{
  "index": "1986-B-3",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $\\Gamma$ consist of all polynomials in $x$ with integer\ncoefficients. For $f$ and $g$ in $\\Gamma$ and $m$ a positive integer,\nlet $f \\equiv g \\pmod{m}$ mean that every coefficient of $f-g$ is an\nintegral multiple of $m$. Let $n$ and $p$ be positive integers with\n$p$ prime. Given that $f,g,h,r$ and $s$ are in $\\Gamma$ with\n$rf+sg\\equiv 1 \\pmod{p}$ and $fg \\equiv h \\pmod{p}$, prove that there\nexist $F$ and $G$ in $\\Gamma$ with $F \\equiv f \\pmod{p}$, $G \\equiv g\n\\pmod{p}$, and $FG \\equiv h \\pmod{p^n}$.",
  "solution": "Solution. We prove by induction on \\( k \\) that there exist polynomials \\( F_{k}, G_{k} \\in \\Gamma \\) such that \\( F_{k} \\equiv f(\\bmod p), G_{k} \\equiv g(\\bmod p) \\), and \\( F_{k} G_{k} \\equiv h\\left(\\bmod p^{k}\\right) \\). For the base case \\( k=1 \\), we take \\( F_{1}=f, G_{1}=g \\).\n\nFor the inductive step, we assume the existence \\( F_{k}, G_{k} \\) as above, and try to construct \\( F_{k+1}, G_{k+1} \\). By assumption, \\( h-F_{k} G_{k}=p^{k} t \\), for some \\( t \\in \\Gamma \\). We will \\( \\operatorname{try} F_{k+1}=F_{k}+p^{k} \\Delta_{1} \\) and \\( G_{k+1}=G_{k}+p^{k} \\Delta_{2} \\), where \\( \\Delta_{1}, \\Delta_{2} \\in \\Gamma \\) are yet to be chosen. Then \\( F_{k+1} \\equiv F_{k} \\equiv f(\\bmod p), G_{k+1} \\equiv G_{k} \\equiv g(\\bmod p) \\), and\n\\[\n\\begin{aligned}\nF_{k+1} G_{k+1} & =F_{k} G_{k}+p^{k}\\left(\\Delta_{2} F_{k}+\\Delta_{1} G_{k}\\right)+p^{2 k} \\Delta_{1} \\Delta_{2} \\\\\n& \\equiv F_{k} G_{k}+p^{k}\\left(\\Delta_{2} F_{k}+\\Delta_{1} G_{k}\\right) \\quad\\left(\\bmod p^{k+1}\\right)\n\\end{aligned}\n\\]\n\nIf we choose \\( \\Delta_{2}=t r \\) and \\( \\Delta_{1}=t s \\), then\n\\[\n\\Delta_{2} F_{k}+\\Delta_{1} G_{k} \\equiv \\operatorname{tr} f+t s g=t(r f+s g) \\equiv t \\quad(\\bmod p)\n\\]\nso \\( p^{k}\\left(\\Delta_{2} F_{k}+\\Delta_{1} G_{k}\\right) \\equiv p^{k} t\\left(\\bmod p^{k+1}\\right) \\), and \\( F_{k+1} G_{k+1} \\equiv F_{k} G_{k}+p^{k} t=h\\left(\\bmod p^{k+1}\\right) \\), completing the inductive step.\n\nRemark. This problem is a special case of a version of Hensel's Lemma [Ei, p. 208], a fundamental result in number theory. Here is a different version [NZM, Theorem 2.23], which can be thought of as a \\( p \\)-adic analogue of Newton's method for solving polynomial equations via successive approximation:\n\nHensel's Lemma. Suppose that \\( f(x) \\) is a polynomial with integral coefficients. If \\( f(a) \\equiv 0\\left(\\bmod p^{j}\\right) \\) and \\( f^{\\prime}(a) \\not \\equiv 0(\\bmod p) \\), then there is a unique \\( t(\\bmod p) \\) such that \\( f\\left(a+t p^{j}\\right) \\equiv 0\\left(\\bmod p^{j+1}\\right) \\).",
  "vars": [
    "x",
    "f",
    "g",
    "h",
    "r",
    "s",
    "k",
    "t",
    "F",
    "G",
    "F_k",
    "G_k",
    "\\\\Delta_1",
    "\\\\Delta_2",
    "a"
  ],
  "params": [
    "m",
    "n",
    "p",
    "j",
    "\\\\Gamma"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "f": "polynomf",
        "g": "polynomg",
        "h": "polynomh",
        "r": "scalarer",
        "s": "scalars",
        "k": "indexk",
        "t": "polyt",
        "F": "adjustedf",
        "G": "adjustedg",
        "F_k": "sequencef",
        "G_k": "sequenceg",
        "\\Delta_1": "deltaone",
        "\\Delta_2": "deltatwo",
        "a": "pointa",
        "m": "modulusm",
        "n": "exponentn",
        "p": "primep",
        "j": "indexj",
        "\\Gamma": "polyset"
      },
