{
  "index": "2019-A-5",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $p$ be an odd prime number, and let $\\mathbb{F}_p$ denote the field of integers modulo $p$. Let $\\mathbb{F}_p[x]$ be the ring of polynomials over $\\mathbb{F}_p$, and let $q(x) \\in \\mathbb{F}_p[x]$ be given by \n\\[\nq(x) = \\sum_{k=1}^{p-1} a_k x^k,\n\\]\nwhere\n\\[\na_k = k^{(p-1)/2} \\mod{p}. \n\\]\nFind the greatest nonnegative integer $n$ such that $(x-1)^n$ divides $q(x)$ in $\\mathbb{F}_p[x]$.",
  "solution": "The answer is $\\frac{p-1}{2}$. \nDefine the operator $D = x \\frac{d}{dx}$, where $\\frac{d}{dx}$ indicates formal differentiation of polynomials.\nFor $n$ as in the problem statement, we have $q(x) = (x-1)^n r(x)$ for some polynomial $r(x)$ in $\\mathbb{F}_p$ not divisible by $x-1$. For $m=0,\\dots,n$, by the product rule we have\n\\[\n(D^m q)(x) \\equiv n^m x^m (x-1)^{n-m} r(x) \\pmod{(x-1)^{n-m+1}}.\n\\]\nSince $r(1) \\neq 0$ and $n \\not\\equiv 0 \\pmod{p}$ (because $n \\leq \\deg(q) = p-1$), we may identify $n$ as the smallest nonnegative integer for which $(D^n q)(1) \\neq 0$.\n\nNow note that $q = D^{(p-1)/2} s$ for\n\\[\ns(x) = 1 + x + \\cdots + x^{p-1} = \\frac{x^p-1}{x-1} = (x-1)^{p-1}\n\\]\nsince $(x-1)^p = x^p-1$ in $\\mathbb{F}_p[x]$.\nBy the same logic as above, $(D^n s)(1) = 0$ for $n=0,\\dots,p-2$ but not for $n=p-1$.\nThis implies the claimed result.\n\n\\noindent\n\\textbf{Remark.}\nOne may also finish by checking directly that \nfor any positive integer $m$,\n\\[\n\\sum_{k=1}^{p-1} k^m \\equiv \\begin{cases} -1 \\pmod{p} & \\mbox{if $(p-1)|m$} \\\\\n0 \\pmod{p} & \\mbox{otherwise.}\n\\end{cases}\n\\]\nIf $(p-1) | m$, then $k^m \\equiv 1 \\pmod{p}$ by the little Fermat theorem, and so the sum is congruent\nto $p-1 \\equiv -1 \\pmod{p}$. Otherwise, for any primitive root $\\ell$ mod $p$, multiplying the sum by $\\ell^m$ permutes the terms modulo $p$ and hence does not change the sum modulo $p$; since $\\ell^n \\not\\equiv 1 \\pmod{p}$, this is only possible if the sum is zero modulo $p$.",
  "vars": [
    "x",
    "k",
    "m",
    "n",
    "q",
    "r",
    "s",
    "D",
    "l"
  ],
  "params": [
    "p",
    "F_p",
    "a_k"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "indeterminate",
        "k": "summationindex",
        "m": "derivationorder",
        "n": "multiplicity",
        "q": "targetpoly",
        "r": "residualpoly",
        "s": "basepoly",
        "D": "derivoperator",
        "l": "primitiveroot",
        "p": "primefieldmodulus",
        "F_p": "finitefield",
        "a_k": "coefficientterm"
      },
      "question": "Let $primefieldmodulus$ be an odd prime number, and let \\mathbb{F}_{primefieldmodulus} denote the field of integers modulo $primefieldmodulus$. Let \\mathbb{F}_{primefieldmodulus}[indeterminate] be the ring of polynomials over \\mathbb{F}_{primefieldmodulus}, and let $targetpoly(indeterminate) \\in \\mathbb{F}_{primefieldmodulus}[indeterminate]$ be given by \n\\[\ntargetpoly(indeterminate) = \\sum_{summationindex=1}^{primefieldmodulus-1} coefficientterm \\, indeterminate^{summationindex},\n\\]\nwhere\n\\[\ncoefficientterm = summationindex^{(primefieldmodulus-1)/2} \\mod{primefieldmodulus}.\n\\]\nFind the greatest nonnegative integer $multiplicity$ such that $(indeterminate-1)^{multiplicity}$ divides $targetpoly(indeterminate)$ in \\mathbb{F}_{primefieldmodulus}[indeterminate].",
