{
  "index": "1985-A-6",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "If $p(x)= a_0 + a_1 x + \\cdots + a_m x^m$ is a polynomial with real\ncoefficients $a_i$, then set\n\\[\n\\Gamma(p(x)) = a_0^2 + a_1^2 + \\cdots + a_m^2.\n\\]\nLet $F(x) = 3x^2+7x+2$. Find, with proof, a polynomial $g(x)$ with\nreal coefficients such that\n\\begin{enumerate}\n\\item[(i)] $g(0)=1$, and\n\\item[(ii)] $\\Gamma(f(x)^n) = \\Gamma(g(x)^n)$\n\\end{enumerate}\nfor every integer $n \\geq 1$.",
  "solution": "Solution. For any polynomial \\( p(x) \\), let \\( \\gamma(p(x))=p(x) p\\left(x^{-1}\\right) \\), which is a Laurent polynomial (an expression of the form \\( \\sum_{j=m}^{n} a_{j} x^{j} \\) where \\( a_{j} \\) are constants and \\( m, n \\) are integers, not necessarily nonnegative). Then \\( \\Gamma(p(x)) \\) equals the coefficient of \\( x^{0} \\) in \\( \\gamma(p(x)) \\).\n\nWe have \\( f(x)=(3 x+1)(x+2) \\). Since\n\\[\n\\gamma(x+2)=(x+2)\\left(x^{-1}+2\\right)=\\left(1+2 x^{-1}\\right)(1+2 x)=\\gamma(1+2 x),\n\\]\nand \\( \\gamma(p(x) q(x))=\\gamma(p(x)) \\gamma(q(x)) \\) for any polynomials \\( p(x) \\) and \\( q(x) \\), we find\n\\[\n\\gamma\\left(f(x)^{n}\\right)=\\gamma\\left((3 x+1)^{n}\\right) \\gamma\\left((x+2)^{n}\\right)=\\gamma\\left((3 x+1)^{n}\\right) \\gamma\\left((1+2 x)^{n}\\right)=\\gamma\\left(g(x)^{n}\\right),\n\\]\nwhere \\( g(x)=(3 x+1)(1+2 x)=6 x^{2}+5 x+1 \\). Taking coefficients of \\( x^{0} \\), we obtain \\( \\Gamma\\left(f(x)^{n}\\right)=\\Gamma\\left(g(x)^{n}\\right) \\). Moreover \\( g(0)=1 \\), so we are done.\n\nRemark. One could also show by brute force that \\( \\Gamma\\left(\\left(\\sum a_{i} x^{i}\\right)\\left(\\sum b_{j} x^{j}\\right)\\right) \\) is unchanged by reversing the order of the \\( b_{j} \\), without mentioning \\( \\gamma \\) or Laurent polynomials. The coefficient of \\( x^{0} \\) in a Laurent polynomial can also be expressed as a contour integral, via the residue theorem; hence one could begin a solution with the observation\n\\[\n\\Gamma(p(x))=\\frac{1}{2 \\pi i} \\oint_{|z|=1} p(z) p\\left(z^{-1}\\right) \\frac{d z}{z}\n\\]\n\nRemark. The solution is not unique: for any integer \\( k \\geq 1 \\), the polynomials \\( g(x)=\\left(3 x^{k}+1\\right)\\left(2 x^{k}+1\\right) \\) and \\( g(x)=\\left(3 x^{k}-1\\right)\\left(2 x^{k}-1\\right) \\) also have the desired properties. We do not know if these are the only ones. Using the fact that the ring \\( \\mathbb{R}[x, 1 / x] \\) of Laurent polynomials is a unique factorization domain (UFD), we could prove that these are the only ones if we could prove the following conjecture of Greg Kuperberg (communicated electronically):\n\nSuppose \\( h_{1}, h_{2} \\in \\mathbb{R}[x, 1 / x] \\) satisfy \\( h_{1}(x)=h_{1}(1 / x) \\) and \\( h_{2}(x)=h_{2}(1 / x) \\).\nThen the following are equivalent:\n(a) The coefficient of \\( x^{0} \\) in \\( h_{1}(x)^{n} \\) equals the coefficient of \\( x^{0} \\) of \\( h_{2}(x)^{n} \\) for all positive integers \\( n \\).\n(b) There exists \\( j \\in \\mathbb{R}[x, 1 / x] \\) such that for \\( i=1,2 \\) we have \\( h_{i}(x)=j\\left(\\epsilon_{i} x^{k_{i}}\\right) \\) for some \\( \\epsilon_{i} \\in\\{1,-1\\} \\) and \\( k_{i} \\geq 1 \\).\n(It is easy to prove that (b) implies (a).)",
  "vars": [
    "x",
    "n",
    "k",
    "z"
  ],
  "params": [
    "a_0",
    "a_1",
    "a_m",
    "a_i",
    "p",
    "F",
    "g",
    "f",
    "m",
    "j",
    "h_1",
    "h_2",
    "k_i",
    "\\\\Gamma",
    "\\\\gamma",
    "\\\\epsilon_i"
  ],
  "sci_consts": [
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "n": "exponentcounter",
        "k": "indexfactor",
        "z": "complexvar",
        "a_0": "coeffzero",
        "a_1": "coeffone",
        "a_m": "coeffem",
        "a_i": "coeffvar",
        "p": "polyfun",
        "F": "targetpoly",
        "g": "resultpoly",
        "f": "givenpoly",
        "m": "degreemax",
        "j": "integervar",
        "h_1": "laurone",
        "h_2": "laurtwo",
        "k_i": "indexset",
        "\\Gamma": "sumsquare",
        "\\gamma": "laurop",
        "\\epsilon_i": "signchoice"
      },
      "question": "If $polyfun(variablex)= coeffzero + coeffone\\,variablex + \\cdots + coeffem\\,variablex^{degreemax}$ is a polynomial with real coefficients $coeffvar$, then set\n\\[\nsumsquare(polyfun(variablex)) = coeffzero^{2} + coeffone^{2} + \\cdots + coeffem^{2}.\n\\]\nLet $targetpoly(variablex) = 3\\,variablex^{2}+7\\,variablex+2$. Find, with proof, a polynomial $resultpoly(variablex)$ with\nreal coefficients such that\n\\begin{enumerate}\n\\item[(i)] $resultpoly(0)=1$, and\n\\item[(ii)] $sumsquare(givenpoly(variablex)^{exponentcounter}) = sumsquare(resultpoly(variablex)^{exponentcounter})$\n\\end{enumerate}\nfor every integer $exponentcounter \\ge 1$.",
      "solution": "Solution. For any polynomial \\( polyfun(variablex) \\), let \\( laurop(polyfun(variablex)) = polyfun(variablex)\\, polyfun\\left(variablex^{-1}\\right) \\), which is a Laurent polynomial (an expression of the form \\( \\sum_{integervar=degreemax}^{exponentcounter} coeffvar_{integervar}\\,variablex^{integervar} \\) where the \\( coeffvar_{integervar} \\) are constants and \\( degreemax, exponentcounter \\) are integers, not necessarily non-negative). Then \\( sumsquare(polyfun(variablex)) \\) equals the coefficient of \\( variablex^{0} \\) in \\( laurop(polyfun(variablex)) \\).\n\nWe have \\( givenpoly(variablex) = (3\\,variablex+1)(variablex+2) \\). Since\n\\[\nlaurop(variablex+2) = (variablex+2)\\bigl(variablex^{-1}+2\\bigr)=\\bigl(1+2\\,variablex^{-1}\\bigr)\\bigl(1+2\\,variablex\\bigr)=laurop(1+2\\,variablex),\n\\]\nand \\( laurop(polyfun(variablex)\\,q(variablex)) = laurop(polyfun(variablex))\\,laurop(q(variablex)) \\) for any polynomials \\( polyfun(variablex) \\) and \\( q(variablex) \\), we obtain\n\\[\nlaurop\\bigl(givenpoly(variablex)^{exponentcounter}\\bigr)=laurop\\bigl((3\\,variablex+1)^{exponentcounter}\\bigr)\\,laurop\\bigl((variablex+2)^{exponentcounter}\\bigr)=laurop\\bigl((3\\,variablex+1)^{exponentcounter}\\bigr)\\,laurop\\bigl((1+2\\,variablex)^{exponentcounter}\\bigr)=laurop\\bigl(resultpoly(variablex)^{exponentcounter}\\bigr),\n\\]\nwhere \\( resultpoly(variablex) = (3\\,variablex+1)(1+2\\,variablex)=6\\,variablex^{2}+5\\,variablex+1 \\). Taking coefficients of \\( variablex^{0} \\) gives \\( sumsquare\\bigl(givenpoly(variablex)^{exponentcounter}\\bigr)=sumsquare\\bigl(resultpoly(variablex)^{exponentcounter}\\bigr) \\). Moreover, \\( resultpoly(0)=1 \\), so the required polynomial has been found.