      "question": "Let $polyset$ consist of all polynomials in $variablex$ with integer coefficients. For $polynomf$ and $polynomg$ in $polyset$ and $modulusm$ a positive integer, let $polynomf \\equiv polynomg \\pmod{modulusm}$ mean that every coefficient of $polynomf-polynomg$ is an integral multiple of $modulusm$. Let $exponentn$ and $primep$ be positive integers with $primep$ prime. Given that $polynomf,polynomg,polynomh,scalarer$ and $scalars$ are in $polyset$ with $scalarer polynomf+scalars polynomg\\equiv 1 \\pmod{primep}$ and $polynomf polynomg \\equiv polynomh \\pmod{primep}$, prove that there exist $adjustedf$ and $adjustedg$ in $polyset$ with $adjustedf \\equiv polynomf \\pmod{primep}$, $adjustedg \\equiv polynomg \\pmod{primep}$, and $adjustedf adjustedg \\equiv polynomh \\pmod{primep^{exponentn}}$.",
      "solution": "Solution. We prove by induction on \\( indexk \\) that there exist polynomials \\( sequencef, sequenceg \\in polyset \\) such that \\( sequencef \\equiv polynomf(\\bmod primep), sequenceg \\equiv polynomg(\\bmod primep) \\), and \\( sequencef sequenceg \\equiv polynomh\\left(\\bmod primep^{indexk}\\right) \\). For the base case \\( indexk=1 \\), we take \\( sequencef=polynomf, sequenceg=polynomg \\).\n\nFor the inductive step, we assume the existence \\( sequencef, sequenceg \\) as above, and try to construct \\( sequencef, sequenceg \\). By assumption, \\( polynomh-sequencef sequenceg=primep^{indexk} polyt \\), for some \\( polyt \\in polyset \\). We will \\operatorname{try} sequencef=sequencef+primep^{indexk} deltaone and sequenceg=sequenceg+primep^{indexk} deltatwo, where \\( deltaone, deltatwo \\in polyset \\) are yet to be chosen. Then \\( sequencef \\equiv sequencef \\equiv polynomf(\\bmod primep), sequenceg \\equiv sequenceg \\equiv polynomg(\\bmod primep) \\), and\n\\[\n\\begin{aligned}\nsequencef sequenceg & =sequencef sequenceg+primep^{indexk}\\left(deltatwo sequencef+deltaone sequenceg\\right)+primep^{2 indexk} deltaone deltatwo \\\\\n& \\equiv sequencef sequenceg+primep^{indexk}\\left(deltatwo sequencef+deltaone sequenceg\\right) \\quad\\left(\\bmod primep^{indexk+1}\\right)\n\\end{aligned}\n\\]\n\nIf we choose \\( deltatwo=polyt scalarer \\) and \\( deltaone=polyt scalars \\), then\n\\[\ndeltatwo sequencef+deltaone sequenceg \\equiv polyt scalarer polynomf+polyt scalars polynomg=polyt( scalarer polynomf+scalars polynomg ) \\equiv polyt \\quad(\\bmod primep)\n\\]\nso \\( primep^{indexk}\\left(deltatwo sequencef+deltaone sequenceg\\right) \\equiv primep^{indexk} polyt\\left(\\bmod primep^{indexk+1}\\right) \\), and \\( sequencef sequenceg \\equiv sequencef sequenceg+primep^{indexk} polyt=polynomh\\left(\\bmod primep^{indexk+1}\\right) \\), completing the inductive step.\n\nRemark. This problem is a special case of a version of Hensel's Lemma [Ei, primep. 208], a fundamental result in number theory. Here is a different version [NZM, Theorem 2.23], which can be thought of as a \\( primep \\)-adic analogue of Newton's method for solving polynomial equations via successive approximation:\n\nHensel's Lemma. Suppose that \\( polynomf(variablex) \\) is a polynomial with integral coefficients. If \\( polynomf(pointa) \\equiv 0\\left(\\bmod primep^{indexj}\\right) \\) and \\( polynomf^{\\prime}(pointa) \\not \\equiv 0(\\bmod primep) \\), then there is a unique \\( polyt(\\bmod primep) \\) such that \\( polynomf\\left(pointa+polyt primep^{indexj}\\right) \\equiv 0\\left(\\bmod primep^{indexj+1}\\right) \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "dandelion",
        "f": "melodyline",
        "g": "breadcrumb",
        "h": "starlight",
        "r": "compassio",
        "s": "daydream",
        "k": "sugarcoat",
        "t": "quasarbeam",
        "F": "nightshade",
