{
  "index": "2002-B-6",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $p$ be a prime number. Prove that the determinant of the matrix\n\\[\n\\begin{pmatrix}\nx & y & z \\\\\nx^p & y^p & z^p \\\\\nx^{p^2} & y^{p^2} & z^{p^2}\n\\end{pmatrix}\n\\]\nis congruent modulo $p$ to a product of polynomials of the form\n$ax+by+cz$, where $a,b,c$ are integers. (We say two integer\npolynomials are congruent modulo $p$ if corresponding coefficients\nare congruent modulo $p$.)\n\n\\end{itemize}\n\\end{document}",
  "solution": "We prove that the determinant is congruent modulo $p$ to\n\\begin{equation} \\label{b6eq2}\nx \\prod_{i=0}^{p-1} (y+ix) \\prod_{i,j=0}^{p-1} (z + ix + jy).\n\\end{equation}\nWe first check that\n\\begin{equation} \\label{b6eq1}\n\\prod_{i=0}^{p-1} (y+ix) \\equiv y^p - x^{p-1} y \\pmod{p}.\n\\end{equation}\nSince both sides are homogeneous as polynomials in $x$ and $y$,\nit suffices to check \\eqref{b6eq1}\nfor $x=1$, as a congruence between polynomials. Now note that\nthe right side has $0,1,\\dots,p-1$ as roots modulo $p$, as does\nthe left side. Moreover, both sides have the same leading coefficient.\nSince they both have degree only $p$, they must then coincide.\n\nWe thus have\n\\begin{multline*}\nx \\prod_{i=0}^{p-1} (y+ix) \\prod_{i,j=0}^{p-1} (z + ix + jy) \\\\\n\\begin{aligned}\n&\\equiv x (y^p - x^{p-1}y) \\prod_{j=0}^{p-1} ((z+jy)^p - x^{p-1} (z+jy)) \\\\\n&\\equiv (xy^p - x^p y) \\prod_{j=0}^{p-1} (z^p - x^{p-1} z + j y^p - j x^{p-1}\ny) \\\\\n&\\equiv (xy^p - x^p y) ((z^p - x^{p-1} z)^p \\\\\n&\\quad - (y^p - x^{p-1}y)^{p-1}(z^p\n- x^{p-1}z)) \\\\\n &\\equiv (xy^p - x^p y)(z^{p^2} - x^{p^2 - p}z^p) \\\\\n&\\quad - x(y^p - x^{p-1}y)^p\n(z^p - x^{p-1}z) \\\\\n&\\equiv xy^p z^{p^2} - x^p y z^{p^2} - x^{p^2-p+1} y^p z^p\n+ x^{p^2} y z^p \\\\\n&\\quad - xy^{p^2}z^p + x^{p^2-p+1} y^p z^p\n+ x^py^{p^2}z - x^{p^2} y^p z \\\\\n&\\equiv x y^p z^{p^2} + y z^p x^{p^2} + z x^p y^{p^2} \\\\\n&\\quad\n- x z^p y^{p^2} - y x^p z^{p^2} - z y^p x^{p^2},\n\\end{aligned}\n\\end{multline*}\nwhich is precisely the desired determinant.\n\nNote: a simpler conceptual proof is as follows. (Everything in this\nparagraph will be modulo $p$.) Note that for any\nintegers $a,b,c$, the column vector $[ax + by + cz, (ax + by + cz)^p,\n(ax + by + cz)^{p^2}]$ is a linear combination of the columns of the\ngiven matrix. Thus $ax+by+cz$ divides the determinant.\nIn particular, all of the factors of \\eqref{b6eq2} divide the determinant;\nsince both \\eqref{b6eq2} and the determinant have degree $p^2+p+1$,\nthey agree up to a scalar multiple. Moreover, they have the same\ncoefficient of $z^{p^2} y^p x$ (since this term only appears in\nthe expansion of \\eqref{b6eq2} when you choose the first term in\neach factor). Thus the determinant is congruent to \\eqref{b6eq2}, as desired.\n\nEither argument can be used to generalize to a corresponding $n \\times n$\ndeterminant, called a Moore determinant;\nwe leave the precise formulation to the reader. Note\nthe similarity with the classical Vandermonde determinant: if\n$A$ is the $n \\times n$ matrix with $A_{ij} = x_i^j$ for\n$i,j=0, \\dots, n-1$, then\n\\[\n\\det(A) = \\prod_{1 \\leq i < j \\leq n} (x_j - x_i).\n\\]\n\n\\end{itemize}\n\n\\end{document}",
  "vars": [
    "x",
    "y",
    "z"
  ],
  "params": [
    "p",
    "a",
    "b",
    "c",
    "i",
    "j"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "symbolx",
        "y": "symboly",
        "z": "symbolz",
        "p": "primep",
        "a": "parama",
        "b": "paramb",
        "c": "paramc",
        "i": "indexi",
        "j": "indexj"
      },
      "question": "Let $primep$ be a prime number. Prove that the determinant of the matrix\n\\[\n\\begin{pmatrix}\nsymbolx & symboly & symbolz \\\\\nsymbolx^{primep} & symboly^{primep} & symbolz^{primep} \\\\\nsymbolx^{primep^2} & symboly^{primep^2} & symbolz^{primep^2}\n\\end{pmatrix}\n\\]\nis congruent modulo $primep$ to a product of polynomials of the form\n$parama symbolx+paramb symboly+paramc symbolz$, where $parama,paramb,paramc$ are integers. (We say two integer\npolynomials are congruent modulo $primep$ if corresponding coefficients\nare congruent modulo $primep$.)",
      "solution": "We prove that the determinant is congruent modulo $primep$ to\n\\begin{equation} \\label{b6eq2}\nsymbolx \\prod_{indexi=0}^{primep-1} (symboly+ indexi symbolx) \\prod_{indexi,indexj=0}^{primep-1} (symbolz + indexi symbolx + indexj symboly).\n\\end{equation}\nWe first check that\n\\begin{equation} \\label{b6eq1}\n\\prod_{indexi=0}^{primep-1} (symboly+ indexi symbolx) \\equiv symboly^{primep} - symbolx^{primep-1} symboly \\pmod{primep}.\n\\end{equation}\nSince both sides are homogeneous as polynomials in $symbolx$ and $symboly$,\nit suffices to check \\eqref{b6eq1}\nfor $symbolx=1$, as a congruence between polynomials. Now note that\nthe right side has $0,1,\\dots,primep-1$ as roots modulo $primep$, as does\nthe left side. Moreover, both sides have the same leading coefficient.\nSince they both have degree only $primep$, they must then coincide.