\n\nRemark. One can also check directly that\n\\( sumsquare\\bigl((\\sum coeffvar_{integervar} variablex^{integervar})(\\sum b_{integervar} variablex^{integervar})\\bigr) \\)\nis unchanged when the \\( b_{integervar} \\) are taken in the reverse order, without mentioning \\( laurop \\) or Laurent polynomials. The coefficient of \\( variablex^{0} \\) in a Laurent polynomial can also be written as a contour integral, via the residue theorem; thus one may begin with\n\\[\nsumsquare(polyfun(variablex))=\\frac{1}{2\\pi i}\\oint_{|complexvar|=1} polyfun(complexvar)\\,polyfun\\bigl(complexvar^{-1}\\bigr)\\,\\frac{d complexvar}{complexvar}.\n\\]\n\nRemark. The solution is not unique: for any integer \\( indexfactor \\ge 1 \\) the polynomials\n\\( resultpoly(variablex)=\\bigl(3\\,variablex^{indexfactor}+1\\bigr)\\bigl(2\\,variablex^{indexfactor}+1\\bigr) \\) and\n\\( resultpoly(variablex)=\\bigl(3\\,variablex^{indexfactor}-1\\bigr)\\bigl(2\\,variablex^{indexfactor}-1\\bigr) \\)\nalso satisfy the stated properties. We do not know if these are the only examples. Using that the ring \\( \\mathbb{R}[variablex,1/variablex] \\) of Laurent polynomials is a unique-factorization domain, we could prove they are the only ones if we could establish the following conjecture of Greg Kuperberg:\n\nSuppose \\( laurone, laurtwo \\in \\mathbb{R}[variablex,1/variablex] \\) satisfy \\( laurone(variablex)=laurone(1/variablex) \\) and \\( laurtwo(variablex)=laurtwo(1/variablex) \\). Then the following are equivalent:\n(a) the coefficient of \\( variablex^{0} \\) in \\( laurone(variablex)^{exponentcounter} \\) equals that in \\( laurtwo(variablex)^{exponentcounter} \\) for every positive integer \\( exponentcounter \\);\n(b) there exists \\( integervar \\in \\mathbb{R}[variablex,1/variablex] \\) such that for \\( i=1,2 \\) we have\n\\( h_{i}(variablex)=integervar\\bigl(signchoice_{i}\\,variablex^{indexset}\\bigr) \\) for some \\( signchoice_{i}\\in\\{1,-1\\} \\) and \\( indexset \\ge 1 \\).\n(It is straightforward that (b) implies (a).)"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sunflower",
        "n": "boardwalk",
        "k": "lamborghini",
        "z": "butterscotch",
        "a_0": "goldfinch",
        "a_1": "woodpecker",
        "a_m": "hummingbird",
        "a_i": "nightingale",
        "p": "watermelon",
        "F": "dragonfly",
        "g": "marshmallow",
        "f": "windmill",
        "m": "cheeseball",
        "j": "peppercorn",
        "h_1": "tortoise",
        "h_2": "armadillo",
        "k_i": "orangutan",
        "\\Gamma": "pineapple",
        "\\gamma": "caterpillar",
        "\\epsilon_i": "blueberry"
      },
      "question": "If $watermelon(sunflower)= goldfinch + woodpecker sunflower + \\cdots + hummingbird sunflower^{cheeseball}$ is a polynomial with real\ncoefficients $nightingale$, then set\n\\[\npineapple(watermelon(sunflower)) = goldfinch^2 + woodpecker^2 + \\cdots + hummingbird^2.\n\\]\nLet $dragonfly(sunflower) = 3sunflower^2+7sunflower+2$. Find, with proof, a polynomial $marshmallow(sunflower)$ with\nreal coefficients such that\n\\begin{enumerate}\n\\item[(i)] $marshmallow(0)=1$, and\n\\item[(ii)] $pineapple(windmill(sunflower)^{boardwalk}) = pineapple(marshmallow(sunflower)^{boardwalk})$\n\\end{enumerate}\nfor every integer $boardwalk \\geq 1$.",
      "solution": "Solution. For any polynomial \\( watermelon(sunflower) \\), let \\( caterpillar(watermelon(sunflower))=watermelon(sunflower) watermelon\\left(sunflower^{-1}\\right) \\), which is a Laurent polynomial (an expression of the form \\( \\sum_{peppercorn=cheeseball}^{boardwalk} a_{\\peppercorn} sunflower^{\\peppercorn} \\) where \\( a_{\\peppercorn} \\) are constants and \\( cheeseball, boardwalk \\) are integers, not necessarily nonnegative). Then \\( pineapple(watermelon(sunflower)) \\) equals the coefficient of \\( sunflower^{0} \\) in \\( caterpillar(watermelon(sunflower)) \\).\n\nWe have \\( windmill(sunflower)=(3 sunflower+1)(sunflower+2) \\). Since\n\\[\ncaterpillar(sunflower+2)=(sunflower+2)\\left(sunflower^{-1}+2\\right)=\\left(1+2 sunflower^{-1}\\right)(1+2 sunflower)=caterpillar(1+2 sunflower),\n\\]\nand \\( caterpillar(watermelon(sunflower) q(sunflower))=caterpillar(watermelon(sunflower)) caterpillar(q(sunflower)) \\) for any polynomials \\( watermelon(sunflower) \\) and \\( q(sunflower) \\), we find\n\\[\ncaterpillar\\left(windmill(sunflower)^{boardwalk}\\right)=caterpillar\\left((3 sunflower+1)^{boardwalk}\\right) caterpillar\\left((sunflower+2)^{boardwalk}\\right)=caterpillar\\left((3 sunflower+1)^{boardwalk}\\right) caterpillar\\left((1+2 sunflower)^{boardwalk}\\right)=caterpillar\\left(marshmallow(sunflower)^{boardwalk}\\right),\n\\]\nwhere \\( marshmallow(sunflower)=(3 sunflower+1)(1+2 sunflower)=6 sunflower^{2}+5 sunflower+1 \\). Taking coefficients of \\( sunflower^{0} \\), we obtain \\( pineapple\\left(windmill(sunflower)^{boardwalk}\\right)=pineapple\\left(marshmallow(sunflower)^{boardwalk}\\right) \\). Moreover \\( marshmallow(0)=1 \\), so we are done.\n\nRemark. One could also show by brute force that \\( pineapple\\left(\\left(\\sum nightingale sunflower^{peppercorn}\\right)\\left(\\sum b_{\\peppercorn} sunflower^{peppercorn}\\right)\\right) \\) is unchanged by reversing the order of the \\( b_{\\peppercorn} \\), without mentioning \\( caterpillar \\) or Laurent polynomials. The coefficient of \\( sunflower^{0} \\) in a Laurent polynomial can also be expressed as a contour integral, via the residue theorem; hence one could begin a solution with the observation\n\\[\npineapple(watermelon(sunflower))=\\frac{1}{2 \\pi i} \\oint_{|butterscotch|=1} watermelon(butterscotch) watermelon\\left(butterscotch^{-1}\\right) \\frac{d butterscotch}{butterscotch}\n\\]\n\nRemark. The solution is not unique: for any integer \\( lamborghini \\geq 1 \\), the polynomials \\( marshmallow(sunflower)=\\left(3 sunflower^{lamborghini}+1\\right)\\left(2 sunflower^{lamborghini}+1\\right) \\) and \\( marshmallow(sunflower)=\\left(3 sunflower^{lamborghini}-1\\right)\\left(2 sunflower^{lamborghini}-1\\right) \\) also have the desired properties. We do not know if these are the only ones. Using the fact that the ring \\( \\mathbb{R}[sunflower, 1 / sunflower] \\) of Laurent polynomials is a unique factorization domain (UFD), we could prove that these are the only ones if we could prove the following conjecture of Greg Kuperberg (communicated electronically):\n\nSuppose \\( tortoise, armadillo \\in \\mathbb{R}[sunflower, 1 / sunflower] \\) satisfy \\( tortoise(sunflower)=tortoise(1 / sunflower) \\) and \\( armadillo(sunflower)=armadillo(1 / sunflower) \\).