        "G": "afterglow",
        "F_k": "moonraker",
        "G_k": "honeycomb",
        "\\\\Delta_1": "peregrine",
        "\\\\Delta_2": "wildfire",
        "a": "lighthouse",
        "m": "cyanotype",
        "n": "brickhouse",
        "p": "drumstick",
        "j": "windchime",
        "\\\\Gamma": "tapestry"
      },
      "question": "Let $\\tapestry$ consist of all polynomials in $dandelion$ with integer\ncoefficients. For $\\melodyline$ and $\\breadcrumb$ in $\\tapestry$ and\n$\\cyanotype$ a positive integer, let $\\melodyline \\equiv \\breadcrumb \\pmod{\\cyanotype}$ mean that every coefficient of $\\melodyline-\\breadcrumb$ is an\nintegral multiple of $\\cyanotype$. Let $\\brickhouse$ and $\\drumstick$ be positive integers with\n$\\drumstick$ prime. Given that $\\melodyline,\\breadcrumb,\\starlight,\\compassio$ and $\\daydream$ are in $\\tapestry$ with\n$\\compassio\\melodyline+\\daydream\\breadcrumb\\equiv 1 \\pmod{\\drumstick}$ and $\\melodyline\\breadcrumb \\equiv \\starlight \\pmod{\\drumstick}$, prove that there\nexist $\\nightshade$ and $\\afterglow$ in $\\tapestry$ with $\\nightshade \\equiv \\melodyline \\pmod{\\drumstick}$, $\\afterglow \\equiv \\breadcrumb\n\\pmod{\\drumstick}$, and $\\nightshade\\afterglow \\equiv \\starlight \\pmod{\\drumstick^{\\brickhouse}}$.",
      "solution": "Solution. We prove by induction on $ \\sugarcoat $ that there exist polynomials $ \\nightshade_{\\sugarcoat}, \\afterglow_{\\sugarcoat} \\in \\tapestry $ such that $ \\nightshade_{\\sugarcoat} \\equiv \\melodyline(\\bmod \\drumstick),\\ \\afterglow_{\\sugarcoat} \\equiv \\breadcrumb(\\bmod \\drumstick) $, and $ \\nightshade_{\\sugarcoat}\\afterglow_{\\sugarcoat} \\equiv \\starlight\\left(\\bmod \\drumstick^{\\sugarcoat}\\right) $. For the base case $ \\sugarcoat=1 $, we take $ \\nightshade_{1}=\\melodyline, \\afterglow_{1}=\\breadcrumb $.\n\nFor the inductive step, we assume the existence of $ \\nightshade_{\\sugarcoat}, \\afterglow_{\\sugarcoat} $ as above, and try to construct $ \\nightshade_{\\sugarcoat+1}, \\afterglow_{\\sugarcoat+1} $. By assumption, $ \\starlight-\\nightshade_{\\sugarcoat}\\afterglow_{\\sugarcoat}=\\drumstick^{\\sugarcoat} \\quasarbeam $, for some $ \\quasarbeam \\in \\tapestry $. We will try $ \\nightshade_{\\sugarcoat+1}=\\nightshade_{\\sugarcoat}+\\drumstick^{\\sugarcoat} \\peregrine $ and $ \\afterglow_{\\sugarcoat+1}=\\afterglow_{\\sugarcoat}+\\drumstick^{\\sugarcoat} \\wildfire $, where $ \\peregrine, \\wildfire \\in \\tapestry $ are yet to be chosen. Then $ \\nightshade_{\\sugarcoat+1} \\equiv \\nightshade_{\\sugarcoat} \\equiv \\melodyline(\\bmod \\drumstick),\\ \\afterglow_{\\sugarcoat+1} \\equiv \\afterglow_{\\sugarcoat} \\equiv \\breadcrumb(\\bmod \\drumstick) $, and\n\\[\n\\begin{aligned}\n\\nightshade_{\\sugarcoat+1}\\afterglow_{\\sugarcoat+1} & =\\nightshade_{\\sugarcoat}\\afterglow_{\\sugarcoat}+\\drumstick^{\\sugarcoat}\\left(\\wildfire \\nightshade_{\\sugarcoat}+\\peregrine \\afterglow_{\\sugarcoat}\\right)+\\drumstick^{2 \\sugarcoat} \\peregrine \\wildfire \\\\\n& \\equiv \\nightshade_{\\sugarcoat}\\afterglow_{\\sugarcoat}+\\drumstick^{\\sugarcoat}\\left(\\wildfire \\nightshade_{\\sugarcoat}+\\peregrine \\afterglow_{\\sugarcoat}\\right) \\quad\\left(\\bmod \\drumstick^{\\sugarcoat+1}\\right)\n\\end{aligned}\n\\]\n\nIf we choose $ \\wildfire=\\quasarbeam \\compassio $ and $ \\peregrine=\\quasarbeam \\daydream $, then\n\\[\n\\wildfire \\nightshade_{\\sugarcoat}+\\peregrine \\afterglow_{\\sugarcoat} \\equiv \\quasarbeam \\compassio \\melodyline+\\quasarbeam \\daydream \\breadcrumb=\\quasarbeam(\\compassio \\melodyline+\\daydream \\breadcrumb) \\equiv \\quasarbeam \\quad(\\bmod \\drumstick)\n\\]\nso $ \\drumstick^{\\sugarcoat}\\left(\\wildfire \\nightshade_{\\sugarcoat}+\\peregrine \\afterglow_{\\sugarcoat}\\right) \\equiv \\drumstick^{\\sugarcoat} \\quasarbeam\\left(\\bmod \\drumstick^{\\sugarcoat+1}\\right) $, and $ \\nightshade_{\\sugarcoat+1}\\afterglow_{\\sugarcoat+1} \\equiv \\nightshade_{\\sugarcoat}\\afterglow_{\\sugarcoat}+\\drumstick^{\\sugarcoat} \\quasarbeam=\\starlight\\left(\\bmod \\drumstick^{\\sugarcoat+1}\\right) $, completing the inductive step.