{
  "index": "1968-B-3",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "B-3. Assume that a \\( 60^{\\circ} \\) angle cannot be trisected with ruler and compass alone. Prove that if \\( n \\) is a positive multiple of 3 , then no angle of \\( 360 / n \\) degrees can be trisected with ruler and compass alone.",
  "solution": "B-3 We need to make use of the following facts about fields and constructibility: (1) If \\( Q \\) is the field of rational numbers, the degree of \\( Q \\) extended by \\( \\cos \\left(360^{\\circ} / k\\right) \\), where \\( k \\) is a positive integer, is \\( \\phi(k) \\), where \\( \\phi \\) is the Euler function. (2) If \\( K, L, M \\) are fields with \\( K \\subset L \\subset M \\) and \\( [L: K]<\\infty,[M: L]<\\infty \\) then \\( [M: K]=[M: L] \\cdot[L: K] \\). (3) Given \\( \\cos \\left(360^{\\circ} / k\\right) \\), then \\( \\cos \\left(360^{\\circ} / 3 k\\right) \\) is constructible if and only if\n\\[\n\\left[\\varrho\\left(\\cos \\frac{360^{\\circ}}{3 k}\\right): Q\\left(\\cos \\frac{360^{\\circ}}{k}\\right)\\right]\n\\]\nis a power of 2 .\nConsequently, \\( \\left[Q\\left(\\cos 360^{\\circ} / 3 k\\right): Q\\left(\\cos 360^{\\circ} / k\\right)\\right] \\cdot \\phi(k)=\\phi(3 k) \\). Now\n\\[\n\\phi\\left(3^{a}\\right)=3^{a-1} \\cdot 2=\\left\\{\\begin{array}{ll}\n3 \\phi\\left(3^{a-1}\\right), & \\text { if } a>1 \\\\\n2, & \\text { if } a=1\n\\end{array}\\right.\n\\]\nand, by the multiplicative property of the Euler function,\n\\[\n\\phi(3 k)=\\left\\{\\begin{array}{ll}\n3 \\phi(k), & \\text { if } 3 \\mid k \\\\\n2 \\phi(k), & \\text { if } 3 \\nmid k .\n\\end{array}\\right.\n\\]\n\nTherefore an angle of size \\( 360^{\\circ} / k \\) is trisectible if and only if \\( 3 \\nmid k \\).",
  "vars": [
    "n",
    "k",
    "a"
  ],
  "params": [
    "Q",
    "K",
    "L",
    "M",
    "\\\\phi",
    "\\\\varrho"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "integercount",
        "k": "divisorindex",
        "a": "exponentvalue",
        "Q": "fieldrationals",
        "K": "fieldalpha",
        "L": "fieldbeta",
        "M": "fieldgamma",
        "\\phi": "eulerfunction",
        "\\varrho": "rhosymbol"
      },
      "question": "B-3. Assume that a \\( 60^{\\circ} \\) angle cannot be trisected with ruler and compass alone. Prove that if \\( integercount \\) is a positive multiple of 3 , then no angle of \\( 360 / integercount \\) degrees can be trisected with ruler and compass alone.",
      "solution": "B-3 We need to make use of the following facts about fields and constructibility: (1) If \\( fieldrationals \\) is the field of rational numbers, the degree of \\( fieldrationals \\) extended by \\( \\cos \\left(360^{\\circ} / divisorindex\\right) \\), where \\( divisorindex \\) is a positive integer, is \\( eulerfunction(divisorindex) \\), where \\( eulerfunction \\) is the Euler function. (2) If \\( fieldalpha, fieldbeta, fieldgamma \\) are fields with \\( fieldalpha \\subset fieldbeta \\subset fieldgamma \\) and \\( [fieldbeta: fieldalpha]<\\infty,[fieldgamma: fieldbeta]<\\infty \\) then \\( [fieldgamma: fieldalpha]=[fieldgamma: fieldbeta] \\cdot[fieldbeta: fieldalpha] \\). (3) Given \\( \\cos \\left(360^{\\circ} / divisorindex\\right) \\), then \\( \\cos \\left(360^{\\circ} / 3 divisorindex\\right) \\) is constructible if and only if\n\\[\n\\left[rhosymbol\\left(\\cos \\frac{360^{\\circ}}{3 divisorindex}\\right): fieldrationals\\left(\\cos \\frac{360^{\\circ}}{divisorindex}\\right)\\right]\n\\]\nis a power of 2 .\nConsequently, \\( \\left[fieldrationals\\left(\\cos 360^{\\circ} / 3 divisorindex\\right): fieldrationals\\left(\\cos 360^{\\circ} / divisorindex\\right)\\right] \\cdot eulerfunction(divisorindex)=eulerfunction(3 divisorindex) \\). Now\n\\[\neulerfunction\\left(3^{exponentvalue}\\right)=3^{exponentvalue-1} \\cdot 2=\\left\\{\\begin{array}{ll}\n3 \\, eulerfunction\\left(3^{exponentvalue-1}\\right), & \\text { if } exponentvalue>1 \\\\\n2, & \\text { if } exponentvalue=1\n\\end{array}\\right.\n\\]\nand, by the multiplicative property of the Euler function,\n\\[\neulerfunction(3 divisorindex)=\\left\\{\\begin{array}{ll}\n3 \\, eulerfunction(divisorindex), & \\text { if } 3 \\mid divisorindex \\\\\n2 \\, eulerfunction(divisorindex), & \\text { if } 3 \\nmid divisorindex .\n\\end{array}\\right.\n\\]\n\nTherefore an angle of size \\( 360^{\\circ} / divisorindex \\) is trisectible if and only if \\( 3 \\nmid divisorindex \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "watermelon",
        "k": "blueberry",
        "a": "pineapple",
        "Q": "umbrella",
        "K": "elephantine",
        "L": "hummingbird",
        "M": "toothpick",
        "\\phi": "jellybean",
        "\\varrho": "marshmallow"
      },