\nThen the following are equivalent:\n(a) The coefficient of \\( sunflower^{0} \\) in \\( tortoise(sunflower)^{boardwalk} \\) equals the coefficient of \\( sunflower^{0} \\) of \\( armadillo(sunflower)^{boardwalk} \\) for all positive integers \\( boardwalk \\).\n(b) There exists \\( peppercorn \\in \\mathbb{R}[sunflower, 1 / sunflower] \\) such that for \\( i=1,2 \\) we have \\( h_{i}(sunflower)=peppercorn\\left(blueberry sunflower^{orangutan}\\right) \\) for some \\( blueberry \\in\\{1,-1\\} \\) and \\( orangutan \\geq 1 \\).\n(It is easy to prove that (b) implies (a).)"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "stillpoint",
        "n": "fractional",
        "k": "continuity",
        "z": "realness",
        "a_0": "flowingzero",
        "a_1": "flowingone",
        "a_m": "flowingmax",
        "a_i": "flowingiota",
        "p": "flatfunc",
        "F": "varipoly",
        "g": "fixedpoly",
        "f": "constpoly",
        "m": "baseline",
        "j": "macrofunc",
        "h_1": "shortone",
        "h_2": "shorttwo",
        "k_i": "continuum",
        "\\Gamma": "\\oppositum",
        "\\gamma": "\\stability",
        "\\epsilon_i": "\\heaviness"
      },
      "question": "If $flatfunc(stillpoint)= flowingzero + flowingone stillpoint + \\cdots + flowingmax stillpoint^{baseline}$ is a polynomial with real\ncoefficients $flowingiota$, then set\n\\[\n\\oppositum(flatfunc(stillpoint)) = flowingzero^2 + flowingone^2 + \\cdots + flowingmax^2.\n\\]\nLet $varipoly(stillpoint) = 3stillpoint^2+7stillpoint+2$. Find, with proof, a polynomial $fixedpoly(stillpoint)$ with\nreal coefficients such that\n\\begin{enumerate}\n\\item[(i)] $fixedpoly(0)=1$, and\n\\item[(ii)] $\\oppositum(constpoly(stillpoint)^{fractional}) = \\oppositum(fixedpoly(stillpoint)^{fractional})$\n\\end{enumerate}\nfor every integer $fractional \\geq 1$.",
      "solution": "Solution. For any polynomial \\( flatfunc(stillpoint) \\), let \\( \\stability(flatfunc(stillpoint))=flatfunc(stillpoint) flatfunc\\left(stillpoint^{-1}\\right) \\), which is a Laurent polynomial (an expression of the form \\( \\sum_{macrofunc=baseline}^{fractional} a_{macrofunc} stillpoint^{macrofunc} \\) where \\( a_{macrofunc} \\) are constants and \\( baseline, fractional \\) are integers, not necessarily nonnegative). Then \\( \\oppositum(flatfunc(stillpoint)) \\) equals the coefficient of \\( stillpoint^{0} \\) in \\( \\stability(flatfunc(stillpoint)) \\).\n\nWe have \\( constpoly(stillpoint)=(3 stillpoint+1)(stillpoint+2) \\). Since\n\\[\n\\stability(stillpoint+2)=(stillpoint+2)\\left(stillpoint^{-1}+2\\right)=\\left(1+2 stillpoint^{-1}\\right)(1+2 stillpoint)=\\stability(1+2 stillpoint),\n\\]\nand \\( \\stability(flatfunc(stillpoint) q(stillpoint))=\\stability(flatfunc(stillpoint)) \\stability(q(stillpoint)) \\) for any polynomials \\( flatfunc(stillpoint) \\) and \\( q(stillpoint) \\), we find\n\\[\n\\stability\\left(constpoly(stillpoint)^{fractional}\\right)=\\stability\\left((3 stillpoint+1)^{fractional}\\right) \\stability\\left((stillpoint+2)^{fractional}\\right)=\\stability\\left((3 stillpoint+1)^{fractional}\\right) \\stability\\left((1+2 stillpoint)^{fractional}\\right)=\\stability\\left(fixedpoly(stillpoint)^{fractional}\\right),\n\\]\nwhere \\( fixedpoly(stillpoint)=(3 stillpoint+1)(1+2 stillpoint)=6 stillpoint^{2}+5 stillpoint+1 \\). Taking coefficients of \\( stillpoint^{0} \\), we obtain \\( \\oppositum\\left(constpoly(stillpoint)^{fractional}\\right)=\\oppositum\\left(fixedpoly(stillpoint)^{fractional}\\right) \\). Moreover \\( fixedpoly(0)=1 \\), so we are done.\n\nRemark. One could also show by brute force that \\( \\oppositum\\left(\\left(\\sum flowingiota stillpoint^{i}\\right)\\left(\\sum b_{macrofunc} stillpoint^{macrofunc}\\right)\\right) \\) is unchanged by reversing the order of the \\( b_{macrofunc} \\), without mentioning \\( \\stability \\) or Laurent polynomials. The coefficient of \\( stillpoint^{0} \\) in a Laurent polynomial can also be expressed as a contour integral, via the residue theorem; hence one could begin a solution with the observation\n\\[\n\\oppositum(flatfunc(stillpoint))=\\frac{1}{2 \\pi i} \\oint_{|realness|=1} flatfunc(realness) flatfunc\\left(realness^{-1}\\right) \\frac{d realness}{realness}\n\\]\n\nRemark. The solution is not unique: for any integer \\( continuity \\geq 1 \\), the polynomials \\( fixedpoly(stillpoint)=\\left(3 stillpoint^{continuity}+1\\right)\\left(2 stillpoint^{continuity}+1\\right) \\) and \\( fixedpoly(stillpoint)=\\left(3 stillpoint^{continuity}-1\\right)\\left(2 stillpoint^{continuity}-1\\right) \\) also have the desired properties. We do not know if these are the only ones. Using the fact that the ring \\( \\mathbb{R}[stillpoint, 1 / stillpoint] \\) of Laurent polynomials is a unique factorization domain (UFD), we could prove that these are the only ones if we could prove the following conjecture of Greg Kuperberg (communicated electronically):\n\nSuppose \\( shortone, shorttwo \\in \\mathbb{R}[stillpoint, 1 / stillpoint] \\) satisfy \\( shortone(stillpoint)=shortone(1 / stillpoint) \\) and \\( shorttwo(stillpoint)=shorttwo(1 / stillpoint) \\).\nThen the following are equivalent:\n(a) The coefficient of \\( stillpoint^{0} \\) in \\( shortone(stillpoint)^{fractional} \\) equals the coefficient of \\( stillpoint^{0} \\) of \\( shorttwo(stillpoint)^{fractional} \\) for all positive integers \\( fractional \\).\n(b) There exists \\( macrofunc \\in \\mathbb{R}[stillpoint, 1 / stillpoint] \\) such that for \\( i=1,2 \\) we have \\( h_{i}(stillpoint)=macrofunc\\left(\\heaviness_{i} stillpoint^{continuum}\\right) \\) for some \\( \\heaviness_{i} \\in\\{1,-1\\} \\) and \\( continuum \\geq 1 \\).\n(It is easy to prove that (b) implies (a).)"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "n": "hjgrksla",
        "k": "vbnmrewq",
        "z": "asdfghjk",
        "a_0": "lkjhgfdsa",
        "a_1": "poiuytrew",
        "a_m": "mnbvcxzas",