      "solution": "The answer is $(primefieldmodulus-1)/2$.\nDefine the operator $derivoperator = indeterminate \\frac{d}{d\\!indeterminate}$, where $\\frac{d}{d\\!indeterminate}$ indicates formal differentiation of polynomials.\nFor $multiplicity$ as in the problem statement, we have $targetpoly(indeterminate) = (indeterminate-1)^{multiplicity} residualpoly(indeterminate)$ for some polynomial $residualpoly(indeterminate)$ in $\\mathbb{F}_{primefieldmodulus}$ not divisible by $indeterminate-1$. For $derivationorder = 0,\\dots,multiplicity$, by the product rule we have\n\\[\n(derivoperator^{derivationorder} targetpoly)(indeterminate) \\equiv multiplicity^{derivationorder} \\, indeterminate^{derivationorder} (indeterminate-1)^{multiplicity-derivationorder} residualpoly(indeterminate) \\pmod{(indeterminate-1)^{multiplicity-derivationorder+1}}.\n\\]\nSince $residualpoly(1) \\neq 0$ and $multiplicity \\not\\equiv 0 \\pmod{primefieldmodulus}$ (because $multiplicity \\le \\deg(targetpoly) = primefieldmodulus-1$), we may identify $multiplicity$ as the smallest nonnegative integer for which $(derivoperator^{multiplicity} targetpoly)(1) \\neq 0$.\n\nNow note that \n\\[\ntargetpoly = derivoperator^{(primefieldmodulus-1)/2}\\, basepoly\n\\]\nfor\n\\[\nbasepoly(indeterminate) = 1 + indeterminate + \\cdots + indeterminate^{primefieldmodulus-1} = \\frac{indeterminate^{primefieldmodulus}-1}{indeterminate-1} = (indeterminate-1)^{primefieldmodulus-1}\n\\]\nsince $(indeterminate-1)^{primefieldmodulus} = indeterminate^{primefieldmodulus}-1$ in $\\mathbb{F}_{primefieldmodulus}[indeterminate]$.\nBy the same logic as above, $(derivoperator^{derivationorder} basepoly)(1) = 0$ for $derivationorder=0,\\dots,primefieldmodulus-2$ but not for $derivationorder=primefieldmodulus-1$.\nThis implies the claimed result.\n\nRemark.\nOne may also finish by checking directly that \nfor any positive integer $derivationorder$,\n\\[\n\\sum_{summationindex=1}^{primefieldmodulus-1} summationindex^{derivationorder} \\equiv \n\\begin{cases} \n-1 \\pmod{primefieldmodulus} & \\text{if } (primefieldmodulus-1) \\mid derivationorder, \\\\\n0 \\pmod{primefieldmodulus} & \\text{otherwise.}\n\\end{cases}\n\\]\nIf $(primefieldmodulus-1) \\mid derivationorder$, then $summationindex^{derivationorder} \\equiv 1 \\pmod{primefieldmodulus}$ by Fermat's little theorem, and so the sum is congruent to $primefieldmodulus-1 \\equiv -1 \\pmod{primefieldmodulus}$. Otherwise, for any primitive root $primitiveroot$ mod $primefieldmodulus$, multiplying the sum by $primitiveroot^{derivationorder}$ permutes the terms modulo $primefieldmodulus$ and hence does not change the sum modulo $primefieldmodulus$; since $primitiveroot^{multiplicity} \\not\\equiv 1 \\pmod{primefieldmodulus}$, this is only possible if the sum is zero modulo $primefieldmodulus$."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "grapefruit",
        "k": "shoreline",
        "m": "willowtree",
        "n": "lighthouse",
        "q": "butterfly",
        "r": "sandcastle",
        "s": "paintbrush",
        "D": "waterfall",
        "l": "chandelier",
        "p": "mountain",
        "F_p": "riverfield",
        "a_k": "stardust"
      },
      "question": "Let mountain be an odd prime number, and let \\mathbb{riverfield} denote the field of integers modulo mountain. Let \\mathbb{riverfield}[grapefruit] be the ring of polynomials over \\mathbb{riverfield}, and let butterfly(grapefruit) \\in \\mathbb{riverfield}[grapefruit] be given by\n\\[\nbutterfly(grapefruit) = \\sum_{shoreline=1}^{mountain-1} stardust\\, grapefruit^{shoreline},\n\\]\nwhere\n\\[\nstardust = shoreline^{(mountain-1)/2} \\mod{mountain}.\n\\]\nFind the greatest nonnegative integer lighthouse such that $(grapefruit-1)^{lighthouse}$ divides butterfly(grapefruit) in \\mathbb{riverfield}[grapefruit].",