      "question": "B-3. Assume that a \\( 60^{\\circ} \\) angle cannot be trisected with ruler and compass alone. Prove that if \\( watermelon \\) is a positive multiple of 3 , then no angle of \\( 360 / watermelon \\) degrees can be trisected with ruler and compass alone.",
      "solution": "B-3 We need to make use of the following facts about fields and constructibility: (1) If \\( umbrella \\) is the field of rational numbers, the degree of \\( umbrella \\) extended by \\( \\cos \\left(360^{\\circ} / blueberry\\right) \\), where \\( blueberry \\) is a positive integer, is \\( jellybean(blueberry) \\), where \\( jellybean \\) is the Euler function. (2) If \\( elephantine, hummingbird, toothpick \\) are fields with \\( elephantine \\subset hummingbird \\subset toothpick \\) and \\( [hummingbird: elephantine]<\\infty,[toothpick: hummingbird]<\\infty \\) then \\( [toothpick: elephantine]=[toothpick: hummingbird] \\cdot[hummingbird: elephantine] \\). (3) Given \\( \\cos \\left(360^{\\circ} / blueberry\\right) \\), then \\( \\cos \\left(360^{\\circ} / 3 blueberry\\right) \\) is constructible if and only if\n\\[\n\\left[marshmallow\\left(\\cos \\frac{360^{\\circ}}{3 blueberry}\\right): umbrella\\left(\\cos \\frac{360^{\\circ}}{blueberry}\\right)\\right]\n\\]\nis a power of 2 .\nConsequently, \\( \\left[umbrella\\left(\\cos 360^{\\circ} / 3 blueberry\\right): umbrella\\left(\\cos 360^{\\circ} / blueberry\\right)\\right] \\cdot jellybean(blueberry)=jellybean(3 blueberry) \\). Now\n\\[\njellybean\\left(3^{pineapple}\\right)=3^{pineapple-1} \\cdot 2=\\left\\{\\begin{array}{ll}\n3 \\, jellybean\\left(3^{pineapple-1}\\right), & \\text { if } pineapple>1 \\\\\n2, & \\text { if } pineapple=1\n\\end{array}\\right.\n\\]\nand, by the multiplicative property of the Euler function,\n\\[\njellybean(3 blueberry)=\\left\\{\\begin{array}{ll}\n3 \\, jellybean(blueberry), & \\text { if } 3 \\mid blueberry \\\\\n2 \\, jellybean(blueberry), & \\text { if } 3 \\nmid blueberry .\n\\end{array}\\right.\n\\]\n\nTherefore an angle of size \\( 360^{\\circ} / blueberry \\) is trisectible if and only if \\( 3 \\nmid blueberry \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "negativemultiple",
        "k": "nonmultiple",
        "a": "unboundedexponent",
        "Q": "irrationalfield",
        "K": "oversetfield",
        "L": "exteriorfield",
        "M": "subfield",
        "\\phi": "noncoprimecount",
        "\\varrho": "imprecisionsym"
      },
      "question": "B-3. Assume that a \\( 60^{\\circ} \\) angle cannot be trisected with ruler and compass alone. Prove that if \\( negativemultiple \\) is a positive multiple of 3 , then no angle of \\( 360 / negativemultiple \\) degrees can be trisected with ruler and compass alone.",
      "solution": "B-3 We need to make use of the following facts about fields and constructibility: (1) If \\( irrationalfield \\) is the field of rational numbers, the degree of \\( irrationalfield \\) extended by \\( \\cos \\left(360^{\\circ} / nonmultiple\\right) \\), where \\( nonmultiple \\) is a positive integer, is \\( noncoprimecount(nonmultiple) \\), where \\( noncoprimecount \\) is the Euler function. (2) If \\( oversetfield, exteriorfield, subfield \\) are fields with \\( oversetfield \\subset exteriorfield \\subset subfield \\) and \\( [exteriorfield: oversetfield]<\\infty,[subfield: exteriorfield]<\\infty \\) then \\( [subfield: oversetfield]=[subfield: exteriorfield] \\cdot[exteriorfield: oversetfield] \\). (3) Given \\( \\cos \\left(360^{\\circ} / nonmultiple\\right) \\), then \\( \\cos \\left(360^{\\circ} / 3 nonmultiple\\right) \\) is constructible if and only if\n\\[\n\\left[imprecisionsym\\left(\\cos \\frac{360^{\\circ}}{3 nonmultiple}\\right): irrationalfield\\left(\\cos \\frac{360^{\\circ}}{nonmultiple}\\right)\\right]\n\\]\nis a power of 2 .\nConsequently, \\( \\left[irrationalfield\\left(\\cos 360^{\\circ} / 3 nonmultiple\\right): irrationalfield\\left(\\cos 360^{\\circ} / nonmultiple\\right)\\right] \\cdot noncoprimecount(nonmultiple)=noncoprimecount(3 nonmultiple) \\). Now\n\\[\nnoncoprimecount\\left(3^{unboundedexponent}\\right)=3^{unboundedexponent-1} \\cdot 2=\\left\\{\\begin{array}{ll}\n3 noncoprimecount\\left(3^{unboundedexponent-1}\\right), & \\text { if } unboundedexponent>1 \\\\\n2, & \\text { if } unboundedexponent=1\n\\end{array}\\right.\n\\]\nand, by the multiplicative property of the Euler function,\n\\[\nnoncoprimecount(3 nonmultiple)=\\left\\{\\begin{array}{ll}\n3 noncoprimecount(nonmultiple), & \\text { if } 3 \\mid nonmultiple \\\\\n2 noncoprimecount(nonmultiple), & \\text { if } 3 \\nmid nonmultiple .\n\\end{array}\\right.\n\\]\n\nTherefore an angle of size \\( 360^{\\circ} / nonmultiple \\) is trisectible if and only if \\( 3 \\nmid nonmultiple \\)."