\n\nRemark. This problem is a special case of a version of Hensel's Lemma [Ei, p. 208], a fundamental result in number theory. Here is a different version [NZM, Theorem 2.23], which can be thought of as a $ \\drumstick $-adic analogue of Newton's method for solving polynomial equations via successive approximation:\n\nHensel's Lemma. Suppose that $ \\melodyline(dandelion) $ is a polynomial with integral coefficients. If $ \\melodyline(\\lighthouse) \\equiv 0\\left(\\bmod \\drumstick^{\\windchime}\\right) $ and $ \\melodyline^{\\prime}(\\lighthouse) \\not \\equiv 0(\\bmod \\drumstick) $, then there is a unique $ \\quasarbeam(\\bmod \\drumstick) $ such that $ \\melodyline\\left(\\lighthouse+\\quasarbeam \\drumstick^{\\windchime}\\right) \\equiv 0\\left(\\bmod \\drumstick^{\\windchime+1}\\right) $.",
      "status": "processed"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "invariable",
        "f": "malfunction",
        "g": "inertpoly",
        "h": "quotient",
        "r": "remainder",
        "s": "dividend",
        "k": "terminal",
        "t": "constant",
        "F": "shallower",
        "G": "regressor",
        "F_k": "finalpoly",
        "G_k": "initialpoly",
        "\\\\Delta_1": "\\\\steadiness_{1}",
        "\\\\Delta_2": "\\\\uniformity_{2}",
        "a": "nonfactor",
        "m": "numerator",
        "n": "mantissa",
        "p": "composite",
        "j": "baseline",
        "\\\\Gamma": "\\\\antialgebra"
      },
      "question": "Let $\\antialgebra$ consist of all polynomials in $invariable$ with integer coefficients. For $malfunction$ and $inertpoly$ in $\\antialgebra$ and $numerator$ a positive integer, let $malfunction \\equiv inertpoly \\pmod{numerator}$ mean that every coefficient of $malfunction-inertpoly$ is an integral multiple of $numerator$. Let $mantissa$ and $composite$ be positive integers with $composite$ prime. Given that $malfunction,inertpoly,quotient,remainder$ and $dividend$ are in $\\antialgebra$ with $remainder malfunction+dividend inertpoly\\equiv 1 \\pmod{composite}$ and $malfunction inertpoly \\equiv quotient \\pmod{composite}$, prove that there exist $shallower$ and $regressor$ in $\\antialgebra$ with $shallower \\equiv malfunction \\pmod{composite}$, $regressor \\equiv inertpoly \\pmod{composite}$, and $shallower regressor \\equiv quotient \\pmod{composite^{mantissa}}$.",
      "solution": "Solution. We prove by induction on \\( terminal \\) that there exist polynomials \\( shallower_{terminal}, regressor_{terminal} \\in \\antialgebra \\) such that \\( shallower_{terminal} \\equiv malfunction(\\bmod composite), regressor_{terminal} \\equiv inertpoly(\\bmod composite) \\), and \\( shallower_{terminal} regressor_{terminal} \\equiv quotient\\left(\\bmod composite^{terminal}\\right) \\). For the base case \\( terminal=1 \\), we take \\( shallower_{1}=malfunction, regressor_{1}=inertpoly \\).