        "a_i": "qazwsxedc",
        "p": "rfvtgbyhn",
        "F": "yhnujmiko",
        "g": "olpknjmbv",
        "f": "plokmijnb",
        "m": "wsxedcrfv",
        "j": "uytresdfg",
        "h_1": "xcvbnmlkj",
        "h_2": "zxcvbnmas",
        "k_i": "qwertyuio",
        "\\\\Gamma": "pasdfghjk",
        "\\\\gamma": "lpoiuytre",
        "\\\\epsilon_i": "mkiolpbnv"
      },
      "question": "If $rfvtgbyhn(qzxwvtnp)= lkjhgfdsa + poiuytrew qzxwvtnp + \\cdots + mnbvcxzas qzxwvtnp^{wsxedcrfv}$ is a polynomial with real\ncoefficients $qazwsxedc_i$, then set\n\\[\npasdfghjk(rfvtgbyhn(qzxwvtnp)) = lkjhgfdsa^2 + poiuytrew^2 + \\cdots + mnbvcxzas^2.\n\\]\nLet $yhnujmiko(qzxwvtnp) = 3qzxwvtnp^2+7qzxwvtnp+2$. Find, with proof, a polynomial $olpknjmbv(qzxwvtnp)$ with\nreal coefficients such that\n\\begin{enumerate}\n\\item[(i)] $olpknjmbv(0)=1$, and\n\\item[(ii)] $pasdfghjk(plokmijnb(qzxwvtnp)^{hjgrksla}) = pasdfghjk(olpknjmbv(qzxwvtnp)^{hjgrksla})$\n\\end{enumerate}\nfor every integer $hjgrksla \\geq 1$.",
      "solution": "Solution. For any polynomial \\( rfvtgbyhn(qzxwvtnp) \\), let \\( lpoiuytre(rfvtgbyhn(qzxwvtnp))=rfvtgbyhn(qzxwvtnp) \\, rfvtgbyhn\\left(qzxwvtnp^{-1}\\right) \\), which is a Laurent polynomial (an expression of the form \\( \\sum_{uytresdfg=wsxedcrfv}^{hjgrksla} lkjhgfdsa_{uytresdfg} qzxwvtnp^{uytresdfg} \\) where \\( lkjhgfdsa_{uytresdfg} \\) are constants and \\( wsxedcrfv, hjgrksla \\) are integers, not necessarily non-negative). Then \\( pasdfghjk(rfvtgbyhn(qzxwvtnp)) \\) equals the coefficient of \\( qzxwvtnp^{0} \\) in \\( lpoiuytre(rfvtgbyhn(qzxwvtnp)) \\).\n\nWe have \\( plokmijnb(qzxwvtnp)=(3 qzxwvtnp+1)(qzxwvtnp+2) \\). Since\n\\[\nlpoiuytre(qzxwvtnp+2)=(qzxwvtnp+2)\\left(qzxwvtnp^{-1}+2\\right)=\\left(1+2 qzxwvtnp^{-1}\\right)(1+2 qzxwvtnp)=lpoiuytre(1+2 qzxwvtnp),\n\\]\nand \\( lpoiuytre(rfvtgbyhn(qzxwvtnp)\\, olpknjmbv(qzxwvtnp))=lpoiuytre(rfvtgbyhn(qzxwvtnp))\\, lpoiuytre(olpknjmbv(qzxwvtnp)) \\) for any polynomials \\( rfvtgbyhn(qzxwvtnp) \\) and \\( olpknjmbv(qzxwvtnp) \\), we find\n\\[\nlpoiuytre\\left(plokmijnb(qzxwvtnp)^{hjgrksla}\\right)=lpoiuytre\\left((3 qzxwvtnp+1)^{hjgrksla}\\right)\\, lpoiuytre\\left((qzxwvtnp+2)^{hjgrksla}\\right)=lpoiuytre\\left((3 qzxwvtnp+1)^{hjgrksla}\\right)\\, lpoiuytre\\left((1+2 qzxwvtnp)^{hjgrksla}\\right)=lpoiuytre\\left(olpknjmbv(qzxwvtnp)^{hjgrksla}\\right),\n\\]\nwhere \\( olpknjmbv(qzxwvtnp)=(3 qzxwvtnp+1)(1+2 qzxwvtnp)=6 qzxwvtnp^{2}+5 qzxwvtnp+1 \\). Taking coefficients of \\( qzxwvtnp^{0} \\), we obtain \\( pasdfghjk\\left(plokmijnb(qzxwvtnp)^{hjgrksla}\\right)=pasdfghjk\\left(olpknjmbv(qzxwvtnp)^{hjgrksla}\\right) \\). Moreover \\( olpknjmbv(0)=1 \\), so we are done.\n\nRemark. One could also show by brute force that \\( pasdfghjk\\left(\\left(\\sum qazwsxedc_{uytresdfg} qzxwvtnp^{uytresdfg}\\right)\\left(\\sum poiuytrew_{uytresdfg} qzxwvtnp^{uytresdfg}\\right)\\right) \\) is unchanged by reversing the order of the \\( poiuytrew_{uytresdfg} \\), without mentioning \\( lpoiuytre \\) or Laurent polynomials. The coefficient of \\( qzxwvtnp^{0} \\) in a Laurent polynomial can also be expressed as a contour integral, via the residue theorem; hence one could begin a solution with the observation\n\\[\npasdfghjk(rfvtgbyhn(qzxwvtnp))=\\frac{1}{2 \\pi i} \\oint_{|asdfghjk|=1} rfvtgbyhn(asdfghjk)\\, rfvtgbyhn\\left(asdfghjk^{-1}\\right) \\frac{d asdfghjk}{asdfghjk}\n\\]\n\nRemark. The solution is not unique: for any integer \\( vbnmrewq \\geq 1 \\), the polynomials \\( olpknjmbv(qzxwvtnp)=\\left(3 qzxwvtnp^{vbnmrewq}+1\\right)\\left(2 qzxwvtnp^{vbnmrewq}+1\\right) \\) and \\( olpknjmbv(qzxwvtnp)=\\left(3 qzxwvtnp^{vbnmrewq}-1\\right)\\left(2 qzxwvtnp^{vbnmrewq}-1\\right) \\) also have the desired properties. We do not know if these are the only ones. Using the fact that the ring \\( \\mathbb{R}[qzxwvtnp, 1 / qzxwvtnp] \\) of Laurent polynomials is a unique factorization domain (UFD), we could prove that these are the only ones if we could prove the following conjecture of Greg Kuperberg (communicated electronically):\n\nSuppose \\( xcvbnmlkj(qzxwvtnp), zxcvbnmas(qzxwvtnp) \\in \\mathbb{R}[qzxwvtnp, 1 / qzxwvtnp] \\) satisfy \\( xcvbnmlkj(qzxwvtnp)=xcvbnmlkj(1 / qzxwvtnp) \\) and \\( zxcvbnmas(qzxwvtnp)=zxcvbnmas(1 / qzxwvtnp) \\).\nThen the following are equivalent:\n(a) The coefficient of \\( qzxwvtnp^{0} \\) in \\( xcvbnmlkj(qzxwvtnp)^{hjgrksla} \\) equals the coefficient of \\( qzxwvtnp^{0} \\) of \\( zxcvbnmas(qzxwvtnp)^{hjgrksla} \\) for all positive integers \\( hjgrksla \\).\n(b) There exists \\( uytresdfg \\in \\mathbb{R}[qzxwvtnp, 1 / qzxwvtnp] \\) such that for \\( i=1,2 \\) we have \\( xcvbnmlkj_{i}(qzxwvtnp)=uytresdfg\\left(mkiolpbnv_{i} qzxwvtnp^{qwertyuio_{i}}\\right) \\) for some \\( mkiolpbnv_{i} \\in\\{1,-1\\} \\) and \\( qwertyuio_{i} \\geq 1 \\).\n(It is easy to prove that (b) implies (a).)"
    },
    "kernel_variant": {
      "question": "Let  \n\\[\n\\Gamma\\!\\bigl(p\\bigr)\\;=\\!\n\\sum_{i,j,k\\ge 0}a_{i,j,k}^{2},\\qquad \n\\Bigl(p(x,y,z)=\\!\\sum_{i,j,k\\ge 0}a_{i,j,k}\\,x^{i}y^{j}z^{k}\\Bigr)\n\\]  \ndenote the squared Euclidean norm of the (finite) real coefficient array of a\nthree–variable polynomial.  \nFix  \n\\[\nF(x,y,z)=\\bigl(2x+1\\bigr)\\bigl(6y+1\\bigr)\\bigl(15z+1\\bigr).\n\\]\n\nA.  Prove that there exists a real polynomial $g(x,y,z)$ such that  \n(i) $g(0,0,0)=1$;  \n(ii) $\\Gamma\\!\\bigl(F^{\\,n}\\bigr)=\\Gamma\\!\\bigl(g^{\\,n}\\bigr)$ holds for every\ninteger $n\\ge 1$;  \n(iii) $g\\not\\equiv F$.\n\nB.  Let  \n\\[\nh(x,y,z)=\\bigl(\\alpha_{1}x+1\\bigr)\\bigl(\\alpha_{2}y+1\\bigr)\n         \\bigl(\\alpha_{3}z+1\\bigr),\\qquad \\alpha_{j}\\in\\mathbb{R},\n\\]\nbe any product of three univariate linear factors whose constant terms equal\n$1$.  Determine precisely those triples\n$(\\alpha_{1},\\alpha_{2},\\alpha_{3})$ that satisfy  \n\\[\n\\Gamma\\!\\bigl(h^{\\,n}\\bigr)=\\Gamma\\!\\bigl(F^{\\,n}\\bigr)\\quad\\forall\\,n\\ge 1,\n\\]\nand prove that every solution is obtained from $(2,6,15)$ by an arbitrary\npermutation of the entries together with independent sign changes.\n\nC.  Let $H(x,y,z)$ be \\emph{any} real polynomial of finite total degree such\nthat  \n\\[\nH(0,0,0)=1,\\qquad \n\\Gamma\\!\\bigl(H^{\\,n}\\bigr)=\\Gamma\\!\\bigl(F^{\\,n}\\bigr)\\quad\\forall\\,n\\ge 1.\n\\]\nShow that $H$ must be of the form singled out in~(B):  \nthere exist a permutation $\\sigma\\in\\mathfrak{S}_{3}$ and signs\n$\\varepsilon_{j}\\in\\{\\!+1,-1\\!\\}$ for which  \n\\[\nH(x,y,z)=\n\\bigl(\\varepsilon_{1}2\\,x+1\\bigr)\\,\n\\bigl(\\varepsilon_{2}6\\,y+1\\bigr)\\,\n\\bigl(\\varepsilon_{3}15\\,z+1\\bigr)\n\\]\nafter the variables $(x,y,z)$ have been reordered by~$\\sigma$.",
