a \\in E$ with $a \\notin E'$. By the [[tower law|0A7KJ68XHT]], $[E'(a) : E']$ divides $[E : E'] = 2$ and is not $1$. So $[E'(a) : E'] = 2$, $[E : E'(a)] = 1$, and $E = E'(a)$.\n23. Let $x^2 + bx + c$ be the minimal polynomial of $a$ over $E'$, and let $s = 2a + b$. Then $s^2 = 4(a^2 + ab) + b^2 = b^2 - 4c \\in E' \\subseteq \\mathcal{C}$.\n24. So $s \\in \\mathcal{C}$, as a square root of an element of $\\mathcal{C}$. Then $a = (s - b)/2 \\in \\mathcal{C}$, since $\\mathcal{C}$ is a field containing $b$.\n25. By the degree of a simple extension, every element of $E = E'(a)$ is $u + va$ with $u, v \\in E'$. All of these lie in $\\mathcal{C}$. So $E \\subseteq \\mathcal{C}$, which proves the claim.\n26. By steps 13, 14 and 16, the claim applies to $E = L$. So $\\zeta \\in \\mathcal{C}$, and the regular $n$-gon is constructible.\n\n## Idea\n\nThe regular $n$-gon is constructible exactly when $\\zeta = e^{2\\pi i/n}$ is, and $\\zeta$ has degree $\\varphi(n)$ over $\\Q$. If $\\zeta$ is constructible, $\\varphi(n)$ is a power of $2$, which forces the stated form of $n$. Conversely, if $\\varphi(n)$ is a power of $2$, the abelian Galois group of $\\Q(\\zeta)$ can be cut down by subgroups of order $2$. This gives a chain of quadratic extensions, and each quadratic step only needs a square root.\n",
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