\n\nFor the inductive step, we assume the existence \\( shallower_{terminal}, regressor_{terminal} \\) as above, and try to construct \\( shallower_{terminal+1}, regressor_{terminal+1} \\). By assumption, \\( quotient-shallower_{terminal} regressor_{terminal}=composite^{terminal} constant \\), for some \\( constant \\in \\antialgebra \\). We will \\operatorname{try} \\( shallower_{terminal+1}=shallower_{terminal}+composite^{terminal} \\steadiness_{1} \\) and \\( regressor_{terminal+1}=regressor_{terminal}+composite^{terminal} \\uniformity_{2} \\), where \\( \\steadiness_{1}, \\uniformity_{2} \\in \\antialgebra \\) are yet to be chosen. Then \\( shallower_{terminal+1} \\equiv shallower_{terminal} \\equiv malfunction(\\bmod composite), regressor_{terminal+1} \\equiv regressor_{terminal} \\equiv inertpoly(\\bmod composite) \\), and\n\\[\n\\begin{aligned}\nshallower_{terminal+1} regressor_{terminal+1} & =shallower_{terminal} regressor_{terminal}+composite^{terminal}\\left(\\uniformity_{2} shallower_{terminal}+\\steadiness_{1} regressor_{terminal}\\right)+composite^{2 terminal} \\steadiness_{1} \\uniformity_{2} \\\\\n& \\equiv shallower_{terminal} regressor_{terminal}+composite^{terminal}\\left(\\uniformity_{2} shallower_{terminal}+\\steadiness_{1} regressor_{terminal}\\right) \\quad\\left(\\bmod composite^{terminal+1}\\right)\n\\end{aligned}\n\\]\n\nIf we choose \\( \\uniformity_{2}=constant\\, remainder \\) and \\( \\steadiness_{1}=constant\\, dividend \\), then\n\\[\n\\uniformity_{2} shallower_{terminal}+\\steadiness_{1} regressor_{terminal} \\equiv \\operatorname{constant\\,remainder} malfunction+constant dividend inertpoly=constant(remainder malfunction+dividend inertpoly) \\equiv constant \\quad(\\bmod composite)\n\\]\nso \\( composite^{terminal}\\left(\\uniformity_{2} shallower_{terminal}+\\steadiness_{1} regressor_{terminal}\\right) \\equiv composite^{terminal} constant\\left(\\bmod composite^{terminal+1}\\right) \\), and \\( shallower_{terminal+1} regressor_{terminal+1} \\equiv shallower_{terminal} regressor_{terminal}+composite^{terminal} constant=quotient\\left(\\bmod composite^{terminal+1}\\right) \\), completing the inductive step.\n\nRemark. This problem is a special case of a version of Hensel's Lemma [Ei, p. 208], a fundamental result in number theory. Here is a different version [NZM, Theorem 2.23], which can be thought of as a \\( composite \\)-adic analogue of Newton's method for solving polynomial equations via successive approximation:\n\nHensel's Lemma. Suppose that \\( malfunction(invariable) \\) is a polynomial with integral coefficients. If \\( malfunction(nonfactor) \\equiv 0\\left(\\bmod composite^{baseline}\\right) \\) and \\( malfunction^{\\prime}(nonfactor) \\not \\equiv 0(\\bmod composite) \\), then there is a unique \\( constant(\\bmod composite) \\) such that \\( malfunction\\left(nonfactor+constant\\, composite^{baseline}\\right) \\equiv 0\\left(\\bmod composite^{baseline+1}\\right) \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "f": "hjgrksla",
        "g": "mvdqplko",
        "h": "tczbrysu",
        "r": "oiwemcfa",
        "s": "blyxdrku",
        "k": "vpanqjoi",
        "t": "lfgscmbr",
        "F": "wneurxza",
        "G": "kvodimce",
        "F_k": "ucypzgfb",
        "G_k": "ayjpqrod",
        "\\\\Delta_1": "nvxqlokr",
        "\\\\Delta_2": "rpzetmsh",
        "a": "dacirnwe",
        "m": "yzlkrqwa",
        "n": "fxvtmeoj",
        "p": "sbhuqane",
        "j": "dfkgzwlm",
        "\\\\Gamma": "gstbrpqe"
      },
      "question": "Let $gstbrpqe$ consist of all polynomials in $qzxwvtnp$ with integer\ncoefficients. For $hjgrksla$ and $mvdqplko$ in $gstbrpqe$ and $yzlkrqwa$ a positive integer,\nlet $hjgrksla \\equiv mvdqplko \\pmod{yzlkrqwa}$ mean that every coefficient of $hjgrksla-mvdqplko$ is an\nintegral multiple of $yzlkrqwa$. Let $fxvtmeoj$ and $sbhuqane$ be positive integers with\n$sbhuqane$ prime. Given that $hjgrksla,mvdqplko,tczbrysu,oiwemcfa$ and $blyxdrku$ are in $gstbrpqe$ with\n$oiwemcfa hjgrksla+blyxdrku mvdqplko\\equiv 1 \\pmod{sbhuqane}$ and $hjgrksla mvdqplko \\equiv tczbrysu \\pmod{sbhuqane}$, prove that there\nexist $wneurxza$ and $kvodimce$ in $gstbrpqe$ with $wneurxza \\equiv hjgrksla \\pmod{sbhuqane}$, $kvodimce \\equiv mvdqplko\n\\pmod{sbhuqane}$, and $wneurxza kvodimce \\equiv tczbrysu \\pmod{sbhuqane^{fxvtmeoj}}$.",