    },
    "garbled_string": {
      "map": {
        "n": "qlskvtrm",
        "k": "zpwfrdqc",
        "a": "gmxoehva",
        "Q": "xesmyolp",
        "K": "dfgplunk",
        "L": "hqzovtjr",
        "M": "nbqwirca",
        "\\phi": "uvyhjprs",
        "\\varrho": "bczlwnmd"
      },
      "question": "Assume that a \\( 60^{\\circ} \\) angle cannot be trisected with ruler and compass alone. Prove that if qlskvtrm is a positive multiple of 3 , then no angle of 360 / qlskvtrm degrees can be trisected with ruler and compass alone.",
      "solution": "B-3 We need to make use of the following facts about fields and constructibility: (1) If \\( xesmyolp \\) is the field of rational numbers, the degree of \\( xesmyolp \\) extended by \\( \\cos \\left(360^{\\circ} / zpwfrdqc\\right) \\), where \\( zpwfrdqc \\) is a positive integer, is \\( uvyhjprs(zpwfrdqc) \\), where \\( uvyhjprs \\) is the Euler function. (2) If \\( dfgplunk, hqzovtjr, nbqwirca \\) are fields with \\( dfgplunk \\subset hqzovtjr \\subset nbqwirca \\) and \\( [hqzovtjr: dfgplunk]<\\infty,[nbqwirca: hqzovtjr]<\\infty \\) then \\( [nbqwirca: dfgplunk]=[nbqwirca: hqzovtjr] \\cdot[hqzovtjr: dfgplunk] \\). (3) Given \\( \\cos \\left(360^{\\circ} / zpwfrdqc\\right) \\), then \\( \\cos \\left(360^{\\circ} / 3 zpwfrdqc\\right) \\) is constructible if and only if\n\\[\n\\left[bczlwnmd\\left(\\cos \\frac{360^{\\circ}}{3 zpwfrdqc}\\right): xesmyolp\\left(\\cos \\frac{360^{\\circ}}{zpwfrdqc}\\right)\\right]\n\\]\nis a power of 2 . Consequently, \\( \\left[xesmyolp\\left(\\cos 360^{\\circ} / 3 zpwfrdqc\\right): xesmyolp\\left(\\cos 360^{\\circ} / zpwfrdqc\\right)\\right] \\cdot uvyhjprs(zpwfrdqc)=uvyhjprs(3 zpwfrdqc) \\). Now\n\\[\nuvyhjprs\\left(3^{gmxoehva}\\right)=3^{gmxoehva-1} \\cdot 2=\\left\\{\\begin{array}{ll}\n3 \\, uvyhjprs\\left(3^{gmxoehva-1}\\right), & \\text { if } gmxoehva>1 \\\\\n2, & \\text { if } gmxoehva=1\n\\end{array}\\right.\n\\]\nand, by the multiplicative property of the Euler function,\n\\[\nuvyhjprs(3 zpwfrdqc)=\\left\\{\\begin{array}{ll}\n3 \\, uvyhjprs(zpwfrdqc), & \\text { if } 3 \\mid zpwfrdqc \\\\\n2 \\, uvyhjprs(zpwfrdqc), & \\text { if } 3 \\nmid zpwfrdqc .\n\\end{array}\\right.\n\\]\n\nTherefore an angle of size \\( 360^{\\circ} / zpwfrdqc \\) is trisectible if and only if \\( 3 \\nmid zpwfrdqc \\)."
    },
    "kernel_variant": {
      "question": "Let $p\\ge 5$ be a fixed odd prime.  \nWrite every positive integer $n$ uniquely as $n=p^{k}m$ with $k\\ge 0$ and $p\\nmid m$, and put $\\displaystyle\\theta_{n}:=\\dfrac{2\\pi}{n}$.\n\n1.  Arithmetic of real cyclotomic fields.  \n   For $s\\ge 1$ set  \n   \\[\n        K_{s}:=\\mathbb{Q}\\!\\bigl(\\cos(2\\pi/s)\\bigr)\\subset\\mathbb{R}.\n   \\]\n\n   (a)  Prove that $K_{s}=\\mathbb{Q}$ if and only if $s\\in\\{1,2,3,4,6\\}$.\n\n   (b)  Show that for every $k\\ge 1$\n        \\[\n            \\bigl[K_{p^{k}m}:K_{p^{\\,k-1}m}\\bigr]=\n            \\begin{cases}\n               \\dfrac{p-1}{2}, &\\text{if }k=1\\text{ and }m\\in\\{1,2\\},\\\\[6pt]\n               p-1,            &\\text{if }k=1\\text{ and }m\\notin\\{1,2\\},\\\\[6pt]\n               p,              &\\text{if }k\\ge 2.\n            \\end{cases}\\tag{$\\star$}\n        \\]\n        In particular,\n        \\[\n            \\bigl[K_{p^{k+1}m}:K_{p^{k}m}\\bigr]=p\\qquad(k\\ge 1).\\tag{$\\star\\star$}\n        \\]\n\n2.  $p$-section of the angle $\\theta_{p^{\\,k+1}m}$.  \n\n   (a)  Fix $k\\ge 0$ and set  \n        \\[\n            \\beta:=\\theta_{p^{\\,k+1}m},\\qquad    \n            \\alpha:=\\cos\\!\\Bigl(\\beta/p\\Bigr)=\\cos\\!\\Bigl(\\dfrac{2\\pi}{p^{\\,k+2}m}\\Bigr).\n        \\]\n        Prove that\n        \\[\n            \\bigl[K_{p^{\\,k+1}m}(\\alpha):K_{p^{\\,k+1}m}\\bigr]=p.\\tag{$\\ddagger$}\n        \\]\n\n   (b)  Deduce that $\\beta$ cannot be $p$-sected with straight-edge and compass, because any Euclidean construction produces only towers of quadratic extensions.\n\n   (c)  Show further that for $k\\ge 1$\n        \\[\n            K_{p^{\\,k+1}m}(\\alpha)=K_{p^{\\,k+2}m},\n            \\qquad\n            \\bigl[K_{p^{k}m}(\\alpha):K_{p^{k}m}\\bigr]=p^{2}.\n        \\]\n        Explain why starting the construction from the smaller field $K_{p^{k}m}$ yields the degree $p^{2}$, thereby clarifying the frequent pitfall of using an inappropriate base field.\n\n3.  Galois-theoretic interpretation.    \n\n   (a)  Put $\\zeta_{s}:=e^{2\\pi i/s}$ and identify  \n        \\[\n            G_{k}:=\\operatorname{Gal}\\!\\bigl(\\mathbb{Q}(\\zeta_{p^{k}m})/\\mathbb{Q}\\bigr)\n                  \\simeq(\\mathbb{Z}/p^{k}m\\mathbb{Z})^{\\times}.\n        \\]\n        For $k\\ge 1$ define  \n        \\[\n            H_{k}:=\\bigl\\{x\\in(\\mathbb{Z}/p^{k}m\\mathbb{Z})^{\\times}\\;:\\;\n                      x\\equiv 1\\pmod{p^{\\,k-1}m}\\bigr\\}.\n        \\]\n        Show that $|H_{1}|=p-1$ while $|H_{k}|=p$ for $k\\ge 2$.\n\n   (b)  Prove that for every $k\\ge 1$ the fixed field of $H_{k}$ inside the maximal real subfield $K_{p^{k}m}$ is exactly $K_{p^{\\,k-1}m}$.\n\n   (c)  Using (b) with $k\\mapsto k+2$, confirm $(\\ddagger)$ and the non-constructibility statement in 2(b).\n\n\\bigskip",