      "solution": "Throughout write  \n\\[\n\\Theta_{p}(\\boldsymbol{\\theta})=\np\\!\\bigl(e^{\\mathrm{i}\\theta_{1}},\n         e^{\\mathrm{i}\\theta_{2}},\n         e^{\\mathrm{i}\\theta_{3}}\\bigr),\\qquad\n\\boldsymbol{\\theta}=(\\theta_{1},\\theta_{2},\\theta_{3})\\in\\mathbb{R}^{3},\n\\]\nand denote by  \n\\[\nd\\mu=\\frac{d\\theta_{1}\\,d\\theta_{2}\\,d\\theta_{3}}{(2\\pi)^{3}}\n\\]\nthe normalised Haar measure on the torus $T^{3}=[0,2\\pi]^{3}$.\nOrthogonality of the exponentials gives Parseval’s identity  \n\\[\n\\Gamma\\!\\bigl(p^{\\,n}\\bigr)=\n\\int_{T^{3}}\\!\\bigl|\\Theta_{p}(\\boldsymbol{\\theta})\\bigr|^{2n}\\,d\\mu\n      \\qquad(n\\ge 0).\n\\tag{$\\mathcal{P}$}\n\\]\n\n--------------------------------------------------------------------\nA.  A non–trivial example\n--------------------------------------------------------------------\n\nPermuting $y$ and $z$ in $F$ produces  \n\\[\ng(x,y,z)=\\bigl(2x+1\\bigr)\\bigl(15y+1\\bigr)\\bigl(6z+1\\bigr).\n\\]\nClearly $g(0,0,0)=1$ and $g\\not\\equiv F$.  A permutation of the\nindeterminates merely re–labels the coefficient array, hence preserves the\nEuclidean norm; therefore $\\Gamma\\!\\bigl(g^{\\,n}\\bigr)=\\Gamma\\!\\bigl(F^{\\,n}\\bigr)$\nfor every $n\\ge 1$.\n\n--------------------------------------------------------------------\nB.  Classification inside the linear family\n--------------------------------------------------------------------\n\n--------------------------------------------------------------------\nB.1  The singularity sets that must coincide\n--------------------------------------------------------------------\n\nFor a one–variable linear form $L_{a}(t)=1+at$ put  \n\\[\nS_{n}(a)=\\Gamma\\!\\bigl(L_{a}(t)^{\\,n}\\bigr)=\n        \\sum_{k=0}^{n}\\binom{n}{k}^{2}a^{\\,2k},\n\\qquad \nG_{a}(z)=\\sum_{n=0}^{\\infty}S_{n}(a)z^{n}\n       =\\frac{1}{\\sqrt{\\bigl(1-(1+a)^{2}z\\bigr)\n                       \\bigl(1-(1-a)^{2}z\\bigr)}} .\n\\tag{1}\n\\]\n\nBecause the three variables are independent,  \n\\[\n\\sum_{n\\ge 0}\\Gamma\\!\\bigl(h^{\\,n}\\bigr)z^{n}\n     =\\prod_{j=1}^{3}G_{\\alpha_{j}}(z),\n\\qquad\n\\sum_{n\\ge 0}\\Gamma\\!\\bigl(F^{\\,n}\\bigr)z^{n}\n     =G_{2}(z)G_{6}(z)G_{15}(z).\n\\]\nHence\n\\[\n\\prod_{j=1}^{3}G_{\\alpha_{j}}(z)=G_{2}(z)G_{6}(z)G_{15}(z).\n\\tag{2}\n\\]\n\nEach $G_{a}$ is a square–root of a rational function whose only\nsingularities (branch points) are at  \n\\[\nz=\\frac{1}{(1+a)^{2}},\\qquad z=\\frac{1}{(1-a)^{2}}.\n\\]\nEquality~(2) therefore forces the two multisets of branch points to be\nidentical:\n\\[\n\\bigl\\{\\tfrac{1}{(1+\\alpha_{1})^{2}},\\tfrac{1}{(1-\\alpha_{1})^{2}},\n       \\tfrac{1}{(1+\\alpha_{2})^{2}},\\tfrac{1}{(1-\\alpha_{2})^{2}},\n       \\tfrac{1}{(1+\\alpha_{3})^{2}},\\tfrac{1}{(1-\\alpha_{3})^{2}}\\bigr\\}\n      =\\Bigl\\{\\tfrac{1}{9},1,\\tfrac{1}{25},\\tfrac{1}{49},\n              \\tfrac{1}{196},\\tfrac{1}{256}\\Bigr\\}.\n\\tag{3}\n\\]\n(The right–hand set comes from $a\\in\\{2,6,15\\}$.)\n\nTaking square roots and reciprocals transforms~(3) into the much simpler\nstatement\n\\[\n\\bigl\\{\\lvert1+\\alpha_{1}\\rvert,\\lvert1-\\alpha_{1}\\rvert,\n       \\lvert1+\\alpha_{2}\\rvert,\\lvert1-\\alpha_{2}\\rvert,\n       \\lvert1+\\alpha_{3}\\rvert,\\lvert1-\\alpha_{3}\\rvert\\bigr\\}\n      =\\{1,3,5,7,14,16\\}.\n\\tag{4}\n\\]\n\n--------------------------------------------------------------------\nB.2  Reconstructing $\\lvert\\alpha_{j}\\rvert$ from the six integers\n--------------------------------------------------------------------\n\nFix any real $\\beta\\neq\\pm1$.  The unordered pair\n\\[\n\\bigl\\lvert1+\\beta\\bigr\\rvert,\\;\\bigl\\lvert1-\\beta\\bigr\\rvert\n\\]\nalways consists of two \\emph{distinct} positive numbers whose difference\nis exactly $2$.  Conversely, if $u>v>0$ and $u-v=2$, then  \n\\[\n\\{\\!u,v\\!\\}=\\{\\lvert1+\\beta\\rvert,\\lvert1-\\beta\\rvert\\}\n\\quad\\Longleftrightarrow\\quad\n\\beta=\\tfrac{u-v}{2}\\cdot\\tfrac{u+v}{2}.\n\\tag{5}\n\\]\nHence each $\\beta$ corresponds to a \\emph{pair} of positive numbers that\ndiffer by $2$, and vice versa.\n\nAmong the six positive integers in the right–hand side of~(4) there are\nexactly four pairs whose members differ by $2$,\n\\[\n(1,3),\\;(3,5),\\;(5,7),\\;(14,16).\n\\]\nBecause we must partition the multiset $\\{1,3,5,7,14,16\\}$ into \\emph{three}\ndisjoint pairs each differing by $2$, a simple “smallest‐element’’ argument\nshows that there is a unique perfect matching:\n\\[\n\\{1,3\\}\\cup\\{5,7\\}\\cup\\{14,16\\}.\n\\]\nIndeed, the smallest element $1$ can only be paired with $3$;\nremoving $\\{1,3\\}$ leaves $\\{5,7,14,16\\}$, whose smallest element $5$ forces\nthe pair $\\{5,7\\}$; the remaining two numbers automatically form the third\npair.\n\nApplying the reconstruction formula~(5) to the three pairs yields\n\\[\n\\lvert\\alpha_{1}\\rvert=2,\\qquad\n\\lvert\\alpha_{2}\\rvert=6,\\qquad\n\\lvert\\alpha_{3}\\rvert=15,\n\\tag{6}\n\\]\nup to ordering.\n\n--------------------------------------------------------------------\nB.3  Sign independence and variable permutations\n--------------------------------------------------------------------\n\nBecause $S_{n}(-a)=S_{n}(a)$ for every $n$ and every real $a$\n(cf.\\ the definition of $S_{n}$ above), replacing any $\\alpha_{j}$ by\n$-\\alpha_{j}$ leaves $\\Gamma\\!\\bigl(h^{\\,n}\\bigr)$ unchanged.\nFurthermore, permuting the variables $(x,y,z)$ only re–orders the\ncoefficients, hence also preserves their Euclidean norm.\nCombining these two observations with~(6) gives the full set of solutions:\n\\[\n(\\alpha_{1},\\alpha_{2},\\alpha_{3})\n=\\bigl(\\varepsilon_{1}2,\\varepsilon_{2}6,\\varepsilon_{3}15\\bigr),\n\\quad\\varepsilon_{j}\\in\\{\\!+1,-1\\!\\},\n\\]\nup to an arbitrary permutation of the three entries.  This completes the\nclassification required in~B.\n\n--------------------------------------------------------------------\nC.  