      "solution": "Solution. We prove by induction on \\( vpanqjoi \\) that there exist polynomials \\( wneurxza_{vpanqjoi}, kvodimce_{vpanqjoi} \\in gstbrpqe \\) such that \\( wneurxza_{vpanqjoi} \\equiv hjgrksla(\\bmod sbhuqane),\\; kvodimce_{vpanqjoi} \\equiv mvdqplko(\\bmod sbhuqane) \\), and \\( wneurxza_{vpanqjoi} kvodimce_{vpanqjoi} \\equiv tczbrysu\\left(\\bmod sbhuqane^{vpanqjoi}\\right) \\). For the base case \\( vpanqjoi=1 \\), we take \\( wneurxza_{1}=hjgrksla,\\; kvodimce_{1}=mvdqplko \\).\n\nFor the inductive step, we assume the existence \\( wneurxza_{vpanqjoi}, kvodimce_{vpanqjoi} \\) as above, and try to construct \\( wneurxza_{vpanqjoi+1}, kvodimce_{vpanqjoi+1} \\). By assumption,\n\\( tczbrysu-wneurxza_{vpanqjoi} kvodimce_{vpanqjoi}=sbhuqane^{vpanqjoi} lfgscmbr \\),\nfor some \\( lfgscmbr \\in gstbrpqe \\). We will \\( \\operatorname{try} \\;\nwneurxza_{vpanqjoi+1}=wneurxza_{vpanqjoi}+sbhuqane^{vpanqjoi} nvxqlokr \\) and\n\\( kvodimce_{vpanqjoi+1}=kvodimce_{vpanqjoi}+sbhuqane^{vpanqjoi} rpzetmsh \\), where\n\\( nvxqlokr, rpzetmsh \\in gstbrpqe \\) are yet to be chosen. Then\n\\( wneurxza_{vpanqjoi+1} \\equiv wneurxza_{vpanqjoi} \\equiv hjgrksla(\\bmod sbhuqane),\\;\nkvodimce_{vpanqjoi+1} \\equiv kvodimce_{vpanqjoi} \\equiv mvdqplko(\\bmod sbhuqane) \\), and\n\\[\n\\begin{aligned}\nwneurxza_{vpanqjoi+1} kvodimce_{vpanqjoi+1}\n& =wneurxza_{vpanqjoi} kvodimce_{vpanqjoi}\n +sbhuqane^{vpanqjoi}\\left(rpzetmsh wneurxza_{vpanqjoi}+nvxqlokr kvodimce_{vpanqjoi}\\right)\n +sbhuqane^{2 vpanqjoi} nvxqlokr rpzetmsh \\\\\n& \\equiv wneurxza_{vpanqjoi} kvodimce_{vpanqjoi}\n +sbhuqane^{vpanqjoi}\\left(rpzetmsh wneurxza_{vpanqjoi}+nvxqlokr kvodimce_{vpanqjoi}\\right)\n \\quad\\left(\\bmod sbhuqane^{vpanqjoi+1}\\right)\n\\end{aligned}\n\\]\n\nIf we choose \\( rpzetmsh=lfgscmbr oiwemcfa \\) and \\( nvxqlokr=lfgscmbr blyxdrku \\), then\n\\[\nrpzetmsh wneurxza_{vpanqjoi}+nvxqlokr kvodimce_{vpanqjoi}\n\\equiv \\operatorname{tr} hjgrksla+lfgscmbr blyxdrku mvdqplko\n=lfgscmbr(oiwemcfa hjgrksla+blyxdrku mvdqplko)\n\\equiv lfgscmbr \\quad(\\bmod sbhuqane)\n\\]\nso \\( sbhuqane^{vpanqjoi}\\left(rpzetmsh wneurxza_{vpanqjoi}+nvxqlokr kvodimce_{vpanqjoi}\\right)\n\\equiv sbhuqane^{vpanqjoi} lfgscmbr\\left(\\bmod sbhuqane^{vpanqjoi+1}\\right) \\), and\n\\( wneurxza_{vpanqjoi+1} kvodimce_{vpanqjoi+1} \\equiv wneurxza_{vpanqjoi} kvodimce_{vpanqjoi}\n+sbhuqane^{vpanqjoi} lfgscmbr = tczbrysu\\left(\\bmod sbhuqane^{vpanqjoi+1}\\right) \\), completing the inductive step.\n\nRemark. This problem is a special case of a version of Hensel's Lemma [Ei, p. 208], a fundamental result in number theory. Here is a different version [NZM, Theorem 2.23], which can be thought of as a \\( sbhuqane \\)-adic analogue of Newton's method for solving polynomial equations via successive approximation:\n\nHensel's Lemma. Suppose that \\( hjgrksla(qzxwvtnp) \\) is a polynomial with integral coefficients. If \\( hjgrksla(dacirnwe) \\equiv 0\\left(\\bmod sbhuqane^{dfkgzwlm}\\right) \\) and \\( hjgrksla^{\\prime}(dacirnwe) \\not \\equiv 0(\\bmod sbhuqane) \\), then there is a unique \\( lfgscmbr(\\bmod sbhuqane) \\) such that \\( hjgrksla\\left(dacirnwe+lfgscmbr sbhuqane^{dfkgzwlm}\\right) \\equiv 0\\left(\\bmod sbhuqane^{dfkgzwlm+1}\\right) \\)."