\n\nWe thus have\n\\begin{multline*}\nsymbolx \\prod_{indexi=0}^{primep-1} (symboly+ indexi symbolx) \\prod_{indexi,indexj=0}^{primep-1} (symbolz + indexi symbolx + indexj symboly) \\\\\n\\begin{aligned}\n&\\equiv symbolx (symboly^{primep} - symbolx^{primep-1} symboly) \\prod_{indexj=0}^{primep-1} ((symbolz+ indexj symboly)^{primep} - symbolx^{primep-1} (symbolz+ indexj symboly)) \\\\\n&\\equiv (symbolx symboly^{primep} - symbolx^{primep} symboly) \\prod_{indexj=0}^{primep-1} (symbolz^{primep} - symbolx^{primep-1} symbolz + indexj symboly^{primep} - indexj symbolx^{primep-1}\nsymboly) \\\\\n&\\equiv (symbolx symboly^{primep} - symbolx^{primep} symboly) ((symbolz^{primep} - symbolx^{primep-1} symbolz)^{primep} \\\\\n&\\quad - (symboly^{primep} - symbolx^{primep-1} symboly)^{primep-1}(symbolz^{primep}\n- symbolx^{primep-1} symbolz)) \\\\\n &\\equiv (symbolx symboly^{primep} - symbolx^{primep} symboly)(symbolz^{primep^2} - symbolx^{primep^2 - primep}symbolz^{primep}) \\\\\n&\\quad - symbolx(symboly^{primep} - symbolx^{primep-1} symboly)^{primep}\n(symbolz^{primep} - symbolx^{primep-1}symbolz) \\\\\n&\\equiv symbolx symboly^{primep} symbolz^{primep^2} - symbolx^{primep} symboly symbolz^{primep^2} - symbolx^{primep^2-primep+1} symboly^{primep} symbolz^{primep}\n+ symbolx^{primep^2} symboly symbolz^{primep} \\\\\n&\\quad - symbolx symboly^{primep^2}symbolz^{primep} + symbolx^{primep^2-primep+1} symboly^{primep} symbolz^{primep}\n+ symbolx^{primep}symboly^{primep^2}symbolz - symbolx^{primep^2} symboly^{primep} symbolz \\\\\n&\\equiv symbolx symboly^{primep} symbolz^{primep^2} + symboly symbolz^{primep} symbolx^{primep^2} + symbolz symbolx^{primep} symboly^{primep^2} \\\\\n&\\quad\n- symbolx symbolz^{primep} symboly^{primep^2} - symboly symbolx^{primep} symbolz^{primep^2} - symbolz symboly^{primep} symbolx^{primep^2},\n\\end{aligned}\n\\end{multline*}\nwhich is precisely the desired determinant.\n\nNote: a simpler conceptual proof is as follows. (Everything in this\nparagraph will be modulo $primep$.) Note that for any\nintegers $parama,paramb,paramc$, the column vector $[parama symbolx + paramb symboly + paramc symbolz, (parama symbolx + paramb symboly + paramc symbolz)^{primep},\n(parama symbolx + paramb symboly + paramc symbolz)^{primep^2}]$ is a linear combination of the columns of the\ngiven matrix. Thus $parama symbolx+paramb symboly+paramc symbolz$ divides the determinant.\nIn particular, all of the factors of \\eqref{b6eq2} divide the determinant;\nsince both \\eqref{b6eq2} and the determinant have degree $primep^2+primep+1$,\nthey agree up to a scalar multiple. Moreover, they have the same\ncoefficient of $symbolz^{primep^2} symboly^{primep} symbolx$ (since this term only appears in\nthe expansion of \\eqref{b6eq2} when you choose the first term in\neach factor). Thus the determinant is congruent to \\eqref{b6eq2}, as desired.\n\nEither argument can be used to generalize to a corresponding $n \\times n$\ndeterminant, called a Moore determinant;\nwe leave the precise formulation to the reader. Note\nthe similarity with the classical Vandermonde determinant: if\n$A$ is the $n \\times n$ matrix with $A_{indexi indexj} = symbolx_{indexi}^{indexj}$ for\n$indexi,indexj=0, \\dots, n-1$, then\n\\[\n\\det(A) = \\prod_{1 \\leq indexi < indexj \\leq n} (symbolx_{indexj} - symbolx_{indexi}).\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "compasspoint",
        "y": "maritimefog",
        "z": "lanternpost",
        "p": "thunderclap",
        "a": "sunriseview",
        "b": "driftwood",
        "c": "meadowbloom",
        "i": "pebblesand",
        "j": "sandpiper"
      },
      "question": "Let $thunderclap$ be a prime number. Prove that the determinant of the matrix\n\\[\n\\begin{pmatrix}\ncompasspoint & maritimefog & lanternpost \\\\\ncompasspoint^{thunderclap} & maritimefog^{thunderclap} & lanternpost^{thunderclap} \\\\\ncompasspoint^{thunderclap^2} & maritimefog^{thunderclap^2} & lanternpost^{thunderclap^2}\n\\end{pmatrix}\n\\]\nis congruent modulo $thunderclap$ to a product of polynomials of the form\n$sunriseviewcompasspoint+driftwoodmaritimefog+meadowbloomlanternpost$, where $sunriseview,driftwood,meadowbloom$ are integers. (We say two integer\npolynomials are congruent modulo $thunderclap$ if corresponding coefficients\nare congruent modulo $thunderclap$.)",
      "solution": "We prove that the determinant is congruent modulo $thunderclap$ to\n\\begin{equation} \\label{b6eq2}\ncompasspoint \\prod_{pebblesand=0}^{thunderclap-1} (maritimefog+pebblesandcompasspoint) \\prod_{pebblesand,sandpiper=0}^{thunderclap-1} (lanternpost + pebblesandcompasspoint + sandpipermaritimefog).\n\\end{equation}\nWe first check that\n\\begin{equation} \\label{b6eq1}\n\\prod_{pebblesand=0}^{thunderclap-1} (maritimefog+pebblesandcompasspoint) \\equiv maritimefog^{thunderclap} - compasspoint^{thunderclap-1} maritimefog \\pmod{thunderclap}.\n\\end{equation}\nSince both sides are homogeneous as polynomials in $compasspoint$ and $maritimefog$,\nit suffices to check \\eqref{b6eq1}\nfor $compasspoint=1$, as a congruence between polynomials. Now note that\nthe right side has $0,1,\\dots,thunderclap-1$ as roots modulo $thunderclap$, as does\nthe left side. Moreover, both sides have the same leading coefficient.\nSince they both have degree only $thunderclap$, they must then coincide.\n\nWe thus have\n\\begin{multline*}\ncompasspoint \\prod_{pebblesand=0}^{thunderclap-1} (maritimefog+pebblesandcompasspoint) \\prod_{pebblesand,sandpiper=0}^{thunderclap-1} (lanternpost + pebblesandcompasspoint + sandpipermaritimefog) \\\\\n\\begin{aligned}\n&\\equiv compasspoint (maritimefog^{thunderclap} - compasspoint^{thunderclap-1}maritimefog) \\prod_{sandpiper=0}^{thunderclap-1} ((lanternpost+sandpipermaritimefog)^{thunderclap} - compasspoint^{thunderclap-1} (lanternpost+sandpipermaritimefog)) \\\\\n&\\equiv (compasspointmaritimefog^{thunderclap} - compasspoint^{thunderclap} maritimefog) \\prod_{sandpiper=0}^{thunderclap-1} (lanternpost^{thunderclap} - compasspoint^{thunderclap-1} lanternpost + sandpiper maritimefog^{thunderclap} - sandpiper compasspoint^{thunderclap-1}\nmaritimefog) \\\\\n&\\equiv (compasspointmaritimefog^{thunderclap} - compasspoint^{thunderclap} maritimefog) ((lanternpost^{thunderclap} - compasspoint^{thunderclap-1} lanternpost)^{thunderclap} \\\\\n&\\quad - (maritimefog^{thunderclap} - compasspoint^{thunderclap-1}maritimefog)^{thunderclap-1}(lanternpost^{thunderclap}\n- compasspoint^{thunderclap-1}lanternpost)) \\\\\n &\\equiv (compasspointmaritimefog^{thunderclap} - compasspoint^{thunderclap} maritimefog)(lanternpost^{thunderclap^2} - compasspoint^{thunderclap^2 - thunderclap}lanternpost^{thunderclap}) \\\\\n&\\quad - compasspoint(maritimefog^{thunderclap} - compasspoint^{thunderclap-1}maritimefog)^{thunderclap}\n(lanternpost^{thunderclap} - compasspoint^{thunderclap-1}lanternpost) \\\\\n&\\equiv compasspointmaritimefog^{thunderclap} lanternpost^{thunderclap^2} - compasspoint^{thunderclap} maritimefog lanternpost^{thunderclap^2} - compasspoint^{thunderclap^2-thunderclap+1} maritimefog^{thunderclap} lanternpost^{thunderclap}\n+ compasspoint^{thunderclap^2} maritimefog lanternpost^{thunderclap} \\\\\n&\\quad - compasspointmaritimefog^{thunderclap^2}lanternpost^{thunderclap} + compasspoint^{thunderclap^2-thunderclap+1} maritimefog^{thunderclap} lanternpost^{thunderclap}\n+ compasspoint^{thunderclap}maritimefog^{thunderclap^2}lanternpost - compasspoint^{thunderclap^2} maritimefog^{thunderclap} lanternpost \\\\\n&\\equiv compasspoint maritimefog^{thunderclap} lanternpost^{thunderclap^2} + maritimefog lanternpost^{thunderclap} compasspoint^{thunderclap^2} + lanternpost compasspoint^{thunderclap} maritimefog^{thunderclap^2} \\\\\n&\\quad - compasspoint lanternpost^{thunderclap} maritimefog^{thunderclap^2} - maritimefog compasspoint^{thunderclap} lanternpost^{thunderclap^2} - lanternpost maritimefog^{thunderclap} compasspoint^{thunderclap^2},\n\\end{aligned}\n\\end{multline*}\nwhich is precisely the desired determinant.\n\nNote: a simpler conceptual proof is as follows. (Everything in this\nparagraph will be modulo $thunderclap$.) Note that for any\nintegers $sunriseview,driftwood,meadowbloom$, the column vector $[sunriseviewcompasspoint + driftwoodmaritimefog + meadowbloomlanternpost, (sunriseviewcompasspoint + driftwoodmaritimefog + meadowbloomlanternpost)^{thunderclap},\n(sunriseviewcompasspoint + driftwoodmaritimefog + meadowbloomlanternpost)^{thunderclap^2}]$ is a linear combination of the columns of the\ngiven matrix. Thus $sunriseviewcompasspoint+driftwoodmaritimefog+meadowbloomlanternpost$ divides the determinant.\nIn particular, all of the factors of \\eqref{b6eq2} divide the determinant;\nsince both \\eqref{b6eq2} and the determinant have degree $thunderclap^2+thunderclap+1$,\nthey agree up to a scalar multiple. Moreover, they have the same\ncoefficient of $lanternpost^{thunderclap^2} maritimefog^{thunderclap} compasspoint$ (since this term only appears in\nthe expansion of \\eqref{b6eq2} when you choose the first term in\neach factor). Thus the determinant is congruent to \\eqref{b6eq2}, as desired.\n\nEither argument can be used to generalize to a corresponding $n \\times n$\ndeterminant, called a Moore determinant;\nwe leave the precise formulation to the reader. Note\nthe similarity with the classical Vandermonde determinant: if\n$A$ is the $n \\times n$ matrix with $A_{pebblesandsandpiper} = compasspoint_{pebblesand}^{sandpiper}$ for\n$pebblesand,sandpiper=0, \\dots, n-1$, then\n\\[\n\\det(A) = \\prod_{1 \\leq pebblesand < sandpiper \\leq n} (compasspoint_{sandpiper} - compasspoint_{pebblesand}).\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantalpha",
        "y": "constantbeta",
        "z": "constantgamma",
        "p": "compositeomega",
        "a": "fractionaldelta",
        "b": "fractionalepsilon",
        "c": "fractionalzeta",
        "i": "continuouseta",
        "j": "continuoustheta"
      },
      "question": "Let $compositeomega$ be a prime number. Prove that the determinant of the matrix\n\\[\n\\begin{pmatrix}\nconstantalpha & constantbeta & constantgamma \\\\\nconstantalpha^{compositeomega} & constantbeta^{compositeomega} & constantgamma^{compositeomega} \\\\\nconstantalpha^{compositeomega^2} & constantbeta^{compositeomega^2} & constantgamma^{compositeomega^2}\n\\end{pmatrix}\n\\]\nis congruent modulo $compositeomega$ to a product of polynomials of the form\n$fractionaldelta constantalpha+fractionalepsilon constantbeta+fractionalzeta constantgamma$, where $fractionaldelta,fractionalepsilon,fractionalzeta$ are integers. (We say two integer\npolynomials are congruent modulo $compositeomega$ if corresponding coefficients\nare congruent modulo $compositeomega$.)\n\n\\end{itemize}\n\\end{document}",
      "solution": "We prove that the determinant is congruent modulo $compositeomega$ to\n\\begin{equation} \\label{b6eq2}\nconstantalpha \\prod_{continuouseta=0}^{compositeomega-1} (constantbeta+continuouseta constantalpha) \\prod_{continuouseta,continuoustheta=0}^{compositeomega-1} (constantgamma + continuouseta constantalpha + continuoustheta constantbeta).\n\\end{equation}\nWe first check that\n\\begin{equation} \\label{b6eq1}\n\\prod_{continuouseta=0}^{compositeomega-1} (constantbeta+continuouseta constantalpha) \\equiv constantbeta^{compositeomega} - constantalpha^{compositeomega-1} constantbeta \\pmod{compositeomega}.\n\\end{equation}\nSince both sides are homogeneous as polynomials in $constantalpha$ and $constantbeta$,\nit suffices to check \\eqref{b6eq1}\nfor $constantalpha=1$, as a congruence between polynomials. Now note that\nthe right side has $0,1,\\dots,compositeomega-1$ as roots modulo $compositeomega$, as does\nthe left side. Moreover, both sides have the same leading coefficient.\nSince they both have degree only $compositeomega$, they must then coincide.