      "solution": "The answer is $\\frac{mountain-1}{2}$. \nDefine the operator $waterfall = grapefruit \\frac{d}{d grapefruit}$, where $\\frac{d}{d grapefruit}$ indicates formal differentiation of polynomials.\nFor $lighthouse$ as in the problem statement, we have $butterfly(grapefruit) = (grapefruit-1)^{lighthouse} sandcastle(grapefruit)$ for some polynomial $sandcastle(grapefruit)$ in $\\mathbb{riverfield}$ not divisible by $grapefruit-1$. For $willowtree=0,\\dots,lighthouse$, by the product rule we have\n\\[\n(waterfall^{willowtree} butterfly)(grapefruit) \\equiv lighthouse^{willowtree} grapefruit^{willowtree} (grapefruit-1)^{lighthouse-willowtree} sandcastle(grapefruit) \\pmod{(grapefruit-1)^{lighthouse-willowtree+1}}.\n\\]\nSince $sandcastle(1) \\neq 0$ and $lighthouse \\not\\equiv 0 \\pmod{mountain}$ (because $lighthouse \\leq \\deg(butterfly) = mountain-1$), we may identify $lighthouse$ as the smallest nonnegative integer for which $(waterfall^{lighthouse} butterfly)(1) \\neq 0$.\n\nNow note that $butterfly = waterfall^{(mountain-1)/2} paintbrush$ for\n\\[\npaintbrush(grapefruit) = 1 + grapefruit + \\cdots + grapefruit^{mountain-1} = \\frac{grapefruit^{mountain}-1}{grapefruit-1} = (grapefruit-1)^{mountain-1}\n\\]\nsince $(grapefruit-1)^{mountain} = grapefruit^{mountain}-1$ in $\\mathbb{riverfield}[grapefruit]$.\nBy the same logic as above, $(waterfall^{lighthouse} paintbrush)(1) = 0$ for $lighthouse=0,\\dots,mountain-2$ but not for $lighthouse=mountain-1$.\nThis implies the claimed result.\n\n\\noindent\n\\textbf{Remark.}\nOne may also finish by checking directly that \nfor any positive integer $willowtree$,\n\\[\n\\sum_{shoreline=1}^{mountain-1} shoreline^{willowtree} \\equiv \\begin{cases} -1 \\pmod{mountain} & \\mbox{if $(mountain-1)|willowtree$} \\\\ 0 \\pmod{mountain} & \\mbox{otherwise.}\\end{cases}\n\\]\nIf $(mountain-1) | willowtree$, then $shoreline^{willowtree} \\equiv 1 \\pmod{mountain}$ by the little Fermat theorem, and so the sum is congruent\nto $mountain-1 \\equiv -1 \\pmod{mountain}$. Otherwise, for any primitive root $chandelier$ mod $mountain$, multiplying the sum by $chandelier^{willowtree}$ permutes the terms modulo $mountain$ and hence does not change the sum modulo $mountain$; since $chandelier^{lighthouse} \\not\\equiv 1 \\pmod{mountain}$, this is only possible if the sum is zero modulo $mountain$. "
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "k": "staticindex",
        "m": "lockedtally",
        "n": "fluidinteger",
        "q": "constantvalue",
        "r": "fullmultiple",
        "s": "differencepoly",
        "D": "integralop",
        "l": "nonrootval",
        "p": "compositen",
        "F_p": "infinitering",
        "a_k": "fixedcoeff"
      },
      "question": "Let $compositen$ be an odd prime number, and let $\\mathbb{infinitering}$ denote the field of integers modulo $compositen$. Let $\\mathbb{infinitering}[fixedvalue]$ be the ring of polynomials over $\\mathbb{infinitering}$, and let $constantvalue(fixedvalue) \\in \\mathbb{infinitering}[fixedvalue]$ be given by \n\\[\nconstantvalue(fixedvalue) = \\sum_{staticindex=1}^{compositen-1} fixedcoeff\\,fixedvalue^{staticindex},\n\\]\nwhere\n\\[\nfixedcoeff = staticindex^{(compositen-1)/2} \\mod{compositen}.\n\\]\nFind the greatest nonnegative integer $fluidinteger$ such that $(fixedvalue-1)^{fluidinteger}$ divides $constantvalue(fixedvalue)$ in $\\mathbb{infinitering}[fixedvalue]$.",