    },
    "kernel_variant": {
      "question": "Fix an integer d ≥ 1 and put Γ := ℤ[x₁,…,x_d].  \nFor u,v ∈ Γ and an integer M ≥ 2 we write  \n\n  u ≡ v (mod M)  \n\nwhen every coefficient of u−v is divisible by M.\n\nLet p be an odd prime, let ℓ ≥ 1 and set N := p^ℓ.  \nSuppose that polynomials  \n\n  a , b , h , r , s ∈ Γ                                                   (∗)\n\nsatisfy  \n\n(1)  r a + s b ≡ 1 (mod p)                     (Bézout relation)  \n(2)  a b        ≡ h (mod p)                     (product condition)  \n(3)  r a − s b ≡ c (mod p) with c ∈ (ℤ/pℤ)ˣ; i.e. the reduction of r a−s b\n     is a non–zero constant modulo p (non-degeneracy).\n\nProve that there exist polynomials Â,𝔅 ∈ Γ such that  \n\n(i) Â ≡ a (mod p) and 𝔅 ≡ b (mod p);  \n(ii) r Â + s 𝔅 ≡ 1 (mod N);  \n(iii) Â 𝔅    ≡ h (mod N).\n\nIn particular, under the additional unit condition (3) the Bézout\nidentity and the prescribed value of the product can be lifted\nsimultaneously from modulus p to modulus p^ℓ.",
      "solution": "We imitate a two–variable Hensel lifting, the extra hypothesis (3)\nensuring an invertible Jacobian.\n\nWrite K := 𝔽_p[x₁,…,x_d] and denote reduction modulo p with a bar.\nFor k = 1,2,…,ℓ define the statement  \n\n 𝒫_k : “there exist A_k , B_k ∈ Γ such that  \n (a) A_k ≡ a (mod p), B_k ≡ b (mod p);  \n (b) r A_k + s B_k ≡ 1 (mod p^{\\,k});  \n (c) A_k B_k        ≡ h (mod p^{\\,k}).”\n\nOur goal is to build A_ℓ,B_ℓ; we then set Â:=A_ℓ, 𝔅:=B_ℓ.\n\nBase step k = 1.  \nTake A₁ := a, B₁ := b.  By (1)–(2) the three conditions hold.\n\nInduction step.  \nAssume 1 ≤ k < ℓ and that A_k,B_k meet 𝒫_k.  Write the\n“product” and “Bézout” errors:\n\n h − A_k B_k = p^{\\,k} t_k  (t_k ∈ Γ),                     (3′)  \n 1 −(r A_k+s B_k)= p^{\\,k} v_k  (v_k ∈ Γ).              (4′)\n\nSeek corrections Δ_A,Δ_B ∈ Γ and put  \n\n A_{k+1}:=A_k+p^{\\,k}Δ_A, B_{k+1}:=B_k+p^{\\,k}Δ_B.        (5′)\n\n(i) Trivially A_{k+1}≡a and B_{k+1}≡b (mod p).\n\n(ii) Bézout combination modulo p^{k+1}:\n\n r A_{k+1}+s B_{k+1}  \n ≡ r A_k+s B_k + p^{\\,k}(r Δ_A+s Δ_B)  \n ≡ 1 − p^{\\,k}v_k + p^{\\,k}(r Δ_A+s Δ_B).         (by (4′))\n\nRequiring r Δ_A+s Δ_B ≡ v_k (mod p) guarantees\nr A_{k+1}+s B_{k+1} ≡ 1 (mod p^{k+1}).                (6′)\n\n(iii) Product modulo p^{k+1}:\n\n A_{k+1}B_{k+1}  \n ≡ A_k B_k + p^{\\,k}(Δ_A B_k+A_k Δ_B)             (mod p^{k+1}).  \nUsing (3′) we need\n\n Δ_A B_k + A_k Δ_B ≡ t_k (mod p).                       (7′)\n\nHence we must solve the linear system over K\n\n  [ r s ] [Δ_A] = [v_k]  \n  [ B̄_k Ā_k ] [Δ_B]   [t_k]                            (8′)\n\nThe determinant is\n\n det M_k = r̄ Ā_k − s̄ B̄_k.                           (9′)\n\nModulo p we have Ā_k≡ā, B̄_k≡b̄, so\n\n det M_k ≡ r̄ ā − s̄ b̄ = c ∈ 𝔽_p^×                    (10′)\n\nby the hypothesis (3).  Therefore det M_k is the non-zero constant c\nand is a unit of the polynomial ring K, so M_k is invertible over K.\nConsequently (8′) admits a unique solution (Δ̄_A,Δ̄_B) in K².  Choose\narbitrary lifts Δ_A,Δ_B ∈ Γ of those residue–class solutions and\ndefine A_{k+1},B_{k+1} by (5′).  Then (6′) and (7′) are satisfied, so\n𝒫_{k+1} holds.\n\nCompletion.  \nInduction produces A_ℓ,B_ℓ with 𝒫_ℓ.  Setting Â:=A_ℓ and 𝔅:=B_ℓ we\nobtain\n\n Â ≡ a, 𝔅 ≡ b (mod p)                (by (a)),  \n rÂ+s𝔅 ≡ 1 (mod p^ℓ)                (by (b)),  \n Â 𝔅 ≡ h (mod p^ℓ)                  (by (c)),\n\nwhich completes the proof.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.695185",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional setting.  \n   • The problem works in the multivariate polynomial ring Γ = ℤ[x₁,…,x_d] with d≥1,  \n     rather than the univariate ring of the original statement.  \n\n2. Non-commutative objects.  \n   • We pass from single polynomials to m×m polynomial *matrices*.  \n     All computations must therefore respect non-commutativity, and right/left multiplications\n     have to be distinguished throughout.