Global rigidity\n--------------------------------------------------------------------\n\n--------------------------------------------------------------------\nC.1  From equality of moments to pointwise equality\n--------------------------------------------------------------------\n\nSet  \n\\[\nU(\\boldsymbol{\\theta})=\\bigl|\\Theta_{H}(\\boldsymbol{\\theta})\\bigr|^{2},\n\\qquad \nV(\\boldsymbol{\\theta})=\\bigl|\\Theta_{F}(\\boldsymbol{\\theta})\\bigr|^{2}\n     \\quad(\\boldsymbol{\\theta}\\in T^{3}).\n\\]\nBoth $U$ and $V$ are continuous, non–negative functions on the compact\n$T^{3}$.  By assumption  \n\\[\n\\int_{T^{3}}U(\\boldsymbol{\\theta})^{\\,n}\\,d\\mu\n     =\\Gamma\\!\\bigl(H^{\\,n}\\bigr)\n     =\\Gamma\\!\\bigl(F^{\\,n}\\bigr)\n     =\\int_{T^{3}}V(\\boldsymbol{\\theta})^{\\,n}\\,d\\mu\n     \\qquad\\forall\\,n\\ge 0.\n\\tag{7}\n\\]\n\n\\emph{Lemma.}\nIf two bounded measurable functions $U,V\\ge 0$ satisfy\n$\\int U^{n}\\,d\\mu=\\int V^{n}\\,d\\mu$ for every $n$, then $U=V$ almost\neverywhere.\n\n\\emph{Proof of the Lemma.}\nPut $W=U-V$.  If $W\\not\\equiv 0$, the set\n$E_{+}=\\{\\boldsymbol{\\theta}:W(\\boldsymbol{\\theta})>0\\}$\nhas positive measure.\nChoose $\\varepsilon>0$ such that \n$E_{\\varepsilon}=\\{\\boldsymbol{\\theta}:W(\\boldsymbol{\\theta})\\ge \\varepsilon\\}$\nstill has positive measure.\nOn $E_{\\varepsilon}$ we have $U\\ge V+\\varepsilon$, hence\n$U^{n}\\ge V^{n}+n\\varepsilon V^{\\,n-1}$.\nIntegrating and using (7) with $n$ large gives a contradiction.\n$\\hfill\\square$\n\nApplying the Lemma we obtain $U=V$ almost everywhere, hence everywhere\nbecause the difference $U-V$ is a trigonometric polynomial and therefore\nreal–analytic on $T^{3}$.  Consequently\n\\[\nH(x,y,z)\\,H\\!\\bigl(x^{-1},y^{-1},z^{-1}\\bigr)=\nF(x,y,z)\\,F\\!\\bigl(x^{-1},y^{-1},z^{-1}\\bigr)\n\\quad\\text{as Laurent polynomials.}\n\\tag{8}\n\\]\n\n--------------------------------------------------------------------\nC.2  Unique factorisation of Laurent polynomials\n--------------------------------------------------------------------\n\nPut  \n\\[\nR=\\mathbb{R}[X^{\\pm1},Y^{\\pm1},Z^{\\pm1}],\\qquad\n\\star:\\,R\\longrightarrow R,\\;\n\\bigl(f^{\\star}\\bigr)(X,Y,Z)=f(X^{-1},Y^{-1},Z^{-1}).\n\\]\nThe ring $R$ is a unique factorisation domain (localisation of the UFD\n$\\mathbb{R}[X,Y,Z]$), and $\\star$ is an involutive automorphism.\nEquation~(8) can be rewritten as  \n\\[\nH\\,H^{\\star}=F\\,F^{\\star}\\qquad\\text{in }R.\n\\tag{9}\n\\]\n\nFactorisation of the right–hand side is explicit:\n\\[\nF\\,F^{\\star}=\n\\bigl(1+2X\\bigr)\\bigl(1+2X^{-1}\\bigr)\\,\n\\bigl(1+6Y\\bigr)\\bigl(1+6Y^{-1}\\bigr)\\,\n\\bigl(1+15Z\\bigr)\\bigl(1+15Z^{-1}\\bigr).\n\\tag{10}\n\\]\nEach linear factor is irreducible in $R$, and any two distinct factors are\ncoprime.  \n\n--------------------------------------------------------------------\nC.3  Locating the factors that divide $H$\n--------------------------------------------------------------------\n\nWrite  \n\\[\nH=\\sum_{i,j,k\\ge 0}c_{i,j,k}X^{i}Y^{j}Z^{k},\\qquad c_{0,0,0}=1.\n\\tag{11}\n\\]\nThus $H$ is a genuine polynomial (no negative exponents).  \nSuppose, for instance, that the factor $1+2X^{-1}$ from~(10) divided $H$:\nthen $H=(1+2X^{-1})Q$ with $Q\\in R$.  \nExpanding, the lowest $X$–exponent occurring in $H$ would be \\emph{negative}\n(unless $Q$ contained a compensating $X$ factor, in which case the constant\nterm of $H$ would change from $1$).  Either alternative contradicts (11).\nHence $1+2X^{-1}$ cannot divide $H$, and the same argument applies to\n$1+6Y^{-1}$ and $1+15Z^{-1}$.\n\nTherefore, for each variable, exactly one of the two factors in the\ncorresponding pair in~(10) divides $H$, and the multiplicity must be~$1$\nbecause it is~$1$ in~(10).  Thus\n\\[\nH=\\varepsilon\\,\n(1+\\varepsilon_{1}2X)\\,(1+\\varepsilon_{2}6Y)\\,(1+\\varepsilon_{3}15Z),\n\\tag{12}\n\\]\nwhere $\\varepsilon,\\varepsilon_{1},\\varepsilon_{2},\\varepsilon_{3}\n      \\in\\{\\!+1,-1\\!\\}$.\nThe constant term condition $c_{0,0,0}=1$ forces $\\varepsilon=1$.\nFinally, permuting $(X,Y,Z)$ corresponds to permuting $(x,y,z)$,\nso~(12) translates back to the claimed form of $H$.\n\n--------------------------------------------------------------------\nC.4  Conclusion\n--------------------------------------------------------------------\n\nCombining~(12) with the independent sign choices\n$\\varepsilon_{j}\\in\\{\\!+1,-1\\!\\}$ and an arbitrary permutation\n$\\sigma\\in\\mathfrak{S}_{3}$ we arrive precisely at the family singled\nout in~B, completing the proof of~C.\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.684367",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension & more variables:  We moved from univariate to\n  trivariate polynomials, turning a one-dimensional coefficient array\n  into a three-dimensional one.\n\n• Additional constraints:  Besides finding one suitable g, the solver\n  must now classify all polynomials h fulfilling the quadratic-norm\n  equalities, not merely exhibit one example.\n\n• Deeper theory:  The solution demands familiarity with\n  Laurent-polynomial rings in several variables, their unique\n  factorisation, and the behaviour of quadratic norms under variable\n  inversion.  These topics do not appear in the original problem.\n\n• Interacting concepts:  The argument blends algebraic factorisation,\n  symmetries of Euclidean norms, invariance under coefficient\n  permutation, and multiplicativity of the γ-map.\n\n• More steps & insight:  Establishing γ for three variables, proving its\n  multiplicativity, analysing constant terms, and executing the UFD\n  uniqueness argument collectively require significantly more work and\n  conceptual depth than the original single-variable manipulation."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\n\\Gamma\\!\\bigl(p\\bigr)\\;=\\!\n\\sum_{i,j,k\\ge 0}a_{i,j,k}^{2},\\qquad \n\\Bigl(p(x,y,z)=\\!\\sum_{i,j,k\\ge 0}a_{i,j,k}\\,x^{i}y^{j}z^{k}\\Bigr)\n\\]  \ndenote the squared Euclidean norm of the (finite) real coefficient array of a\nthree–variable polynomial.  \nFix  \n\\[\nF(x,y,z)=\\bigl(2x+1\\bigr)\\bigl(6y+1\\bigr)\\bigl(15z+1\\bigr).\n\\]\n\nA.  Prove that there exists a real polynomial $g(x,y,z)$ such that  \n(i) $g(0,0,0)=1$;  \n(ii) $\\Gamma\\!\\bigl(F^{\\,n}\\bigr)=\\Gamma\\!\\bigl(g^{\\,n}\\bigr)$ holds for every\ninteger $n\\ge 1$;  \n(iii) $g\\not\\equiv F$.\n\nB.  Let  \n\\[\nh(x,y,z)=\\bigl(\\alpha_{1}x+1\\bigr)\\bigl(\\alpha_{2}y+1\\bigr)\n         \\bigl(\\alpha_{3}z+1\\bigr),\\qquad \\alpha_{j}\\in\\mathbb{R},\n\\]\nbe any product of three univariate linear factors whose constant terms equal\n$1$.  