      "solution": "The answer is $\\frac{compositen-1}{2}$. \n\nDefine the operator $integralop = fixedvalue \\frac{d}{d fixedvalue}$, where $\\frac{d}{d fixedvalue}$ indicates formal differentiation of polynomials.\n\nFor $fluidinteger$ as in the problem statement, we have $constantvalue(fixedvalue) = (fixedvalue-1)^{fluidinteger} \\, fullmultiple(fixedvalue)$ for some polynomial $fullmultiple(fixedvalue)$ in $\\mathbb{infinitering}$ not divisible by $fixedvalue-1$. For $lockedtally = 0, \\dots, fluidinteger$, by the product rule we have\n\\[\n(integralop^{lockedtally} \\, constantvalue)(fixedvalue) \\equiv fluidinteger^{lockedtally} \\, fixedvalue^{lockedtally} \\, (fixedvalue-1)^{fluidinteger-lockedtally} \\, fullmultiple(fixedvalue) \\pmod{(fixedvalue-1)^{fluidinteger-lockedtally+1}}.\n\\]\nSince $fullmultiple(1) \\neq 0$ and $fluidinteger \\not\\equiv 0 \\pmod{compositen}$ (because $fluidinteger \\le \\deg(constantvalue) = compositen-1$), we may identify $fluidinteger$ as the smallest nonnegative integer for which $(integralop^{fluidinteger} \\, constantvalue)(1) \\neq 0$.\n\nNow note that $constantvalue = integralop^{(compositen-1)/2}\\,differencepoly$ for\n\\[\ndifferencepoly(fixedvalue) = 1 + fixedvalue + \\cdots + fixedvalue^{compositen-1} = \\frac{fixedvalue^{compositen}-1}{fixedvalue-1} = (fixedvalue-1)^{compositen-1},\n\\]\nsince $(fixedvalue-1)^{compositen} = fixedvalue^{compositen}-1$ in $\\mathbb{infinitering}[fixedvalue]$.\nBy the same logic as above, $(integralop^{lockedtally}\\,differencepoly)(1) = 0$ for $lockedtally = 0, \\dots, compositen-2$ but not for $lockedtally = compositen-1$.\nThis implies the claimed result.\n\nRemark.\nOne may also finish by checking directly that for any positive integer $lockedtally$,\n\\[\n\\sum_{staticindex=1}^{compositen-1} staticindex^{lockedtally} \\equiv \\begin{cases}-1 \\pmod{compositen} & \\text{if $(compositen-1)|lockedtally$}\\\\0 \\pmod{compositen} & \\text{otherwise.}\\end{cases}\n\\]\nIf $(compositen-1) \\mid lockedtally$, then $staticindex^{lockedtally} \\equiv 1 \\pmod{compositen}$ by the little Fermat theorem, and so the sum is congruent to $compositen-1 \\equiv -1 \\pmod{compositen}$. Otherwise, for any primitive root $nonrootval$ mod $compositen$, multiplying the sum by $nonrootval^{lockedtally}$ permutes the terms modulo $compositen$ and hence does not change the sum modulo $compositen$; since $nonrootval^{lockedtally} \\not\\equiv 1 \\pmod{compositen}$, this is only possible if the sum is zero modulo $compositen$. "
    },
    "garbled_string": {
      "map": {
        "x": "zagmireth",
        "k": "fubszalor",
        "m": "quoplenix",
        "n": "jibwacton",
        "q": "mergustip",
        "r": "flondexar",
        "s": "vordalemp",
        "D": "clypetron",
        "l": "spignarok",
        "p": "dulcibraz",
        "F_p": "brangolix",
        "a_k": "travincor"
      },
      "question": "Let $dulcibraz$ be an odd prime number, and let $\\mathbb{F}_{dulcibraz}$ denote the field of integers modulo $dulcibraz$. Let $\\mathbb{F}_{dulcibraz}[zagmireth]$ be the ring of polynomials over $\\mathbb{F}_{dulcibraz}$, and let $mergustip(zagmireth) \\in \\mathbb{F}_{dulcibraz}[zagmireth]$ be given by \n\\[\nmergustip(zagmireth) = \\sum_{fubszalor=1}^{dulcibraz-1} travincor zagmireth^{fubszalor},\n\\]\nwhere\n\\[\ntravincor = fubszalor^{(dulcibraz-1)/2} \\mod{dulcibraz}.\n\\]\nFind the greatest nonnegative integer $jibwacton$ such that $(zagmireth-1)^{jibwacton}$ divides $mergustip(zagmireth)$ in $\\mathbb{F}_{dulcibraz}[zagmireth]$.",