      "solution": "Throughout write $n=p^{k}m$ with $p\\nmid m$, put  \n$\\zeta_{s}=e^{2\\pi i/s}$, and recall  \n\\[\n   K_{s}=\\mathbb{Q}(\\zeta_{s}+\\zeta_{s}^{-1})\n        \\subset\\mathbb{Q}(\\zeta_{s}),\\qquad\n   \\bigl[\\mathbb{Q}(\\zeta_{s}):\\mathbb{Q}\\bigr]=\\varphi(s).\n\\]\nBecause complex conjugation has fixed field $K_{s}$,  \n\\[\n   [K_{s}:\\mathbb{Q}]\n        =\n   \\begin{cases}\n        \\dfrac{\\varphi(s)}{2},&s\\notin\\{1,2,3,4,6\\},\\\\[4pt]\n        1,&s\\in\\{1,2,3,4,6\\}.\n   \\end{cases}\\tag{1}\n\\]\n\n--------------------------------------------------------------------\n1.  Proof of $(\\star)$ and $(\\star\\star)$.\n\nSince $(p,m)=1$ we have  \n\\[\n   \\varphi(p^{k}m)=p^{k-1}(p-1)\\varphi(m).\n\\]\nInsert this into (1) for $k\\ge 1$ to obtain  \n\\[\n   [K_{p^{k}m}:\\mathbb{Q}]\n        =\\dfrac{p^{\\,k-1}(p-1)\\varphi(m)}{2}.\\tag{2}\n\\]\n\n(a)  The case $k=1$.  \n\nIf $m\\in\\{1,2\\}$ then $\\varphi(m)=1$ and $K_{m}=\\mathbb{Q}$, so  \n\\[\n   [K_{pm}:K_{m}]\n      =[K_{pm}:\\mathbb{Q}]\n      =\\dfrac{p-1}{2}.\n\\]\n\nIf $m\\notin\\{1,2\\}$, then either $m\\in\\{3,4,6\\}$ (for which $K_{m}=\\mathbb{Q}$ and $\\varphi(m)=2$) or $m\\ge 5$ (for which $[K_{m}:\\mathbb{Q}]=\\varphi(m)/2$).  \nIn both sub-cases\n\\[\n   [K_{pm}:K_{m}]\n      =\\dfrac{\\tfrac12(p-1)\\varphi(m)}{[K_{m}:\\mathbb{Q}]}\n      =p-1.\n\\]\n\n(b)  The case $k\\ge 2$.  \n\nDividing consecutive degrees given by (2) we obtain\n\\[\n   [K_{p^{k}m}:K_{p^{\\,k-1}m}]\n      =\\dfrac{p^{\\,k-1}(p-1)}{2}\\bigg/\\dfrac{p^{\\,k-2}(p-1)}{2}=p,\n\\]\nwhich is the third line of $(\\star)$.  With $k\\ge 1$ this yields\n\\[\n   [K_{p^{k+1}m}:K_{p^{k}m}]=p,\\tag{$\\star\\star$}\n\\]\nestablishing $(\\star\\star)$ in its correct range.  \n\\qed\n\n--------------------------------------------------------------------\n2.  The $p$-section problem.\n\nFix $k\\ge 0$ and put  \n\\[\n   \\beta:=\\theta_{p^{\\,k+1}m},\\qquad\n   \\alpha:=\\cos\\!\\Bigl(\\beta/p\\Bigr)=\\cos\\!\\Bigl(\\dfrac{2\\pi}{p^{\\,k+2}m}\\Bigr)\n   \\in K_{p^{\\,k+2}m}.\n\\]\n\n----------------------------------------------------------------\n2(a).  Degree computation over $K_{p^{\\,k+1}m}$.\n\nBecause $k+1\\ge 1$, $(\\star\\star)$ gives  \n\\[\n   \\bigl[K_{p^{\\,k+2}m}:K_{p^{\\,k+1}m}\\bigr]=p.\\tag{3}\n\\]\n\nWe prove that $\\alpha\\notin K_{p^{\\,k+1}m}$ by exhibiting an automorphism\nof $K_{p^{\\,k+2}m}$ that fixes $K_{p^{\\,k+1}m}$ but moves $\\alpha$.\n\nLet  \n\\[\n   \\sigma:\\zeta_{p^{\\,k+2}m}\\longmapsto\n          \\zeta_{p^{\\,k+2}m}^{\\,1+p^{\\,k+1}m}.\n\\]\nBecause $1+p^{\\,k+1}m\\equiv 1\\pmod{p^{\\,k+1}m}$, the automorphism $\\sigma$\nacts trivially on $K_{p^{\\,k+1}m}$.\n\nHowever,\n\\[\n   \\sigma(\\alpha)\n      =\\cos\\!\\Bigl(\\tfrac{1+p^{\\,k+1}m}{p^{\\,k+2}m}\\,2\\pi\\Bigr)\n      =\\cos\\!\\Bigl(\\dfrac{2\\pi}{p^{\\,k+2}m}+\\dfrac{2\\pi}{p}\\Bigr)\n      \\neq\\cos\\!\\Bigl(\\dfrac{2\\pi}{p^{\\,k+2}m}\\Bigr)=\\alpha,\n\\]\nbecause $p\\ge 5$ implies $\\dfrac{2\\pi}{p}\\not\\equiv 0\\pmod{2\\pi}$.\nThus $\\alpha$ is not fixed by $\\sigma$ and therefore\n$\\alpha\\notin K_{p^{\\,k+1}m}$.\n\nConsequently the minimal polynomial of $\\alpha$ over $K_{p^{\\,k+1}m}$\nhas degree divisible by\n\\[\n   |\\langle\\sigma\\rangle|=p.\n\\]\nSince the whole extension in (3) already has degree $p$, the degree equals\n$p$, whence\n\\[\n   [K_{p^{\\,k+1}m}(\\alpha):K_{p^{\\,k+1}m}]=p,\n\\]\nestablishing $(\\ddagger)$.\n\n----------------------------------------------------------------\n2(b).  Non-constructibility.\n\nAny straight-edge-and-compass construction starting from\n$K_{p^{\\,k+1}m}$ yields fields obtained by successive quadratic\nextensions.  Hence every constructible element lies in a tower whose\ntotal degree over $K_{p^{\\,k+1}m}$ is a power of $2$.\nBecause $(\\ddagger)$ furnishes an odd prime divisor $p$ of that degree,\n$\\alpha$ (equivalently $\\beta/p$) is not constructible, and the angle\n$\\beta$ cannot be $p$-sected.