Determine precisely those triples\n$(\\alpha_{1},\\alpha_{2},\\alpha_{3})$ that satisfy  \n\\[\n\\Gamma\\!\\bigl(h^{\\,n}\\bigr)=\\Gamma\\!\\bigl(F^{\\,n}\\bigr)\\quad\\forall\\,n\\ge 1,\n\\]\nand prove that every solution is obtained from $(2,6,15)$ by an arbitrary\npermutation of the entries together with independent sign changes.\n\nC.  Let $H(x,y,z)$ be \\emph{any} real polynomial of finite total degree such\nthat  \n\\[\nH(0,0,0)=1,\\qquad \n\\Gamma\\!\\bigl(H^{\\,n}\\bigr)=\\Gamma\\!\\bigl(F^{\\,n}\\bigr)\\quad\\forall\\,n\\ge 1.\n\\]\nShow that $H$ must be of the form singled out in~(B):  \nthere exist a permutation $\\sigma\\in\\mathfrak{S}_{3}$ and signs\n$\\varepsilon_{j}\\in\\{\\!+1,-1\\!\\}$ for which  \n\\[\nH(x,y,z)=\n\\bigl(\\varepsilon_{1}2\\,x+1\\bigr)\\,\n\\bigl(\\varepsilon_{2}6\\,y+1\\bigr)\\,\n\\bigl(\\varepsilon_{3}15\\,z+1\\bigr)\n\\]\nafter the variables $(x,y,z)$ have been reordered by~$\\sigma$.",
      "solution": "Throughout write  \n\\[\n\\Theta_{p}(\\boldsymbol{\\theta})=\np\\!\\bigl(e^{\\mathrm{i}\\theta_{1}},\n         e^{\\mathrm{i}\\theta_{2}},\n         e^{\\mathrm{i}\\theta_{3}}\\bigr),\\qquad\n\\boldsymbol{\\theta}=(\\theta_{1},\\theta_{2},\\theta_{3})\\in\\mathbb{R}^{3},\n\\]\nand denote by  \n\\[\nd\\mu=\\frac{d\\theta_{1}\\,d\\theta_{2}\\,d\\theta_{3}}{(2\\pi)^{3}}\n\\]\nthe normalised Haar measure on the torus $T^{3}=[0,2\\pi]^{3}$.\nOrthogonality of the exponentials gives Parseval’s identity  \n\\[\n\\Gamma\\!\\bigl(p^{\\,n}\\bigr)=\n\\int_{T^{3}}\\!\\bigl|\\Theta_{p}(\\boldsymbol{\\theta})\\bigr|^{2n}\\,d\\mu\n      \\qquad(n\\ge 0).\n\\tag{$\\mathcal{P}$}\n\\]\n\n--------------------------------------------------------------------\nA.  A non–trivial example\n--------------------------------------------------------------------\n\nPermuting $y$ and $z$ in $F$ produces  \n\\[\ng(x,y,z)=\\bigl(2x+1\\bigr)\\bigl(15y+1\\bigr)\\bigl(6z+1\\bigr).\n\\]\nClearly $g(0,0,0)=1$ and $g\\not\\equiv F$.  A permutation of the\nindeterminates merely re–labels the coefficient array, hence preserves the\nEuclidean norm; therefore $\\Gamma\\!\\bigl(g^{\\,n}\\bigr)=\\Gamma\\!\\bigl(F^{\\,n}\\bigr)$\nfor every $n\\ge 1$.\n\n--------------------------------------------------------------------\nB.  Classification inside the linear family\n--------------------------------------------------------------------\n\n--------------------------------------------------------------------\nB.1  The singularity sets that must coincide\n--------------------------------------------------------------------\n\nFor a one–variable linear form $L_{a}(t)=1+at$ put  \n\\[\nS_{n}(a)=\\Gamma\\!\\bigl(L_{a}(t)^{\\,n}\\bigr)=\n        \\sum_{k=0}^{n}\\binom{n}{k}^{2}a^{\\,2k},\n\\qquad \nG_{a}(z)=\\sum_{n=0}^{\\infty}S_{n}(a)z^{n}\n       =\\frac{1}{\\sqrt{\\bigl(1-(1+a)^{2}z\\bigr)\n                       \\bigl(1-(1-a)^{2}z\\bigr)}} .\n\\tag{1}\n\\]\n\nBecause the three variables are independent,  \n\\[\n\\sum_{n\\ge 0}\\Gamma\\!\\bigl(h^{\\,n}\\bigr)z^{n}\n     =\\prod_{j=1}^{3}G_{\\alpha_{j}}(z),\n\\qquad\n\\sum_{n\\ge 0}\\Gamma\\!\\bigl(F^{\\,n}\\bigr)z^{n}\n     =G_{2}(z)G_{6}(z)G_{15}(z).\n\\]\nHence\n\\[\n\\prod_{j=1}^{3}G_{\\alpha_{j}}(z)=G_{2}(z)G_{6}(z)G_{15}(z).\n\\tag{2}\n\\]\n\nEach $G_{a}$ is a square–root of a rational function whose only\nsingularities (branch points) are at  \n\\[\nz=\\frac{1}{(1+a)^{2}},\\qquad z=\\frac{1}{(1-a)^{2}}.\n\\]\nEquality~(2) therefore forces the two multisets of branch points to be\nidentical:\n\\[\n\\bigl\\{\\tfrac{1}{(1+\\alpha_{1})^{2}},\\tfrac{1}{(1-\\alpha_{1})^{2}},\n       \\tfrac{1}{(1+\\alpha_{2})^{2}},\\tfrac{1}{(1-\\alpha_{2})^{2}},\n       \\tfrac{1}{(1+\\alpha_{3})^{2}},\\tfrac{1}{(1-\\alpha_{3})^{2}}\\bigr\\}\n      =\\Bigl\\{\\tfrac{1}{9},1,\\tfrac{1}{25},\\tfrac{1}{49},\n              \\tfrac{1}{196},\\tfrac{1}{256}\\Bigr\\}.\n\\tag{3}\n\\]\n(The right–hand set comes from $a\\in\\{2,6,15\\}$.)\n\nTaking square roots and reciprocals transforms~(3) into the much simpler\nstatement\n\\[\n\\bigl\\{\\lvert1+\\alpha_{1}\\rvert,\\lvert1-\\alpha_{1}\\rvert,\n       \\lvert1+\\alpha_{2}\\rvert,\\lvert1-\\alpha_{2}\\rvert,\n       \\lvert1+\\alpha_{3}\\rvert,\\lvert1-\\alpha_{3}\\rvert\\bigr\\}\n      =\\{1,3,5,7,14,16\\}.\n\\tag{4}\n\\]\n\n--------------------------------------------------------------------\nB.2  Reconstructing $\\lvert\\alpha_{j}\\rvert$ from the six integers\n--------------------------------------------------------------------\n\nFix any real $\\beta\\neq\\pm1$.  The unordered pair\n\\[\n\\bigl\\lvert1+\\beta\\bigr\\rvert,\\;\\bigl\\lvert1-\\beta\\bigr\\rvert\n\\]\nalways consists of two \\emph{distinct} positive numbers whose difference\nis exactly $2$.  Conversely, if $u>v>0$ and $u-v=2$, then  \n\\[\n\\{\\!u,v\\!\\}=\\{\\lvert1+\\beta\\rvert,\\lvert1-\\beta\\rvert\\}\n\\quad\\Longleftrightarrow\\quad\n\\beta=\\tfrac{u-v}{2}\\cdot\\tfrac{u+v}{2}.\n\\tag{5}\n\\]\nHence each $\\beta$ corresponds to a \\emph{pair} of positive numbers that\ndiffer by $2$, and vice versa.\n\nAmong the six positive integers in the right–hand side of~(4) there are\nexactly four pairs whose members differ by $2$,\n\\[\n(1,3),\\;(3,5),\\;(5,7),\\;(14,16).\n\\]\nBecause we must partition the multiset $\\{1,3,5,7,14,16\\}$ into \\emph{three}\ndisjoint pairs each differing by $2$, a simple “smallest‐element’’ argument\nshows that there is a unique perfect matching:\n\\[\n\\{1,3\\}\\cup\\{5,7\\}\\cup\\{14,16\\}.\n\\]\nIndeed, the smallest element $1$ can only be paired with $3$;\nremoving $\\{1,3\\}$ leaves $\\{5,7,14,16\\}$, whose smallest element $5$ forces\nthe pair $\\{5,7\\}$; the remaining two numbers automatically form the third\npair.\n\nApplying the reconstruction formula~(5) to the three pairs yields\n\\[\n\\lvert\\alpha_{1}\\rvert=2,\\qquad\n\\lvert\\alpha_{2}\\rvert=6,\\qquad\n\\lvert\\alpha_{3}\\rvert=15,\n\\tag{6}\n\\]\nup to ordering.\n\n--------------------------------------------------------------------\nB.3  Sign independence and variable permutations\n--------------------------------------------------------------------\n\nBecause $S_{n}(-a)=S_{n}(a)$ for every $n$ and every real $a$\n(cf.\\ the definition of $S_{n}$ above), replacing any $\\alpha_{j}$ by\n$-\\alpha_{j}$ leaves $\\Gamma\\!\\bigl(h^{\\,n}\\bigr)$ unchanged.