\n\nWe thus have\n\\begin{multline*}\nconstantalpha \\prod_{continuouseta=0}^{compositeomega-1} (constantbeta+continuouseta constantalpha) \\prod_{continuouseta,continuoustheta=0}^{compositeomega-1} (constantgamma + continuouseta constantalpha + continuoustheta constantbeta) \\\\\n\\begin{aligned}\n&\\equiv constantalpha (constantbeta^{compositeomega} - constantalpha^{compositeomega-1}constantbeta) \\prod_{continuoustheta=0}^{compositeomega-1} ((constantgamma+continuoustheta constantbeta)^{compositeomega} - constantalpha^{compositeomega-1} (constantgamma+continuoustheta constantbeta)) \\\\\n&\\equiv (constantalpha constantbeta^{compositeomega} - constantalpha^{compositeomega} constantbeta) \\prod_{continuoustheta=0}^{compositeomega-1} (constantgamma^{compositeomega} - constantalpha^{compositeomega-1} constantgamma + continuoustheta constantbeta^{compositeomega} - continuoustheta constantalpha^{compositeomega-1}\nconstantbeta) \\\\\n&\\equiv (constantalpha constantbeta^{compositeomega} - constantalpha^{compositeomega} constantbeta) ((constantgamma^{compositeomega} - constantalpha^{compositeomega-1} constantgamma)^{compositeomega} \\\\\n&\\quad - (constantbeta^{compositeomega} - constantalpha^{compositeomega-1}constantbeta)^{compositeomega-1}(constantgamma^{compositeomega}\n- constantalpha^{compositeomega-1}constantgamma)) \\\\\n &\\equiv (constantalpha constantbeta^{compositeomega} - constantalpha^{compositeomega} constantbeta)(constantgamma^{compositeomega^2} - constantalpha^{compositeomega^2 - compositeomega}constantgamma^{compositeomega}) \\\\\n&\\quad - constantalpha(constantbeta^{compositeomega} - constantalpha^{compositeomega-1}constantbeta)^{compositeomega}\n(constantgamma^{compositeomega} - constantalpha^{compositeomega-1}constantgamma) \\\\\n&\\equiv constantalpha constantbeta^{compositeomega} constantgamma^{compositeomega^2} - constantalpha^{compositeomega} constantbeta constantgamma^{compositeomega^2} - constantalpha^{compositeomega^2-compositeomega+1} constantbeta^{compositeomega} constantgamma^{compositeomega}\n+ constantalpha^{compositeomega^2} constantbeta constantgamma^{compositeomega} \\\\\n&\\quad - constantalpha constantbeta^{compositeomega^2}constantgamma^{compositeomega} + constantalpha^{compositeomega^2-compositeomega+1} constantbeta^{compositeomega} constantgamma^{compositeomega}\n+ constantalpha^{compositeomega}constantbeta^{compositeomega^2}constantgamma - constantalpha^{compositeomega^2} constantbeta^{compositeomega} constantgamma \\\\\n&\\equiv constantalpha constantbeta^{compositeomega} constantgamma^{compositeomega^2} + constantbeta constantgamma^{compositeomega} constantalpha^{compositeomega^2} + constantgamma constantalpha^{compositeomega} constantbeta^{compositeomega^2} \\\\\n&\\quad\n- constantalpha constantgamma^{compositeomega} constantbeta^{compositeomega^2} - constantbeta constantalpha^{compositeomega} constantgamma^{compositeomega^2} - constantgamma constantbeta^{compositeomega} constantalpha^{compositeomega^2},\n\\end{aligned}\n\\end{multline*}\nwhich is precisely the desired determinant.\n\nNote: a simpler conceptual proof is as follows. (Everything in this\nparagraph will be modulo $compositeomega$.) Note that for any\nintegers $fractionaldelta,fractionalepsilon,fractionalzeta$, the column vector $[fractionaldelta constantalpha + fractionalepsilon constantbeta + fractionalzeta constantgamma, (fractionaldelta constantalpha + fractionalepsilon constantbeta + fractionalzeta constantgamma)^{compositeomega},\n(fractionaldelta constantalpha + fractionalepsilon constantbeta + fractionalzeta constantgamma)^{compositeomega^2}]$ is a linear combination of the columns of the\ngiven matrix. Thus $fractionaldelta constantalpha+fractionalepsilon constantbeta+fractionalzeta constantgamma$ divides the determinant.\nIn particular, all of the factors of \\eqref{b6eq2} divide the determinant;\nsince both \\eqref{b6eq2} and the determinant have degree $compositeomega^2+compositeomega+1$,\nthey agree up to a scalar multiple. Moreover, they have the same\ncoefficient of $constantgamma^{compositeomega^2} constantbeta^{compositeomega} constantalpha$ (since this term only appears in\nthe expansion of \\eqref{b6eq2} when you choose the first term in\neach factor). Thus the determinant is congruent to \\eqref{b6eq2}, as desired.\n\nEither argument can be used to generalize to a corresponding $n \\times n$\ndeterminant, called a Moore determinant;\nwe leave the precise formulation to the reader. Note\nthe similarity with the classical Vandermonde determinant: if\n$A$ is the $n \\times n$ matrix with $A_{continuouseta continuoustheta} = constantalpha_{continuouseta}^{continuoustheta}$ for\n$continuouseta,continuoustheta=0, \\dots, n-1$, then\n\\[\n\\det(A) = \\prod_{1 \\leq continuouseta < continuoustheta \\leq n} (constantalpha_{continuoustheta} - constantalpha_{continuouseta}).\n\\]\n\n\\end{itemize}\n\n\\end{document}"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mpdlqvce",
        "p": "tznbrfsa",
        "a": "gmqscvlj",
        "b": "krnvdxoe",
        "c": "fplwzhtu",
        "i": "dsbcamre",
        "j": "vxqltgpu"
      },
      "question": "Let $tznbrfsa$ be a prime number. Prove that the determinant of the matrix\n\\[\n\\begin{pmatrix}\nqzxwvtnp & hjgrksla & mpdlqvce \\\\\nqzxwvtnp^{tznbrfsa} & hjgrksla^{tznbrfsa} & mpdlqvce^{tznbrfsa} \\\\\nqzxwvtnp^{tznbrfsa^{2}} & hjgrksla^{tznbrfsa^{2}} & mpdlqvce^{tznbrfsa^{2}}\n\\end{pmatrix}\n\\]\nis congruent modulo $tznbrfsa$ to a product of polynomials of the form $gmqscvlj\\, qzxwvtnp + krnvdxoe\\, hjgrksla + fplwzhtu\\, mpdlqvce$, where $gmqscvlj,krnvdxoe,fplwzhtu$ are integers.",