      "solution": "The answer is $\\frac{dulcibraz-1}{2}$. \nDefine the operator $clypetron = zagmireth \\frac{d}{dzagmireth}$, where $\\frac{d}{dzagmireth}$ indicates formal differentiation of polynomials.\nFor $jibwacton$ as in the problem statement, we have $mergustip(zagmireth) = (zagmireth-1)^{jibwacton} flondexar(zagmireth)$ for some polynomial $flondexar(zagmireth)$ in $\\mathbb{F}_{dulcibraz}$ not divisible by $zagmireth-1$. For $quoplenix=0,\\dots,jibwacton$, by the product rule we have\n\\[\n(clypetron^{quoplenix} mergustip)(zagmireth) \\equiv jibwacton^{quoplenix} zagmireth^{quoplenix} (zagmireth-1)^{jibwacton-quoplenix} flondexar(zagmireth) \\pmod{(zagmireth-1)^{jibwacton-quoplenix+1}}.\n\\]\nSince $flondexar(1) \\neq 0$ and $jibwacton \\not\\equiv 0 \\pmod{dulcibraz}$ (because $jibwacton \\leq \\deg(mergustip) = dulcibraz-1$), we may identify $jibwacton$ as the smallest nonnegative integer for which $(clypetron^{jibwacton} mergustip)(1) \\neq 0$.\n\nNow note that $mergustip = clypetron^{(dulcibraz-1)/2} vordalemp$ for\n\\[\nvordalemp(zagmireth) = 1 + zagmireth + \\cdots + zagmireth^{dulcibraz-1} = \\frac{zagmireth^{dulcibraz}-1}{zagmireth-1} = (zagmireth-1)^{dulcibraz-1}\n\\]\nsince $(zagmireth-1)^{dulcibraz} = zagmireth^{dulcibraz}-1$ in $\\mathbb{F}_{dulcibraz}[zagmireth]$.\nBy the same logic as above, $(clypetron^{jibwacton} vordalemp)(1) = 0$ for $jibwacton=0,\\dots,dulcibraz-2$ but not for $jibwacton=dulcibraz-1$.\nThis implies the claimed result.\n\n\\noindent\n\\textbf{Remark.}\nOne may also finish by checking directly that \nfor any positive integer $quoplenix$,\n\\[\n\\sum_{fubszalor=1}^{dulcibraz-1} fubszalor^{quoplenix} \\equiv \\begin{cases} -1 \\pmod{dulcibraz} & \\mbox{if $(dulcibraz-1)|quoplenix$} \\\\ 0 \\pmod{dulcibraz} & \\mbox{otherwise.}\\end{cases}\n\\]\nIf $(dulcibraz-1) | quoplenix$, then $fubszalor^{quoplenix} \\equiv 1 \\pmod{dulcibraz}$ by the little Fermat theorem, and so the sum is congruent\nto $dulcibraz-1 \\equiv -1 \\pmod{dulcibraz}$. Otherwise, for any primitive root $spignarok$ mod $dulcibraz$, multiplying the sum by $spignarok^{quoplenix}$ permutes the terms modulo $dulcibraz$ and hence does not change the sum modulo $dulcibraz$; since $spignarok^{jibwacton} \\not\\equiv 1 \\pmod{dulcibraz}$, this is only possible if the sum is zero modulo $dulcibraz$. "
    },
    "kernel_variant": {
      "question": "Let \\(p\\) be an odd prime with \\(p\\equiv 1\\pmod{12}\\) and fix a primitive root \\(g\\) modulo \\(p\\).\n\n•  For \\(k\\in\\mathbb F_p^{\\times}\\) write  \n  \\(\\chi_2(k):=k^{\\,(p-1)/2}\\in\\{\\;1,-1\\}\\)  (the quadratic character),  \n  \\(\\chi_3(k):=k^{\\,(p-1)/3}\\in\\{1,\\zeta,\\zeta^{2}\\}\\)  (the cubic character), where  \n  \\(\\zeta:=g^{(p-1)/3}\\) is a fixed primitive third root of unity in \\(\\mathbb F_p^{\\times}\\).\n\n•  Define the polynomial  \n\\[\nq(x)=\\sum_{k=0}^{p-2}\\Bigl(\\chi_2\\!\\bigl(g^{k}\\bigr)+2\\,\\chi_3\\!\\bigl(g^{k}\\bigr)\\Bigr)\\;x^{\\,g^{2k}}\n\\in\\mathbb F_p[x].\n\\]\n\n(Thus the summation runs over all non–zero residues; the exponent of \\(x\\) is the square of the residue, written as the power \\(g^{2k}\\).)\n\nDetermine the largest integer \\(n\\ge 0\\) such that \\((x-1)^n\\) divides \\(q(x)\\) in \\(\\mathbb F_p[x]\\).\n\n--------------------------------------------------------------------",
      "solution": "Step 0.  A differentiational criterion  \nWrite \\(D:=x\\dfrac{d}{dx}\\).  For any polynomial \\(F\\in\\mathbb F_p[x]\\) one has  \n\\[\n(x-1)^n\\mid F\\quad\\Longleftrightarrow\\quad \n(D^{m}F)(1)=0\\ \\text{ for }m=0,1,\\dots ,n-1.\n\\tag{A}\n\\]\nConsequently\n\\[\nn=\\min\\bigl\\{m\\ge 0:\\;(D^{m}q)(1)\\ne 0\\bigr\\}.\n\\tag{B}\n\\]\n\nStep 1.  Writing down \\((D^{m}q)(1)\\)  \nBecause \\(D^{m}x^{\\,g^{2k}}=g^{2km}\\,x^{g^{2k}}\\),  \n\\[\n(D^{m}q)(1)=\\sum_{t=0}^{p-2}\\Bigl(\\chi_2(g^{t})+2\\,\\chi_3(g^{t})\\Bigr)\\,g^{2mt}.\n\\tag{1}\n\\]\nPut  \n\\[\nr:=g^{2m}\\in\\mathbb F_p^{\\times}.\n\\]\n\nStep 2.  Orthogonality via finite geometric series  \nIntroduce\n\\[\nS_2(m):=\\sum_{t=0}^{p-2}\\chi_2\\!\\bigl(g^{t}\\bigr)r^{\\,t}, \n\\qquad\nS_3(m):=\\sum_{t=0}^{p-2}\\chi_3\\!