\nFurthermore, permuting the variables $(x,y,z)$ only re–orders the\ncoefficients, hence also preserves their Euclidean norm.\nCombining these two observations with~(6) gives the full set of solutions:\n\\[\n(\\alpha_{1},\\alpha_{2},\\alpha_{3})\n=\\bigl(\\varepsilon_{1}2,\\varepsilon_{2}6,\\varepsilon_{3}15\\bigr),\n\\quad\\varepsilon_{j}\\in\\{\\!+1,-1\\!\\},\n\\]\nup to an arbitrary permutation of the three entries.  This completes the\nclassification required in~B.\n\n--------------------------------------------------------------------\nC.  Global rigidity\n--------------------------------------------------------------------\n\n--------------------------------------------------------------------\nC.1  From equality of moments to pointwise equality\n--------------------------------------------------------------------\n\nSet  \n\\[\nU(\\boldsymbol{\\theta})=\\bigl|\\Theta_{H}(\\boldsymbol{\\theta})\\bigr|^{2},\n\\qquad \nV(\\boldsymbol{\\theta})=\\bigl|\\Theta_{F}(\\boldsymbol{\\theta})\\bigr|^{2}\n     \\quad(\\boldsymbol{\\theta}\\in T^{3}).\n\\]\nBoth $U$ and $V$ are continuous, non–negative functions on the compact\n$T^{3}$.  By assumption  \n\\[\n\\int_{T^{3}}U(\\boldsymbol{\\theta})^{\\,n}\\,d\\mu\n     =\\Gamma\\!\\bigl(H^{\\,n}\\bigr)\n     =\\Gamma\\!\\bigl(F^{\\,n}\\bigr)\n     =\\int_{T^{3}}V(\\boldsymbol{\\theta})^{\\,n}\\,d\\mu\n     \\qquad\\forall\\,n\\ge 0.\n\\tag{7}\n\\]\n\n\\emph{Lemma.}\nIf two bounded measurable functions $U,V\\ge 0$ satisfy\n$\\int U^{n}\\,d\\mu=\\int V^{n}\\,d\\mu$ for every $n$, then $U=V$ almost\neverywhere.\n\n\\emph{Proof of the Lemma.}\nPut $W=U-V$.  If $W\\not\\equiv 0$, the set\n$E_{+}=\\{\\boldsymbol{\\theta}:W(\\boldsymbol{\\theta})>0\\}$\nhas positive measure.\nChoose $\\varepsilon>0$ such that \n$E_{\\varepsilon}=\\{\\boldsymbol{\\theta}:W(\\boldsymbol{\\theta})\\ge \\varepsilon\\}$\nstill has positive measure.\nOn $E_{\\varepsilon}$ we have $U\\ge V+\\varepsilon$, hence\n$U^{n}\\ge V^{n}+n\\varepsilon V^{\\,n-1}$.\nIntegrating and using (7) with $n$ large gives a contradiction.\n$\\hfill\\square$\n\nApplying the Lemma we obtain $U=V$ almost everywhere, hence everywhere\nbecause the difference $U-V$ is a trigonometric polynomial and therefore\nreal–analytic on $T^{3}$.  Consequently\n\\[\nH(x,y,z)\\,H\\!\\bigl(x^{-1},y^{-1},z^{-1}\\bigr)=\nF(x,y,z)\\,F\\!\\bigl(x^{-1},y^{-1},z^{-1}\\bigr)\n\\quad\\text{as Laurent polynomials.}\n\\tag{8}\n\\]\n\n--------------------------------------------------------------------\nC.2  Unique factorisation of Laurent polynomials\n--------------------------------------------------------------------\n\nPut  \n\\[\nR=\\mathbb{R}[X^{\\pm1},Y^{\\pm1},Z^{\\pm1}],\\qquad\n\\star:\\,R\\longrightarrow R,\\;\n\\bigl(f^{\\star}\\bigr)(X,Y,Z)=f(X^{-1},Y^{-1},Z^{-1}).\n\\]\nThe ring $R$ is a unique factorisation domain (localisation of the UFD\n$\\mathbb{R}[X,Y,Z]$), and $\\star$ is an involutive automorphism.\nEquation~(8) can be rewritten as  \n\\[\nH\\,H^{\\star}=F\\,F^{\\star}\\qquad\\text{in }R.\n\\tag{9}\n\\]\n\nFactorisation of the right–hand side is explicit:\n\\[\nF\\,F^{\\star}=\n\\bigl(1+2X\\bigr)\\bigl(1+2X^{-1}\\bigr)\\,\n\\bigl(1+6Y\\bigr)\\bigl(1+6Y^{-1}\\bigr)\\,\n\\bigl(1+15Z\\bigr)\\bigl(1+15Z^{-1}\\bigr).\n\\tag{10}\n\\]\nEach linear factor is irreducible in $R$, and any two distinct factors are\ncoprime.  \n\n--------------------------------------------------------------------\nC.3  Locating the factors that divide $H$\n--------------------------------------------------------------------\n\nWrite  \n\\[\nH=\\sum_{i,j,k\\ge 0}c_{i,j,k}X^{i}Y^{j}Z^{k},\\qquad c_{0,0,0}=1.\n\\tag{11}\n\\]\nThus $H$ is a genuine polynomial (no negative exponents).  \nSuppose, for instance, that the factor $1+2X^{-1}$ from~(10) divided $H$:\nthen $H=(1+2X^{-1})Q$ with $Q\\in R$.  \nExpanding, the lowest $X$–exponent occurring in $H$ would be \\emph{negative}\n(unless $Q$ contained a compensating $X$ factor, in which case the constant\nterm of $H$ would change from $1$).  Either alternative contradicts (11).\nHence $1+2X^{-1}$ cannot divide $H$, and the same argument applies to\n$1+6Y^{-1}$ and $1+15Z^{-1}$.\n\nTherefore, for each variable, exactly one of the two factors in the\ncorresponding pair in~(10) divides $H$, and the multiplicity must be~$1$\nbecause it is~$1$ in~(10).  Thus\n\\[\nH=\\varepsilon\\,\n(1+\\varepsilon_{1}2X)\\,(1+\\varepsilon_{2}6Y)\\,(1+\\varepsilon_{3}15Z),\n\\tag{12}\n\\]\nwhere $\\varepsilon,\\varepsilon_{1},\\varepsilon_{2},\\varepsilon_{3}\n      \\in\\{\\!+1,-1\\!\\}$.\nThe constant term condition $c_{0,0,0}=1$ forces $\\varepsilon=1$.\nFinally, permuting $(X,Y,Z)$ corresponds to permuting $(x,y,z)$,\nso~(12) translates back to the claimed form of $H$.\n\n--------------------------------------------------------------------\nC.4  Conclusion\n--------------------------------------------------------------------\n\nCombining~(12) with the independent sign choices\n$\\varepsilon_{j}\\in\\{\\!+1,-1\\!\\}$ and an arbitrary permutation\n$\\sigma\\in\\mathfrak{S}_{3}$ we arrive precisely at the family singled\nout in~B, completing the proof of~C.\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.536543",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension & more variables:  We moved from univariate to\n  trivariate polynomials, turning a one-dimensional coefficient array\n  into a three-dimensional one.\n\n• Additional constraints:  Besides finding one suitable g, the solver\n  must now classify all polynomials h fulfilling the quadratic-norm\n  equalities, not merely exhibit one example.\n\n• Deeper theory:  The solution demands familiarity with\n  Laurent-polynomial rings in several variables, their unique\n  factorisation, and the behaviour of quadratic norms under variable\n  inversion.  These topics do not appear in the original problem.\n\n• Interacting concepts:  The argument blends algebraic factorisation,\n  symmetries of Euclidean norms, invariance under coefficient\n  permutation, and multiplicativity of the γ-map.\n\n• More steps & insight:  Establishing γ for three variables, proving its\n  multiplicativity, analysing constant terms, and executing the UFD\n  uniqueness argument collectively require significantly more work and\n  conceptual depth than the original single-variable manipulation."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}