\n\n----------------------------------------------------------------\n2(c).  Working over the smaller field $K_{p^{k}m}$ ($k\\ge 1$).\n\nUsing $(\\star\\star)$ twice we get\n\\[\n   [K_{p^{\\,k+2}m}:K_{p^{k}m}]\n      =[K_{p^{\\,k+2}m}:K_{p^{\\,k+1}m}]\\,\n       [K_{p^{\\,k+1}m}:K_{p^{k}m}]\n      =p\\cdot p=p^{2}.\n\\]\nBecause $\\alpha\\notin K_{p^{\\,k+1}m}$ it certainly is not in\n$K_{p^{k}m}$, so its degree over $K_{p^{k}m}$ is the full $p^{2}$ and\n\\[\n   K_{p^{k}m}(\\alpha)=K_{p^{\\,k+2}m}.\n\\]\nThis confirms the necessity of choosing the correct base field when\ncomputing degrees in constructibility questions.\n\\qed\n\n--------------------------------------------------------------------\n3.  Galois-theoretic explanation.\n\nLet\n\\[\n   L_{k}:=\\mathbb{Q}(\\zeta_{p^{k}m}),\\qquad\n   G_{k}:=\\operatorname{Gal}(L_{k}/\\mathbb{Q})\n         \\simeq(\\mathbb{Z}/p^{k}m\\mathbb{Z})^{\\times},\n\\]\nand denote complex conjugation by $\\iota$.\n\n----------------------------------------------------------------\n3(a).  The subgroups $H_{k}$.\n\nReduction modulo $p^{\\,k-1}m$ gives a surjection\n\\[\n   \\rho_{k}:G_{k}\\longrightarrow(\\mathbb{Z}/p^{\\,k-1}m\\mathbb{Z})^{\\times}\n\\]\nwhose kernel is\n\\[\n   H_{k}=\\ker\\rho_{k}\n        =\\{x\\equiv 1\\pmod{p^{\\,k-1}m}\\}.\n\\]\nA direct count shows $|H_{1}|=p-1$ and $|H_{k}|=p$ for $k\\ge 2$.\n\n----------------------------------------------------------------\n3(b).  Fixed fields.\n\nInside the maximal real subfield\n$K_{p^{k}m}=L_{k}^{\\langle\\iota\\rangle}$ consider the subgroup\n\\[\n   H_{k}\\langle\\iota\\rangle/\\langle\\iota\\rangle\n      \\;\\le\\;\\operatorname{Gal}(K_{p^{k}m}/\\mathbb{Q}).\n\\]\nFor $k\\ge 2$ this subgroup has order $p$, so its fixed field is an\nindex-$p$ subfield of $K_{p^{k}m}$; by $(\\star)$ this fixed field is\n$K_{p^{\\,k-1}m}$.  For $k=1$ an analogous argument yields the two\npossible indices $\\dfrac{p-1}{2}$ and $p-1$.\n\n----------------------------------------------------------------\n3(c).  Verification of $(\\ddagger)$.\n\nReplacing $k$ by $k+2$ in (b) shows that\n$K_{p^{\\,k+2}m}$ is the unique index-$p$ extension of $K_{p^{\\,k+1}m}$\ninside $L_{k+2}$.  Since $\\alpha$ is moved by the automorphism $\\sigma$\nfrom Section 2(a), it lies outside $K_{p^{\\,k+1}m}$ and therefore\ngenerates that unique degree-$p$ extension, confirming $(\\ddagger)$ and\nagain precluding constructibility.\n\n----------------------------------------------------------------\nConsequences.\n\nIf $p\\ge 5$ divides $n$, then the angle $\\theta_{n}=2\\pi/n$ is not\n$p$-sectible.  Multiplying $n$ by any power of $2$ does not affect this\nobstruction, because it is encoded in the $p$-primary part of the\nassociated real cyclotomic field.  \n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.580064",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher-order prime powers.  \n    The original problems dealt with a single prime divisor of n.  Here the statement must hold simultaneously for every power pᵏ dividing n, and the proof must track how the obstruction propagates up the tower n↦pn↦p²n↦… .\n\n2.  Precise index computation.  \n    One must compute the exact degree of the field extension (★), not merely show that it is divisible by p.  This requires explicit use of the Euler-totient formula and of cyclotomic-field Galois theory rather than the elementary degree-counting used in the original solution.\n\n3.  Explicit Galois-group subgroup.  \n    The solver must identify the subgroup Gal(L_k/L_{k−1}) inside (ℤ/pᵏmℤ)×, relate it to the real subfield via complex conjugation, and prove that its order forbids constructibility.  Thus the argument demands familiarity with the interplay between complex conjugation, real subfields, and subgroup indices—topics absent from the original problem.\n\n4.  Robust obstruction.  \n    The enhanced statement shows that adding any number of quadratic extensions (i.e., adjoining as many “power-of-2 steps” as one wishes) can never overcome the prime obstruction.  This calls for a deeper appreciation of constructibility than the binary “possible/impossible” conclusion of the original problem.\n\n5.  Multiple interacting concepts.  \n    The solution intertwines cyclotomic fields, Euler’s totient function, real subfields, constructibility criteria, subgroup indices, and the incompatibility of odd primes with powers of two, thereby demanding a broader arsenal of algebraic-number-theoretic tools than the original or kernel variants."