\\bigl(g^{t}\\bigr)r^{\\,t}.\n\\tag{2}\n\\]\nBecause \\(g\\) is primitive, \\(\\chi_2(g^{t})=(-1)^t\\) and \\(\\chi_3(g^{t})=\\zeta^{t}\\).  Hence  \n\\[\nS_2(m)=\\sum_{t=0}^{p-2}(-r)^t,\\qquad\nS_3(m)=\\sum_{t=0}^{p-2}(\\zeta r)^t.\n\\]\nFor every \\(u\\in\\mathbb F_p^{\\times}\\),\n\\[\n\\sum_{t=0}^{p-2}u^{\\,t}=\n\\begin{cases}\n-1 &\\text{if }u=1,\\\\[2pt]\n0 &\\text{if }u\\ne 1,\n\\end{cases}\n\\tag{3}\n\\]\nso that\n\\[\nS_2(m)=\\begin{cases}-1 &\\text{if }r=-1,\\\\ 0&\\text{otherwise,}\\end{cases}\\qquad\nS_3(m)=\\begin{cases}-1 &\\text{if }r=\\zeta^{-1},\\\\ 0&\\text{otherwise.}\\end{cases}\n\\tag{4}\n\\]\n\nStep 3.  Translating the conditions \\(r=-1\\) and \\(r=\\zeta^{-1}\\)\n\nBecause \\(r=g^{2m}\\) and \\(g\\) has order \\(p-1\\),\n\\[\n\\begin{aligned}\nr=-1=g^{(p-1)/2}&\\quad\\Longleftrightarrow\\quad\n2m\\equiv\\dfrac{p-1}{2}\\pmod{p-1},\\\\[4pt]\nr=\\zeta^{-1}=g^{2(p-1)/3}&\\quad\\Longleftrightarrow\\quad\n2m\\equiv\\dfrac{2(p-1)}{3}\\pmod{p-1}.\n\\end{aligned}\n\\]\nBecause \\(\\gcd(2,\\,p-1)=2\\), each congruence has two distinct solutions modulo \\(p-1\\).  Choosing representatives in the interval \\(0\\le m\\le p-2\\) gives  \n\\[\n\\boxed{\\;\n\\begin{aligned}\nm_1&=\\dfrac{p-1}{4}, &\nm_1'&=\\dfrac{3(p-1)}{4},\\\\[6pt]\nm_2&=\\dfrac{p-1}{3}, &\nm_2'&=\\dfrac{5(p-1)}{6}.\n\\end{aligned}}\n\\tag{5}\n\\]\n\nStep 4.  Evaluation of \\((D^{m}q)(1)\\)  \nInsert (4) into (1):\n\\[\n(D^{m}q)(1)=S_2(m)+2S_3(m)=\n\\begin{cases}\n-1,& m\\in\\{m_1,m_1'\\},\\\\[4pt]\n-2,& m\\in\\{m_2,m_2'\\},\\\\[4pt]\n0,& \\text{otherwise}.\n\\end{cases}\n\\tag{6}\n\\]\nHence (B) shows that the smallest index for which the derivative is non–zero is \\(m_1=(p-1)/4\\).\n\nTherefore\n\\[\n\\boxed{\\;n=\\dfrac{p-1}{4}\\;}\n\\]\nis the greatest power of \\((x-1)\\) dividing \\(q(x)\\) in \\(\\mathbb F_p[x]\\).\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.860380",
        "was_fixed": false,
        "difficulty_analysis": "1.  Multiple interacting characters.  \n    The coefficients involve both a quadratic and a cubic character; showing when the corresponding twisted power sums vanish forces the solver to juggle two independent characters simultaneously.\n\n2.  Non-linear exponents in \\(x\\).  \n    The exponent \\(g^{2k}\\) (a square in the multiplicative group) makes the standard “replace \\(k\\) by the derivative index” trick impossible; one must compute \\(D^{m}x^{g^{2k}}\\bigl|_{x=1}=g^{2km}\\), introducing an additional multiplicative parameter \\(r=g^{2m}\\).\n\n3.  Group-theoretic congruences.  \n    Determining when \\(g^{2m}\\) equals \\(-1\\) or \\(\\zeta^{-1}\\) entails solving congruences in \\(\\mathbb Z/(p-1)\\mathbb Z\\).  Unlike the original problem, this requires working inside the cyclic multiplicative group and using the fact that \\(p\\equiv1\\pmod{12}\\).\n\n4.  Simultaneous cancellations.  \n    Even after the vanishing pattern of each character sum is understood, the two sums are coupled by the coefficient “\\(1+2\\)”.  One has to check that no accidental cancellation occurs at \\(m=m_1\\) and that the first non-zero derivative indeed arises there.\n\n5.  Larger prerequisite toolkit.  \n    The solver must be comfortable with primitive roots, multiplicative characters of various orders, geometric-series evaluations modulo a prime, derivative operators on formal power series, and valuation arguments—considerably broader than the purely “power-sum modulo \\(p\\)” technique sufficient for the original problem.\n\nHence the enhanced variant is substantially more intricate than both the original and the earlier kernel variant, satisfying all requested escalation criteria."