      "solution": "We prove that the determinant is congruent modulo $tznbrfsa$ to\n\\begin{equation} \\label{b6eq2}\nqzxwvtnp \\prod_{dsbcamre=0}^{tznbrfsa-1} (hjgrksla + dsbcamre\\, qzxwvtnp) \\prod_{dsbcamre,vxqltgpu=0}^{tznbrfsa-1} (mpdlqvce + dsbcamre\\, qzxwvtnp + vxqltgpu\\, hjgrksla).\n\\end{equation}\nWe first check that\n\\begin{equation} \\label{b6eq1}\n\\prod_{dsbcamre=0}^{tznbrfsa-1} (hjgrksla + dsbcamre\\, qzxwvtnp) \\equiv hjgrksla^{tznbrfsa} - qzxwvtnp^{tznbrfsa-1} hjgrksla \\pmod{tznbrfsa}.\n\\end{equation}\nSince both sides are homogeneous as polynomials in $qzxwvtnp$ and $hjgrksla$, it suffices to check \\eqref{b6eq1} for $qzxwvtnp=1$, as a congruence between polynomials. Now note that the right side has $0,1,\\dots ,tznbrfsa-1$ as roots modulo $tznbrfsa$, as does the left side. Moreover, both sides have the same leading coefficient. Since they both have degree only $tznbrfsa$, they must then coincide.\n\nWe thus have\n\\begin{multline*}\nqzxwvtnp \\prod_{dsbcamre=0}^{tznbrfsa-1} (hjgrksla + dsbcamre\\, qzxwvtnp) \\prod_{dsbcamre,vxqltgpu=0}^{tznbrfsa-1} (mpdlqvce + dsbcamre\\, qzxwvtnp + vxqltgpu\\, hjgrksla) \\\\\n\\begin{aligned}\n&\\equiv qzxwvtnp (hjgrksla^{tznbrfsa} - qzxwvtnp^{tznbrfsa-1} hjgrksla) \\prod_{vxqltgpu=0}^{tznbrfsa-1} ((mpdlqvce + vxqltgpu\\, hjgrksla)^{tznbrfsa} - qzxwvtnp^{tznbrfsa-1} (mpdlqvce + vxqltgpu\\, hjgrksla)) \\\\\n&\\equiv (qzxwvtnp hjgrksla^{tznbrfsa} - qzxwvtnp^{tznbrfsa} hjgrksla) \\prod_{vxqltgpu=0}^{tznbrfsa-1} (mpdlqvce^{tznbrfsa} - qzxwvtnp^{tznbrfsa-1} mpdlqvce + vxqltgpu\\, hjgrksla^{tznbrfsa} - vxqltgpu\\, qzxwvtnp^{tznbrfsa-1} hjgrksla) \\\\\n&\\equiv (qzxwvtnp hjgrksla^{tznbrfsa} - qzxwvtnp^{tznbrfsa} hjgrksla) ((mpdlqvce^{tznbrfsa} - qzxwvtnp^{tznbrfsa-1} mpdlqvce)^{tznbrfsa} \\\\\n&\\quad - (hjgrksla^{tznbrfsa} - qzxwvtnp^{tznbrfsa-1} hjgrksla)^{tznbrfsa-1}(mpdlqvce^{tznbrfsa} - qzxwvtnp^{tznbrfsa-1} mpdlqvce)) \\\\\n&\\equiv (qzxwvtnp hjgrksla^{tznbrfsa} - qzxwvtnp^{tznbrfsa} hjgrksla)(mpdlqvce^{tznbrfsa^{2}} - qzxwvtnp^{tznbrfsa^{2} - tznbrfsa} mpdlqvce^{tznbrfsa}) \\\\\n&\\quad - qzxwvtnp(hjgrksla^{tznbrfsa} - qzxwvtnp^{tznbrfsa-1} hjgrksla)^{tznbrfsa}(mpdlqvce^{tznbrfsa} - qzxwvtnp^{tznbrfsa-1} mpdlqvce) \\\\\n&\\equiv qzxwvtnp hjgrksla^{tznbrfsa} mpdlqvce^{tznbrfsa^{2}} - qzxwvtnp^{tznbrfsa} hjgrksla mpdlqvce^{tznbrfsa^{2}} - qzxwvtnp^{tznbrfsa^{2}-tznbrfsa+1} hjgrksla^{tznbrfsa} mpdlqvce^{tznbrfsa} \\\\\n&\\quad + qzxwvtnp^{tznbrfsa^{2}} hjgrksla mpdlqvce^{tznbrfsa} - qzxwvtnp hjgrksla^{tznbrfsa^{2}} mpdlqvce^{tznbrfsa} + qzxwvtnp^{tznbrfsa^{2}-tznbrfsa+1} hjgrksla^{tznbrfsa} mpdlqvce^{tznbrfsa} \\\\\n&\\quad + qzxwvtnp^{tznbrfsa} hjgrksla^{tznbrfsa^{2}} mpdlqvce - qzxwvtnp^{tznbrfsa^{2}} hjgrksla^{tznbrfsa} mpdlqvce \\\\\n&\\equiv qzxwvtnp hjgrksla^{tznbrfsa} mpdlqvce^{tznbrfsa^{2}} + hjgrksla mpdlqvce^{tznbrfsa} qzxwvtnp^{tznbrfsa^{2}} + mpdlqvce qzxwvtnp^{tznbrfsa} hjgrksla^{tznbrfsa^{2}} \\\\\n&\\quad - qzxwvtnp mpdlqvce^{tznbrfsa} hjgrksla^{tznbrfsa^{2}} - hjgrksla qzxwvtnp^{tznbrfsa} mpdlqvce^{tznbrfsa^{2}} - mpdlqvce hjgrksla^{tznbrfsa} qzxwvtnp^{tznbrfsa^{2}},\n\\end{aligned}\n\\end{multline*}\nwhich is precisely the desired determinant.\n\nNote: a simpler conceptual proof is as follows. (Everything in this paragraph will be modulo $tznbrfsa$.) Note that for any integers $gmqscvlj,krnvdxoe,fplwzhtu$, the column vector $[gmqscvlj\\, qzxwvtnp + krnvdxoe\\, hjgrksla + fplwzhtu\\, mpdlqvce, (gmqscvlj\\, qzxwvtnp + krnvdxoe\\, hjgrksla + fplwzhtu\\, mpdlqvce)^{tznbrfsa}, (gmqscvlj\\, qzxwvtnp + krnvdxoe\\, hjgrksla + fplwzhtu\\, mpdlqvce)^{tznbrfsa^{2}}]$ is a linear combination of the columns of the given matrix. Thus $gmqscvlj\\, qzxwvtnp + krnvdxoe\\, hjgrksla + fplwzhtu\\, mpdlqvce$ divides the determinant. In particular, all of the factors of \\eqref{b6eq2} divide the determinant; since both \\eqref{b6eq2} and the determinant have degree $tznbrfsa^{2}+tznbrfsa+1$, they agree up to a scalar multiple. Moreover, they have the same coefficient of $mpdlqvce^{tznbrfsa^{2}} hjgrksla^{tznbrfsa} qzxwvtnp$ (since this term only appears in the expansion of \\eqref{b6eq2} when you choose the first term in each factor). Thus the determinant is congruent to \\eqref{b6eq2}, as desired.\n\nEither argument can be used to generalize to a corresponding $n \\times n$ determinant, called a Moore determinant; we leave the precise formulation to the reader. Note the similarity with the classical Vandermonde determinant: if $A$ is the $n \\times n$ matrix with $A_{dsbcamre vxqltgpu} = qzxwvtnp_{dsbcamre}^{vxqltgpu}$ for $dsbcamre,vxqltgpu=0, \\dots , n-1$, then\n\\[\n\\det(A) = \\prod_{1 \\leq dsbcamre < vxqltgpu \\leq n} (qzxwvtnp_{vxqltgpu} - qzxwvtnp_{dsbcamre}).\n\\]"
    },
    "kernel_variant": {
      "question": "Let $q$ be an odd prime and $n\\ge 1$ an integer.\n\n1.  (The two $(n+1)$–tuples of indeterminates)  \n    Introduce two independent ordered families\n    \\[\n        A=(a_{0},a_{1},\\dots ,a_{n}), \\qquad \n        B=(b_{0},b_{1},\\dots ,b_{n}).\n    \\]\n\n2.  (Concatenation)  \n    Form the $(2n+2)$–tuple\n    \\[\n        C=(c_{0},c_{1},\\dots ,c_{2n+1})\n        :=(a_{0},a_{1},\\dots ,a_{n},\\,b_{0},b_{1},\\dots ,b_{n}),\n    \\]\n    i.e.\n    \\[\n        c_{k}=\n        \\begin{cases}\n            a_{k}, & 0\\le k\\le n,\\\\[4pt]\n            b_{k-n-1}, & n+1\\le k\\le 2n+1 .\n        \\end{cases}\n    \\]\n\n3.  (Moore matrix and determinant)  \n    Let\n    \\[\n        \\mathcal M_{q,n}(A,B):=\n        \\bigl(c_{j}^{\\,q^{\\,i}}\\bigr)_{0\\le i,j\\le 2n+1},\n        \\qquad \n        \\Delta_{n}(A,B):=\\det\\mathcal M_{q,n}(A,B)\\in\\mathbb Z[A,B].\n    \\]\n\nProve the following statements.\n\n(1)  Complete factorisation modulo $q$.  \n     In the polynomial ring $\\mathbb F_{q}[A,B]$\n     \\[\n         \\boxed{\\;\n         \\Delta_{n}(A,B)\\equiv\n         \\,c_{0}\\prod_{k=1}^{2n+1}\\;\n         \\prod_{(i_{1},\\dots ,i_{k})\\in\\mathbb F_{q}^{\\,k}}\n         F_{k}(i_{1},\\dots ,i_{k})\n         \\pmod q\\;},\n     \\]\n     where, for $k\\ge 0$ and $(i_{1},\\dots ,i_{k})\\in\\mathbb F_{q}^{\\,k}$,\n     \\[\n          F_{k}(i_{1},\\dots ,i_{k})\n          :=c_{k}+i_{1}c_{k-1}+\\dots +i_{k}c_{0}.\n     \\]\n     Writing every $c_{j}$ in terms of the $a_{\\bullet}$ and $b_{\\bullet}$\n     yields the explicit piece-wise description\n     \\[\n     \\Delta_{n}(A,B)\\equiv \n     a_{0}\n     \\Bigl[\n     \\prod_{k=1}^{n}\\;\n     \\prod_{(i_{1},\\dots ,i_{k})\\in\\mathbb F_{q}^{\\,k}}\n     \\Bigl(a_{k}+\\sum_{t=1}^{k} i_{t}a_{k-t}\\Bigr)\n     \\Bigr]\n     \\Bigl[\n     \\prod_{r=0}^{n} \\;\n     \\prod_{(i_{1},\\dots ,i_{n+1+r})\\in\\mathbb F_{q}^{\\,n+1+r}}\n     \\Bigl(\n           b_{r}\n           +\\sum_{t=1}^{r} i_{t} b_{\\,r-t}\n           +\\sum_{t=r+1}^{n+1+r} i_{t} a_{\\,n+1+r-t}\n     \\Bigr)\n     \\Bigr]\n     \\pmod q.\n     \\]\n\n(2)  Multiplicity one and uniqueness.  \n     Show that every linear factor in the product of (1) occurs\n     exactly once and that this product is the unique factorisation of\n     $\\Delta_{n}(A,B)$ into linear factors in $\\mathbb F_{q}[A,B]$.\n\n(3)  Square-freeness.  \n     Prove that $\\Delta_{n}(A,B)$ is square-free modulo $q$, i.e.\n     $\\gcd(\\Delta_{n},\\partial\\Delta_{n}/\\partial x)=1$\n     in $\\mathbb F_{q}[A,B]$ for every indeterminate $x$ from\n     $A\\cup B$.\n\n(4)  A specific coefficient.  \n     Determine, in closed form, the coefficient of the monomial\n     \\[\n         a_{0}\\,a_{1}^{\\,q}\\,a_{2}^{\\,q^{2}}\\dotsm a_{n}^{\\,q^{\\,n}}\\;b_{0}\n     \\]\n     in the expansion of $\\Delta_{n}(A,B)$.",
      "solution": "Throughout “$\\equiv$’’ denotes equality in $\\mathbb F_{q}[A,B]$,\nand indices $i,j,k$ start at $0$ unless stated otherwise.\nPut\n\\[\n  F_{0}:=c_{0},\n  \\qquad\n  F_{k}(i_{1},\\dots ,i_{k})\n  :=c_{k}+i_{1}c_{k-1}+\\dots +i_{k}c_{0}\\quad(k\\ge 1),\n\\]\nand write\n\\[\n  s_{m}:=1+q+\\dots +q^{\\,m}.\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{Step 1: Each listed linear form divides $\\Delta_{n}$.}\n\nFix $k$ with $1\\le k\\le 2n+1$ and a tuple\n$\\boldsymbol i=(i_{1},\\dots ,i_{k})\\in\\mathbb F_{q}^{\\,k}$.\nReplace the $k$-th column of $\\mathcal M_{q,n}(A,B)$\nby the $\\mathbb F_{q}$–linear combination\n\\[\n  \\bigl(F_{k}(\\boldsymbol i),F_{k}(\\boldsymbol i)^{q},\n  \\dots ,F_{k}(\\boldsymbol i)^{q^{2n+1}}\\bigr)^{\\!\\mathsf T}.\n\\]\nBecause the Frobenius map $x\\mapsto x^{q}$ is $\\mathbb F_{q}$–linear,\nthis new column is an $\\mathbb F_{q}$–linear combination of the first\n$k$ columns.  Hence, if $F_{k}(\\boldsymbol i)=0$, the modified matrix\nhas two $\\mathbb F_{q}$–dependent columns and therefore determinant~$0$.\nThus $F_{k}(\\boldsymbol i)$ divides $\\Delta_{n}$ in\n$\\mathbb F_{q}[A,B]$.  For $k=0$ the same argument\ngives the factor $c_{0}=a_{0}$.  \\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{Step 2: Equality of the two sides in (1).}\n\nThe Moore determinant of size $2n+2$ satisfies\n\\[\n  \\deg \\Delta_{n}=s_{2n+1}=1+q+\\dots +q^{\\,2n+1}.\n\\]\nOn the right-hand side of (1) every factor is linear,\nand their number is exactly $s_{2n+1}$; the two polynomials\nhave the same total degree.\nSince each factor divides $\\Delta_{n}$, their product\n$\\Phi$ divides $\\Delta_{n}$, so $\\Delta_{n}=\\lambda\\Phi$\nfor some $\\lambda\\in\\mathbb F_{q}$.\n\nConsider the monomial\n\\[\n  M:=c_{0}\\,c_{1}^{\\,q}\\,c_{2}^{\\,q^{2}}\\dotsm c_{2n+1}^{\\,q^{\\,2n+1}}.\n\\]\nIn $\\Phi$ there is exactly one way to obtain $M$:\none must pick the leading term $c_{k}$ from the factor\n$F_{k}(\\boldsymbol i)$ for every $k$, and these leading terms\nare pairwise distinct.  Hence the coefficient of $M$ in $\\Phi$ is~$1$.\n\nFor the determinant $\\Delta_{n}$ a Laplace expansion along the first\nrow shows that the only permutation contributing to $M$ is the\nidentity; its sign is $+1$, hence the coefficient of $M$ in\n$\\Delta_{n}$ is also~$1$.  Consequently $\\lambda=1$ and\n$\\Delta_{n}\\equiv\\Phi$, proving (1).  \\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{Step 3: Multiplicity one, square-freeness and uniqueness.}\n\nFix a linear factor\n$L:=F_{k}(i_{1},\\dots ,i_{k})$ and write\n\\[\n   \\Delta_{n}=L\\cdot G,\n\\]\nwhere $G$ denotes the product of all other linear factors in~(1).\nDifferentiate with respect to $c_{k}$\n(which occurs in $L$ with coefficient~$1$):\n\\[\n   \\frac{\\partial\\Delta_{n}}{\\partial c_{k}}\n   \\;=\\;\n   G\\;+\\;L\\;\\frac{\\partial G}{\\partial c_{k}}.\n\\]\nReducing the above congruence modulo $L$ gives\n$(\\partial\\Delta_{n}/\\partial c_{k})\\equiv G$, which is non-zero\nbecause no other factor vanishes identically modulo $L$.\nHence $L\\nmid(\\partial\\Delta_{n}/\\partial c_{k})$ and so $L^{2}\\nmid\n\\Delta_{n}$.  Therefore every linear factor occurs with multiplicity~$1$\nand $\\Delta_{n}$ is square-free modulo~$q$.\n\nBecause the factors are pair-wise distinct and each appears once,\nany factorisation of $\\Delta_{n}$ into linear factors must contain\nexactly the same multiset of factors; Step~2 forces the same scalar\nmultiple, so the factorisation given in (1) is unique.\n\\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{Step 4: The coefficient of\n$a_{0}\\,a_{1}^{\\,q}\\dotsm a_{n}^{\\,q^{\\,n}}\\,b_{0}$.}\n\nView $\\Delta_{n}$ as a polynomial in the variables\n$c_{0},c_{1},\\dots ,c_{2n+1}$.\nA monomial in its expansion is uniquely determined by a\npermutation $\\sigma$ of $\\{0,1,\\dots ,2n+1\\}$:\nthe exponent of $c_{j}$ equals\n$\\sum_{\\sigma(i)=j} q^{\\,i}$.\nBecause the powers $q^{\\,0},q^{\\,1},\\dots ,q^{\\,2n+1}$\nare distinct, this representation is unique.