      }
    },
    "original_kernel_variant": {
      "question": "Let $p\\ge 5$ be a fixed odd prime.  \nWrite every positive integer $n$ uniquely as $n=p^{k}m$ with $k\\ge 0$ and $p\\nmid m$, and put $\\displaystyle\\theta_{n}:=\\dfrac{2\\pi}{n}$.\n\n1.  Arithmetic of real cyclotomic fields.  \n   For $s\\ge 1$ set  \n   \\[\n        K_{s}:=\\mathbb{Q}\\!\\bigl(\\cos(2\\pi/s)\\bigr)\\subset\\mathbb{R}.\n   \\]\n\n   (a)  Prove that $K_{s}=\\mathbb{Q}$ if and only if $s\\in\\{1,2,3,4,6\\}$.\n\n   (b)  Show that for every $k\\ge 1$\n        \\[\n            \\bigl[K_{p^{k}m}:K_{p^{\\,k-1}m}\\bigr]=\n            \\begin{cases}\n               \\dfrac{p-1}{2}, &\\text{if }k=1\\text{ and }m\\in\\{1,2\\},\\\\[6pt]\n               p-1,            &\\text{if }k=1\\text{ and }m\\notin\\{1,2\\},\\\\[6pt]\n               p,              &\\text{if }k\\ge 2.\n            \\end{cases}\\tag{$\\star$}\n        \\]\n        In particular,\n        \\[\n            \\bigl[K_{p^{k+1}m}:K_{p^{k}m}\\bigr]=p\\qquad(k\\ge 1).\\tag{$\\star\\star$}\n        \\]\n\n2.  $p$-section of the angle $\\theta_{p^{\\,k+1}m}$.  \n\n   (a)  Fix $k\\ge 0$ and set  \n        \\[\n            \\beta:=\\theta_{p^{\\,k+1}m},\\qquad    \n            \\alpha:=\\cos\\!\\Bigl(\\beta/p\\Bigr)=\\cos\\!\\Bigl(\\dfrac{2\\pi}{p^{\\,k+2}m}\\Bigr).\n        \\]\n        Prove that\n        \\[\n            \\bigl[K_{p^{\\,k+1}m}(\\alpha):K_{p^{\\,k+1}m}\\bigr]=p.\\tag{$\\ddagger$}\n        \\]\n\n   (b)  Deduce that $\\beta$ cannot be $p$-sected with straight-edge and compass, because any Euclidean construction produces only towers of quadratic extensions.\n\n   (c)  Show further that for $k\\ge 1$\n        \\[\n            K_{p^{\\,k+1}m}(\\alpha)=K_{p^{\\,k+2}m},\n            \\qquad\n            \\bigl[K_{p^{k}m}(\\alpha):K_{p^{k}m}\\bigr]=p^{2}.\n        \\]\n        Explain why starting the construction from the smaller field $K_{p^{k}m}$ yields the degree $p^{2}$, thereby clarifying the frequent pitfall of using an inappropriate base field.\n\n3.  Galois-theoretic interpretation.    \n\n   (a)  Put $\\zeta_{s}:=e^{2\\pi i/s}$ and identify  \n        \\[\n            G_{k}:=\\operatorname{Gal}\\!\\bigl(\\mathbb{Q}(\\zeta_{p^{k}m})/\\mathbb{Q}\\bigr)\n                  \\simeq(\\mathbb{Z}/p^{k}m\\mathbb{Z})^{\\times}.\n        \\]\n        For $k\\ge 1$ define  \n        \\[\n            H_{k}:=\\bigl\\{x\\in(\\mathbb{Z}/p^{k}m\\mathbb{Z})^{\\times}\\;:\\;\n                      x\\equiv 1\\pmod{p^{\\,k-1}m}\\bigr\\}.\n        \\]\n        Show that $|H_{1}|=p-1$ while $|H_{k}|=p$ for $k\\ge 2$.\n\n   (b)  Prove that for every $k\\ge 1$ the fixed field of $H_{k}$ inside the maximal real subfield $K_{p^{k}m}$ is exactly $K_{p^{\\,k-1}m}$.\n\n   (c)  Using (b) with $k\\mapsto k+2$, confirm $(\\ddagger)$ and the non-constructibility statement in 2(b).\n\n\\bigskip",
      "solution": "Throughout write $n=p^{k}m$ with $p\\nmid m$, put  \n$\\zeta_{s}=e^{2\\pi i/s}$, and recall  \n\\[\n   K_{s}=\\mathbb{Q}(\\zeta_{s}+\\zeta_{s}^{-1})\n        \\subset\\mathbb{Q}(\\zeta_{s}),\\qquad\n   \\bigl[\\mathbb{Q}(\\zeta_{s}):\\mathbb{Q}\\bigr]=\\varphi(s).\n\\]\nBecause complex conjugation has fixed field $K_{s}$,  \n\\[\n   [K_{s}:\\mathbb{Q}]\n        =\n   \\begin{cases}\n        \\dfrac{\\varphi(s)}{2},&s\\notin\\{1,2,3,4,6\\},\\\\[4pt]\n        1,&s\\in\\{1,2,3,4,6\\}.\n   \\end{cases}\\tag{1}\n\\]\n\n--------------------------------------------------------------------\n1.  Proof of $(\\star)$ and $(\\star\\star)$.\n\nSince $(p,m)=1$ we have  \n\\[\n   \\varphi(p^{k}m)=p^{k-1}(p-1)\\varphi(m).\n\\]\nInsert this into (1) for $k\\ge 1$ to obtain  \n\\[\n   [K_{p^{k}m}:\\mathbb{Q}]\n        =\\dfrac{p^{\\,k-1}(p-1)\\varphi(m)}{2}.\\tag{2}\n\\]\n\n(a)  The case $k=1$.  \n\nIf $m\\in\\{1,2\\}$ then $\\varphi(m)=1$ and $K_{m}=\\mathbb{Q}$, so  \n\\[\n   [K_{pm}:K_{m}]\n      =[K_{pm}:\\mathbb{Q}]\n      =\\dfrac{p-1}{2}.\n\\]\n\nIf $m\\notin\\{1,2\\}$, then either $m\\in\\{3,4,6\\}$ (for which $K_{m}=\\mathbb{Q}$ and $\\varphi(m)=2$) or $m\\ge 5$ (for which $[K_{m}:\\mathbb{Q}]=\\varphi(m)/2$).  \nIn both sub-cases\n\\[\n   [K_{pm}:K_{m}]\n      =\\dfrac{\\tfrac12(p-1)\\varphi(m)}{[K_{m}:\\mathbb{Q}]}\n      =p-1.\n\\]\n\n(b)  The case $k\\ge 2$.  \n\nDividing consecutive degrees given by (2) we obtain\n\\[\n   [K_{p^{k}m}:K_{p^{\\,k-1}m}]\n      =\\dfrac{p^{\\,k-1}(p-1)}{2}\\bigg/\\dfrac{p^{\\,k-2}(p-1)}{2}=p,\n\\]\nwhich is the third line of $(\\star)$.  With $k\\ge 1$ this yields\n\\[\n   [K_{p^{k+1}m}:K_{p^{k}m}]=p,\\tag{$\\star\\star$}\n\\]\nestablishing $(\\star\\star)$ in its correct range.  \n\\qed\n\n--------------------------------------------------------------------\n2.  The $p$-section problem.\n\nFix $k\\ge 0$ and put  \n\\[\n   \\beta:=\\theta_{p^{\\,k+1}m},\\qquad\n   \\alpha:=\\cos\\!\\Bigl(\\beta/p\\Bigr)=\\cos\\!\\Bigl(\\dfrac{2\\pi}{p^{\\,k+2}m}\\Bigr)\n   \\in K_{p^{\\,k+2}m}.\n\\]\n\n----------------------------------------------------------------\n2(a).  Degree computation over $K_{p^{\\,k+1}m}$.\n\nBecause $k+1\\ge 1$, $(\\star\\star)$ gives  \n\\[\n   \\bigl[K_{p^{\\,k+2}m}:K_{p^{\\,k+1}m}\\bigr]=p.\\tag{3}\n\\]\n\nWe prove that $\\alpha\\notin K_{p^{\\,k+1}m}$ by exhibiting an automorphism\nof $K_{p^{\\,k+2}m}$ that fixes $K_{p^{\\,k+1}m}$ but moves $\\alpha$.\n\nLet  \n\\[\n   \\sigma:\\zeta_{p^{\\,k+2}m}\\longmapsto\n          \\zeta_{p^{\\,k+2}m}^{\\,1+p^{\\,k+1}m}.\n\\]\nBecause $1+p^{\\,k+1}m\\equiv 1\\pmod{p^{\\,k+1}m}$, the automorphism $\\sigma$\nacts trivially on $K_{p^{\\,k+1}m}$.\n\nHowever,\n\\[\n   \\sigma(\\alpha)\n      =\\cos\\!\\Bigl(\\tfrac{1+p^{\\,k+1}m}{p^{\\,k+2}m}\\,2\\pi\\Bigr)\n      =\\cos\\!\\Bigl(\\dfrac{2\\pi}{p^{\\,k+2}m}+\\dfrac{2\\pi}{p}\\Bigr)\n      \\neq\\cos\\!\\Bigl(\\dfrac{2\\pi}{p^{\\,k+2}m}\\Bigr)=\\alpha,\n\\]\nbecause $p\\ge 5$ implies $\\dfrac{2\\pi}{p}\\not\\equiv 0\\pmod{2\\pi}$.\nThus $\\alpha$ is not fixed by $\\sigma$ and therefore\n$\\alpha\\notin K_{p^{\\,k+1}m}$.\n\nConsequently the minimal polynomial of $\\alpha$ over $K_{p^{\\,k+1}m}$\nhas degree divisible by\n\\[\n   |\\langle\\sigma\\rangle|=p.\n\\]\nSince the whole extension in (3) already has degree $p$, the degree equals\n$p$, whence\n\\[\n   [K_{p^{\\,k+1}m}(\\alpha):K_{p^{\\,k+1}m}]=p,\n\\]\nestablishing $(\\ddagger)$.\n\n----------------------------------------------------------------\n2(b).  Non-constructibility.\n\nAny straight-edge-and-compass construction starting from\n$K_{p^{\\,k+1}m}$ yields fields obtained by successive quadratic\nextensions.  Hence every constructible element lies in a tower whose\ntotal degree over $K_{p^{\\,k+1}m}$ is a power of $2$.\nBecause $(\\ddagger)$ furnishes an odd prime divisor $p$ of that degree,\n$\\alpha$ (equivalently $\\beta/p$) is not constructible, and the angle\n$\\beta$ cannot be $p$-sected.\n\n----------------------------------------------------------------\n2(c).  Working over the smaller field $K_{p^{k}m}$ ($k\\ge 1$).\n\nUsing $(\\star\\star)$ twice we get\n\\[\n   [K_{p^{\\,k+2}m}:K_{p^{k}m}]\n      =[K_{p^{\\,k+2}m}:K_{p^{\\,k+1}m}]\\,\n       [K_{p^{\\,k+1}m}:K_{p^{k}m}]\n      =p\\cdot p=p^{2}.\n\\]\nBecause $\\alpha\\notin K_{p^{\\,k+1}m}$ it certainly is not in\n$K_{p^{k}m}$, so its degree over $K_{p^{k}m}$ is the full $p^{2}$ and\n\\[\n   K_{p^{k}m}(\\alpha)=K_{p^{\\,k+2}m}.\n\\]\nThis confirms the necessity of choosing the correct base field when\ncomputing degrees in constructibility questions.\n\\qed\n\n--------------------------------------------------------------------\n3.  Galois-theoretic explanation.