      }
    },
    "original_kernel_variant": {
      "question": "Let \\(p\\) be an odd prime with \\(p\\equiv 1\\pmod{12}\\) and fix a primitive root \\(g\\) modulo \\(p\\).\n\n•  For \\(k\\in\\mathbb F_p^{\\times}\\) write  \n  \\(\\chi_2(k):=k^{\\,(p-1)/2}\\in\\{\\;1,-1\\}\\)  (the quadratic character),  \n  \\(\\chi_3(k):=k^{\\,(p-1)/3}\\in\\{1,\\zeta,\\zeta^{2}\\}\\)  (the cubic character), where  \n  \\(\\zeta:=g^{(p-1)/3}\\) is a fixed primitive third root of unity in \\(\\mathbb F_p^{\\times}\\).\n\n•  Define the polynomial  \n\\[\nq(x)=\\sum_{k=0}^{p-2}\\Bigl(\\chi_2\\!\\bigl(g^{k}\\bigr)+2\\,\\chi_3\\!\\bigl(g^{k}\\bigr)\\Bigr)\\;x^{\\,g^{2k}}\n\\in\\mathbb F_p[x].\n\\]\n\n(Thus the summation runs over all non–zero residues; the exponent of \\(x\\) is the square of the residue, written as the power \\(g^{2k}\\).)\n\nDetermine the largest integer \\(n\\ge 0\\) such that \\((x-1)^n\\) divides \\(q(x)\\) in \\(\\mathbb F_p[x]\\).\n\n--------------------------------------------------------------------",
      "solution": "Step 0.  A differentiational criterion  \nWrite \\(D:=x\\dfrac{d}{dx}\\).  For any polynomial \\(F\\in\\mathbb F_p[x]\\) one has  \n\\[\n(x-1)^n\\mid F\\quad\\Longleftrightarrow\\quad \n(D^{m}F)(1)=0\\ \\text{ for }m=0,1,\\dots ,n-1.\n\\tag{A}\n\\]\nConsequently\n\\[\nn=\\min\\bigl\\{m\\ge 0:\\;(D^{m}q)(1)\\ne 0\\bigr\\}.\n\\tag{B}\n\\]\n\nStep 1.  Writing down \\((D^{m}q)(1)\\)  \nBecause \\(D^{m}x^{\\,g^{2k}}=g^{2km}\\,x^{g^{2k}}\\),  \n\\[\n(D^{m}q)(1)=\\sum_{t=0}^{p-2}\\Bigl(\\chi_2(g^{t})+2\\,\\chi_3(g^{t})\\Bigr)\\,g^{2mt}.\n\\tag{1}\n\\]\nPut  \n\\[\nr:=g^{2m}\\in\\mathbb F_p^{\\times}.\n\\]\n\nStep 2.  Orthogonality via finite geometric series  \nIntroduce\n\\[\nS_2(m):=\\sum_{t=0}^{p-2}\\chi_2\\!\\bigl(g^{t}\\bigr)r^{\\,t}, \n\\qquad\nS_3(m):=\\sum_{t=0}^{p-2}\\chi_3\\!\\bigl(g^{t}\\bigr)r^{\\,t}.\n\\tag{2}\n\\]\nBecause \\(g\\) is primitive, \\(\\chi_2(g^{t})=(-1)^t\\) and \\(\\chi_3(g^{t})=\\zeta^{t}\\).  Hence  \n\\[\nS_2(m)=\\sum_{t=0}^{p-2}(-r)^t,\\qquad\nS_3(m)=\\sum_{t=0}^{p-2}(\\zeta r)^t.\n\\]\nFor every \\(u\\in\\mathbb F_p^{\\times}\\),\n\\[\n\\sum_{t=0}^{p-2}u^{\\,t}=\n\\begin{cases}\n-1 &\\text{if }u=1,\\\\[2pt]\n0 &\\text{if }u\\ne 1,\n\\end{cases}\n\\tag{3}\n\\]\nso that\n\\[\nS_2(m)=\\begin{cases}-1 &\\text{if }r=-1,\\\\ 0&\\text{otherwise,}\\end{cases}\\qquad\nS_3(m)=\\begin{cases}-1 &\\text{if }r=\\zeta^{-1},\\\\ 0&\\text{otherwise.}\\end{cases}\n\\tag{4}\n\\]\n\nStep 3.  Translating the conditions \\(r=-1\\) and \\(r=\\zeta^{-1}\\)\n\nBecause \\(r=g^{2m}\\) and \\(g\\) has order \\(p-1\\),\n\\[\n\\begin{aligned}\nr=-1=g^{(p-1)/2}&\\quad\\Longleftrightarrow\\quad\n2m\\equiv\\dfrac{p-1}{2}\\pmod{p-1},\\\\[4pt]\nr=\\zeta^{-1}=g^{2(p-1)/3}&\\quad\\Longleftrightarrow\\quad\n2m\\equiv\\dfrac{2(p-1)}{3}\\pmod{p-1}.