\n\nSuppose the monomial\n\\[\n  N:=a_{0}\\,a_{1}^{\\,q}\\dotsm a_{n}^{\\,q^{\\,n}}\\,b_{0}\n      =c_{0}\\,c_{1}^{\\,q}\\dotsm c_{n}^{\\,q^{\\,n}}\\;c_{\\,n+1}\n\\]\nappears.  Translating into $c$-variables,\n\\[\n   \\deg_{c_{0}}N=1,\\;\n   \\deg_{c_{1}}N=q,\\;\n   \\dots ,\n   \\deg_{c_{n}}N=q^{\\,n},\\;\n   \\deg_{c_{n+1}}N=1,\\;\n   \\deg_{c_{j}}N=0\\;(j>n+1).\n\\]\nBecause $\\deg_{c_{n+1}}N=1$, exactly one row index $r$\nsatisfies $\\sigma(r)=n+1$ and $q^{\\,r}=1$, hence $r=0$.\nBut $\\deg_{c_{0}}N=1$ imposes simultaneously $\\sigma(0)=0$,\nwhich contradicts $\\sigma(0)=n+1$.  Therefore the monomial\n$N$ \\emph{cannot} occur in $\\Delta_{n}(A,B)$, and its coefficient is\n\\[\n  0.\n  \\qquad\\qquad\\square\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{Answer to (4):} the sought coefficient equals $0$.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.777011",
        "was_fixed": false,
        "difficulty_analysis": "1. Dimension doubled: the matrix has 2n + 2 columns/rows while the\n   current kernel variant has only four.  The presence of two\n   distinct blocks (A and B) makes column-dependence arguments\n   more intricate.\n\n2. Additional structural layer: the student must recognise and\n   exploit the relation between the Moore determinant and the\n   Galois-theoretic norm from 𝔽_{q^{\\,n+1}} to 𝔽_q, a viewpoint\n   absent from the original problem.\n\n3. Multi-part requirement:\n   • Part (1) generalises the triangular product to arbitrary n;\n   • Part (2) demands familiarity with finite-field extensions and\n     conjugacy orbits;\n   • Part (3) introduces square-freeness and uniqueness arguments\n     via the Jacobian criterion;\n   • Part (4) is a non-trivial coefficient extraction exercise.\n\n4. The solution must juggle several advanced techniques:\n   determinant manipulations, combinatorial degree counting,\n   Frobenius automorphism orbits, field norms, and algebraic\n   geometry (Jacobian criterion), moving far beyond the pattern\n   matching that suffices for the 3×3 or 4×4 cases.\n\nHence the enhanced variant is substantively harder, richer in theory,\nand requires a deeper, multi-faceted proof than either the\noriginal or the current kernel versions."
      }
    },
    "original_kernel_variant": {
      "question": "Let $q$ be an odd prime and $n\\ge 1$ an integer.\n\n1.  (The two $(n+1)$–tuples of indeterminates)  \n    Introduce two independent ordered families\n    \\[\n        A=(a_{0},a_{1},\\dots ,a_{n}), \\qquad \n        B=(b_{0},b_{1},\\dots ,b_{n}).\n    \\]\n\n2.  (Concatenation)  \n    Form the $(2n+2)$–tuple\n    \\[\n        C=(c_{0},c_{1},\\dots ,c_{2n+1})\n        :=(a_{0},a_{1},\\dots ,a_{n},\\,b_{0},b_{1},\\dots ,b_{n}),\n    \\]\n    i.e.\n    \\[\n        c_{k}=\n        \\begin{cases}\n            a_{k}, & 0\\le k\\le n,\\\\[4pt]\n            b_{k-n-1}, & n+1\\le k\\le 2n+1 .\n        \\end{cases}\n    \\]\n\n3.  (Moore matrix and determinant)  \n    Let\n    \\[\n        \\mathcal M_{q,n}(A,B):=\n        \\bigl(c_{j}^{\\,q^{\\,i}}\\bigr)_{0\\le i,j\\le 2n+1},\n        \\qquad \n        \\Delta_{n}(A,B):=\\det\\mathcal M_{q,n}(A,B)\\in\\mathbb Z[A,B].\n    \\]\n\nProve the following statements.\n\n(1)  Complete factorisation modulo $q$.  \n     In the polynomial ring $\\mathbb F_{q}[A,B]$\n     \\[\n         \\boxed{\\;\n         \\Delta_{n}(A,B)\\equiv\n         \\,c_{0}\\prod_{k=1}^{2n+1}\\;\n         \\prod_{(i_{1},\\dots ,i_{k})\\in\\mathbb F_{q}^{\\,k}}\n         F_{k}(i_{1},\\dots ,i_{k})\n         \\pmod q\\;},\n     \\]\n     where, for $k\\ge 0$ and $(i_{1},\\dots ,i_{k})\\in\\mathbb F_{q}^{\\,k}$,\n     \\[\n          F_{k}(i_{1},\\dots ,i_{k})\n          :=c_{k}+i_{1}c_{k-1}+\\dots +i_{k}c_{0}.\n     \\]\n     Writing every $c_{j}$ in terms of the $a_{\\bullet}$ and $b_{\\bullet}$\n     yields the explicit piece-wise description\n     \\[\n     \\Delta_{n}(A,B)\\equiv \n     a_{0}\n     \\Bigl[\n     \\prod_{k=1}^{n}\\;\n     \\prod_{(i_{1},\\dots ,i_{k})\\in\\mathbb F_{q}^{\\,k}}\n     \\Bigl(a_{k}+\\sum_{t=1}^{k} i_{t}a_{k-t}\\Bigr)\n     \\Bigr]\n     \\Bigl[\n     \\prod_{r=0}^{n} \\;\n     \\prod_{(i_{1},\\dots ,i_{n+1+r})\\in\\mathbb F_{q}^{\\,n+1+r}}\n     \\Bigl(\n           b_{r}\n           +\\sum_{t=1}^{r} i_{t} b_{\\,r-t}\n           +\\sum_{t=r+1}^{n+1+r} i_{t} a_{\\,n+1+r-t}\n     \\Bigr)\n     \\Bigr]\n     \\pmod q.\n     \\]\n\n(2)  Multiplicity one and uniqueness.  \n     Show that every linear factor in the product of (1) occurs\n     exactly once and that this product is the unique factorisation of\n     $\\Delta_{n}(A,B)$ into linear factors in $\\mathbb F_{q}[A,B]$.\n\n(3)  Square-freeness.  \n     Prove that $\\Delta_{n}(A,B)$ is square-free modulo $q$, i.e.\n     $\\gcd(\\Delta_{n},\\partial\\Delta_{n}/\\partial x)=1$\n     in $\\mathbb F_{q}[A,B]$ for every indeterminate $x$ from\n     $A\\cup B$.\n\n(4)  A specific coefficient.  \n     Determine, in closed form, the coefficient of the monomial\n     \\[\n         a_{0}\\,a_{1}^{\\,q}\\,a_{2}^{\\,q^{2}}\\dotsm a_{n}^{\\,q^{\\,n}}\\;b_{0}\n     \\]\n     in the expansion of $\\Delta_{n}(A,B)$.",