\n\nLet\n\\[\n   L_{k}:=\\mathbb{Q}(\\zeta_{p^{k}m}),\\qquad\n   G_{k}:=\\operatorname{Gal}(L_{k}/\\mathbb{Q})\n         \\simeq(\\mathbb{Z}/p^{k}m\\mathbb{Z})^{\\times},\n\\]\nand denote complex conjugation by $\\iota$.\n\n----------------------------------------------------------------\n3(a).  The subgroups $H_{k}$.\n\nReduction modulo $p^{\\,k-1}m$ gives a surjection\n\\[\n   \\rho_{k}:G_{k}\\longrightarrow(\\mathbb{Z}/p^{\\,k-1}m\\mathbb{Z})^{\\times}\n\\]\nwhose kernel is\n\\[\n   H_{k}=\\ker\\rho_{k}\n        =\\{x\\equiv 1\\pmod{p^{\\,k-1}m}\\}.\n\\]\nA direct count shows $|H_{1}|=p-1$ and $|H_{k}|=p$ for $k\\ge 2$.\n\n----------------------------------------------------------------\n3(b).  Fixed fields.\n\nInside the maximal real subfield\n$K_{p^{k}m}=L_{k}^{\\langle\\iota\\rangle}$ consider the subgroup\n\\[\n   H_{k}\\langle\\iota\\rangle/\\langle\\iota\\rangle\n      \\;\\le\\;\\operatorname{Gal}(K_{p^{k}m}/\\mathbb{Q}).\n\\]\nFor $k\\ge 2$ this subgroup has order $p$, so its fixed field is an\nindex-$p$ subfield of $K_{p^{k}m}$; by $(\\star)$ this fixed field is\n$K_{p^{\\,k-1}m}$.  For $k=1$ an analogous argument yields the two\npossible indices $\\dfrac{p-1}{2}$ and $p-1$.\n\n----------------------------------------------------------------\n3(c).  Verification of $(\\ddagger)$.\n\nReplacing $k$ by $k+2$ in (b) shows that\n$K_{p^{\\,k+2}m}$ is the unique index-$p$ extension of $K_{p^{\\,k+1}m}$\ninside $L_{k+2}$.  Since $\\alpha$ is moved by the automorphism $\\sigma$\nfrom Section 2(a), it lies outside $K_{p^{\\,k+1}m}$ and therefore\ngenerates that unique degree-$p$ extension, confirming $(\\ddagger)$ and\nagain precluding constructibility.\n\n----------------------------------------------------------------\nConsequences.\n\nIf $p\\ge 5$ divides $n$, then the angle $\\theta_{n}=2\\pi/n$ is not\n$p$-sectible.  Multiplying $n$ by any power of $2$ does not affect this\nobstruction, because it is encoded in the $p$-primary part of the\nassociated real cyclotomic field.  \n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.468255",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher-order prime powers.  \n    The original problems dealt with a single prime divisor of n.  Here the statement must hold simultaneously for every power pᵏ dividing n, and the proof must track how the obstruction propagates up the tower n↦pn↦p²n↦… .\n\n2.  Precise index computation.  \n    One must compute the exact degree of the field extension (★), not merely show that it is divisible by p.  This requires explicit use of the Euler-totient formula and of cyclotomic-field Galois theory rather than the elementary degree-counting used in the original solution.\n\n3.  Explicit Galois-group subgroup.  \n    The solver must identify the subgroup Gal(L_k/L_{k−1}) inside (ℤ/pᵏmℤ)×, relate it to the real subfield via complex conjugation, and prove that its order forbids constructibility.  Thus the argument demands familiarity with the interplay between complex conjugation, real subfields, and subgroup indices—topics absent from the original problem.\n\n4.  Robust obstruction.  \n    The enhanced statement shows that adding any number of quadratic extensions (i.e., adjoining as many “power-of-2 steps” as one wishes) can never overcome the prime obstruction.  This calls for a deeper appreciation of constructibility than the binary “possible/impossible” conclusion of the original problem.\n\n5.  Multiple interacting concepts.  \n    The solution intertwines cyclotomic fields, Euler’s totient function, real subfields, constructibility criteria, subgroup indices, and the incompatibility of odd primes with powers of two, thereby demanding a broader arsenal of algebraic-number-theoretic tools than the original or kernel variants."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}