\n\\end{aligned}\n\\]\nBecause \\(\\gcd(2,\\,p-1)=2\\), each congruence has two distinct solutions modulo \\(p-1\\).  Choosing representatives in the interval \\(0\\le m\\le p-2\\) gives  \n\\[\n\\boxed{\\;\n\\begin{aligned}\nm_1&=\\dfrac{p-1}{4}, &\nm_1'&=\\dfrac{3(p-1)}{4},\\\\[6pt]\nm_2&=\\dfrac{p-1}{3}, &\nm_2'&=\\dfrac{5(p-1)}{6}.\n\\end{aligned}}\n\\tag{5}\n\\]\n\nStep 4.  Evaluation of \\((D^{m}q)(1)\\)  \nInsert (4) into (1):\n\\[\n(D^{m}q)(1)=S_2(m)+2S_3(m)=\n\\begin{cases}\n-1,& m\\in\\{m_1,m_1'\\},\\\\[4pt]\n-2,& m\\in\\{m_2,m_2'\\},\\\\[4pt]\n0,& \\text{otherwise}.\n\\end{cases}\n\\tag{6}\n\\]\nHence (B) shows that the smallest index for which the derivative is non–zero is \\(m_1=(p-1)/4\\).\n\nTherefore\n\\[\n\\boxed{\\;n=\\dfrac{p-1}{4}\\;}\n\\]\nis the greatest power of \\((x-1)\\) dividing \\(q(x)\\) in \\(\\mathbb F_p[x]\\).\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.655031",
        "was_fixed": false,
        "difficulty_analysis": "1.  Multiple interacting characters.  \n    The coefficients involve both a quadratic and a cubic character; showing when the corresponding twisted power sums vanish forces the solver to juggle two independent characters simultaneously.\n\n2.  Non-linear exponents in \\(x\\).  \n    The exponent \\(g^{2k}\\) (a square in the multiplicative group) makes the standard “replace \\(k\\) by the derivative index” trick impossible; one must compute \\(D^{m}x^{g^{2k}}\\bigl|_{x=1}=g^{2km}\\), introducing an additional multiplicative parameter \\(r=g^{2m}\\).\n\n3.  Group-theoretic congruences.  \n    Determining when \\(g^{2m}\\) equals \\(-1\\) or \\(\\zeta^{-1}\\) entails solving congruences in \\(\\mathbb Z/(p-1)\\mathbb Z\\).  Unlike the original problem, this requires working inside the cyclic multiplicative group and using the fact that \\(p\\equiv1\\pmod{12}\\).\n\n4.  Simultaneous cancellations.  \n    Even after the vanishing pattern of each character sum is understood, the two sums are coupled by the coefficient “\\(1+2\\)”.  One has to check that no accidental cancellation occurs at \\(m=m_1\\) and that the first non-zero derivative indeed arises there.\n\n5.  Larger prerequisite toolkit.  \n    The solver must be comfortable with primitive roots, multiplicative characters of various orders, geometric-series evaluations modulo a prime, derivative operators on formal power series, and valuation arguments—considerably broader than the purely “power-sum modulo \\(p\\)” technique sufficient for the original problem.\n\nHence the enhanced variant is substantially more intricate than both the original and the earlier kernel variant, satisfying all requested escalation criteria."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}