      "solution": "Throughout “$\\equiv$’’ denotes equality in $\\mathbb F_{q}[A,B]$,\nand indices $i,j,k$ start at $0$ unless stated otherwise.\nPut\n\\[\n  F_{0}:=c_{0},\n  \\qquad\n  F_{k}(i_{1},\\dots ,i_{k})\n  :=c_{k}+i_{1}c_{k-1}+\\dots +i_{k}c_{0}\\quad(k\\ge 1),\n\\]\nand write\n\\[\n  s_{m}:=1+q+\\dots +q^{\\,m}.\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{Step 1: Each listed linear form divides $\\Delta_{n}$.}\n\nFix $k$ with $1\\le k\\le 2n+1$ and a tuple\n$\\boldsymbol i=(i_{1},\\dots ,i_{k})\\in\\mathbb F_{q}^{\\,k}$.\nReplace the $k$-th column of $\\mathcal M_{q,n}(A,B)$\nby the $\\mathbb F_{q}$–linear combination\n\\[\n  \\bigl(F_{k}(\\boldsymbol i),F_{k}(\\boldsymbol i)^{q},\n  \\dots ,F_{k}(\\boldsymbol i)^{q^{2n+1}}\\bigr)^{\\!\\mathsf T}.\n\\]\nBecause the Frobenius map $x\\mapsto x^{q}$ is $\\mathbb F_{q}$–linear,\nthis new column is an $\\mathbb F_{q}$–linear combination of the first\n$k$ columns.  Hence, if $F_{k}(\\boldsymbol i)=0$, the modified matrix\nhas two $\\mathbb F_{q}$–dependent columns and therefore determinant~$0$.\nThus $F_{k}(\\boldsymbol i)$ divides $\\Delta_{n}$ in\n$\\mathbb F_{q}[A,B]$.  For $k=0$ the same argument\ngives the factor $c_{0}=a_{0}$.  \\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{Step 2: Equality of the two sides in (1).}\n\nThe Moore determinant of size $2n+2$ satisfies\n\\[\n  \\deg \\Delta_{n}=s_{2n+1}=1+q+\\dots +q^{\\,2n+1}.\n\\]\nOn the right-hand side of (1) every factor is linear,\nand their number is exactly $s_{2n+1}$; the two polynomials\nhave the same total degree.\nSince each factor divides $\\Delta_{n}$, their product\n$\\Phi$ divides $\\Delta_{n}$, so $\\Delta_{n}=\\lambda\\Phi$\nfor some $\\lambda\\in\\mathbb F_{q}$.\n\nConsider the monomial\n\\[\n  M:=c_{0}\\,c_{1}^{\\,q}\\,c_{2}^{\\,q^{2}}\\dotsm c_{2n+1}^{\\,q^{\\,2n+1}}.\n\\]\nIn $\\Phi$ there is exactly one way to obtain $M$:\none must pick the leading term $c_{k}$ from the factor\n$F_{k}(\\boldsymbol i)$ for every $k$, and these leading terms\nare pairwise distinct.  Hence the coefficient of $M$ in $\\Phi$ is~$1$.\n\nFor the determinant $\\Delta_{n}$ a Laplace expansion along the first\nrow shows that the only permutation contributing to $M$ is the\nidentity; its sign is $+1$, hence the coefficient of $M$ in\n$\\Delta_{n}$ is also~$1$.  Consequently $\\lambda=1$ and\n$\\Delta_{n}\\equiv\\Phi$, proving (1).  \\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{Step 3: Multiplicity one, square-freeness and uniqueness.}\n\nFix a linear factor\n$L:=F_{k}(i_{1},\\dots ,i_{k})$ and write\n\\[\n   \\Delta_{n}=L\\cdot G,\n\\]\nwhere $G$ denotes the product of all other linear factors in~(1).\nDifferentiate with respect to $c_{k}$\n(which occurs in $L$ with coefficient~$1$):\n\\[\n   \\frac{\\partial\\Delta_{n}}{\\partial c_{k}}\n   \\;=\\;\n   G\\;+\\;L\\;\\frac{\\partial G}{\\partial c_{k}}.\n\\]\nReducing the above congruence modulo $L$ gives\n$(\\partial\\Delta_{n}/\\partial c_{k})\\equiv G$, which is non-zero\nbecause no other factor vanishes identically modulo $L$.\nHence $L\\nmid(\\partial\\Delta_{n}/\\partial c_{k})$ and so $L^{2}\\nmid\n\\Delta_{n}$.  Therefore every linear factor occurs with multiplicity~$1$\nand $\\Delta_{n}$ is square-free modulo~$q$.\n\nBecause the factors are pair-wise distinct and each appears once,\nany factorisation of $\\Delta_{n}$ into linear factors must contain\nexactly the same multiset of factors; Step~2 forces the same scalar\nmultiple, so the factorisation given in (1) is unique.\n\\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{Step 4: The coefficient of\n$a_{0}\\,a_{1}^{\\,q}\\dotsm a_{n}^{\\,q^{\\,n}}\\,b_{0}$.}\n\nView $\\Delta_{n}$ as a polynomial in the variables\n$c_{0},c_{1},\\dots ,c_{2n+1}$.\nA monomial in its expansion is uniquely determined by a\npermutation $\\sigma$ of $\\{0,1,\\dots ,2n+1\\}$:\nthe exponent of $c_{j}$ equals\n$\\sum_{\\sigma(i)=j} q^{\\,i}$.\nBecause the powers $q^{\\,0},q^{\\,1},\\dots ,q^{\\,2n+1}$\nare distinct, this representation is unique.\n\nSuppose the monomial\n\\[\n  N:=a_{0}\\,a_{1}^{\\,q}\\dotsm a_{n}^{\\,q^{\\,n}}\\,b_{0}\n      =c_{0}\\,c_{1}^{\\,q}\\dotsm c_{n}^{\\,q^{\\,n}}\\;c_{\\,n+1}\n\\]\nappears.  Translating into $c$-variables,\n\\[\n   \\deg_{c_{0}}N=1,\\;\n   \\deg_{c_{1}}N=q,\\;\n   \\dots ,\n   \\deg_{c_{n}}N=q^{\\,n},\\;\n   \\deg_{c_{n+1}}N=1,\\;\n   \\deg_{c_{j}}N=0\\;(j>n+1).\n\\]\nBecause $\\deg_{c_{n+1}}N=1$, exactly one row index $r$\nsatisfies $\\sigma(r)=n+1$ and $q^{\\,r}=1$, hence $r=0$.\nBut $\\deg_{c_{0}}N=1$ imposes simultaneously $\\sigma(0)=0$,\nwhich contradicts $\\sigma(0)=n+1$.  Therefore the monomial\n$N$ \\emph{cannot} occur in $\\Delta_{n}(A,B)$, and its coefficient is\n\\[\n  0.\n  \\qquad\\qquad\\square\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\textbf{Answer to (4):} the sought coefficient equals $0$.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.595052",
        "was_fixed": false,
        "difficulty_analysis": "1. Dimension doubled: the matrix has 2n + 2 columns/rows while the\n   current kernel variant has only four.  The presence of two\n   distinct blocks (A and B) makes column-dependence arguments\n   more intricate.\n\n2. Additional structural layer: the student must recognise and\n   exploit the relation between the Moore determinant and the\n   Galois-theoretic norm from 𝔽_{q^{\\,n+1}} to 𝔽_q, a viewpoint\n   absent from the original problem.\n\n3. Multi-part requirement:\n   • Part (1) generalises the triangular product to arbitrary n;\n   • Part (2) demands familiarity with finite-field extensions and\n     conjugacy orbits;\n   • Part (3) introduces square-freeness and uniqueness arguments\n     via the Jacobian criterion;\n   • Part (4) is a non-trivial coefficient extraction exercise.\n\n4. The solution must juggle several advanced techniques:\n   determinant manipulations, combinatorial degree counting,\n   Frobenius automorphism orbits, field norms, and algebraic\n   geometry (Jacobian criterion), moving far beyond the pattern\n   matching that suffices for the 3×3 or 4×4 cases.\n\nHence the enhanced variant is substantively harder, richer in theory,\nand requires a deeper, multi-faceted proof than either the\noriginal or the current kernel versions."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}