\n\n3. Multiple simultaneous constraints.  \n   • We are asked to lift *three* identities at once: a Bézout combination,\n     equality of the two products AB and BA with a prescribed matrix H, and\n     commutativity of the pair (A,B).  \n   • Each extra constraint introduces an additional linear condition that has\n     to be preserved at every induction step.\n\n4. Surjectivity in a simple Artinian ring.  \n   • Ensuring solvability of the linear system for the corrections\n     (Δ_A,Δ_B) can no longer rely on elementary gcd arguments (which make sense only\n     in commutative rings).  One needs the structure theory of full matrix\n     rings, in particular their simplicity, to prove that a certain Λ̄-linear\n     map is surjective.\n\n5. Depth of induction.  \n   • The proof is an elaborate matrix-valued version of Hensel’s lemma,\n     carried out simultaneously for several conditions and inside a\n     non-commutative ring.  Each induction step requires working with three\n     coupled matrix equations, instead of one scalar equation in the original\n     Olympiad problem.\n\nThese layers of extra dimensions, non-commutativity, and interacting\nconstraints make the enhanced variant substantially harder than both the\ninitial question and the current kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer d ≥ 1 and put Γ := ℤ[x₁,…,x_d].  \nFor u,v ∈ Γ and an integer M ≥ 2 we write  \n\n  u ≡ v (mod M)  \n\nwhen every coefficient of u−v is divisible by M.\n\nLet p be an odd prime, let ℓ ≥ 1 and set N := p^ℓ.  \nSuppose that polynomials  \n\n  a , b , h , r , s ∈ Γ                                                   (∗)\n\nsatisfy  \n\n(1)  r a + s b ≡ 1 (mod p)                     (Bézout relation)  \n(2)  a b        ≡ h (mod p)                     (product condition)  \n(3)  r a − s b ≡ c (mod p) with c ∈ (ℤ/pℤ)ˣ; i.e. the reduction of r a−s b\n     is a non–zero constant modulo p (non-degeneracy).\n\nProve that there exist polynomials Â,𝔅 ∈ Γ such that  \n\n(i) Â ≡ a (mod p) and 𝔅 ≡ b (mod p);  \n(ii) r Â + s 𝔅 ≡ 1 (mod N);  \n(iii) Â 𝔅    ≡ h (mod N).\n\nIn particular, under the additional unit condition (3) the Bézout\nidentity and the prescribed value of the product can be lifted\nsimultaneously from modulus p to modulus p^ℓ.",
      "solution": "We imitate a two–variable Hensel lifting, the extra hypothesis (3)\nensuring an invertible Jacobian.\n\nWrite K := 𝔽_p[x₁,…,x_d] and denote reduction modulo p with a bar.\nFor k = 1,2,…,ℓ define the statement  \n\n 𝒫_k : “there exist A_k , B_k ∈ Γ such that  \n (a) A_k ≡ a (mod p), B_k ≡ b (mod p);  \n (b) r A_k + s B_k ≡ 1 (mod p^{\\,k});  \n (c) A_k B_k        ≡ h (mod p^{\\,k}).”\n\nOur goal is to build A_ℓ,B_ℓ; we then set Â:=A_ℓ, 𝔅:=B_ℓ.\n\nBase step k = 1.  \nTake A₁ := a, B₁ := b.  By (1)–(2) the three conditions hold.\n\nInduction step.  \nAssume 1 ≤ k < ℓ and that A_k,B_k meet 𝒫_k.  Write the\n“product” and “Bézout” errors:\n\n h − A_k B_k = p^{\\,k} t_k  (t_k ∈ Γ),                     (3′)  \n 1 −(r A_k+s B_k)= p^{\\,k} v_k  (v_k ∈ Γ).              (4′)\n\nSeek corrections Δ_A,Δ_B ∈ Γ and put  \n\n A_{k+1}:=A_k+p^{\\,k}Δ_A, B_{k+1}:=B_k+p^{\\,k}Δ_B.        (5′)\n\n(i) Trivially A_{k+1}≡a and B_{k+1}≡b (mod p).\n\n(ii) Bézout combination modulo p^{k+1}:\n\n r A_{k+1}+s B_{k+1}  \n ≡ r A_k+s B_k + p^{\\,k}(r Δ_A+s Δ_B)  \n ≡ 1 − p^{\\,k}v_k + p^{\\,k}(r Δ_A+s Δ_B).         (by (4′))\n\nRequiring r Δ_A+s Δ_B ≡ v_k (mod p) guarantees\nr A_{k+1}+s B_{k+1} ≡ 1 (mod p^{k+1}).                (6′)\n\n(iii) Product modulo p^{k+1}:\n\n A_{k+1}B_{k+1}  \n ≡ A_k B_k + p^{\\,k}(Δ_A B_k+A_k Δ_B)             (mod p^{k+1}).  \nUsing (3′) we need\n\n Δ_A B_k + A_k Δ_B ≡ t_k (mod p).                       (7′)\n\nHence we must solve the linear system over K\n\n  [ r s ] [Δ_A] = [v_k]  \n  [ B̄_k Ā_k ] [Δ_B]   [t_k]                            (8′)\n\nThe determinant is\n\n det M_k = r̄ Ā_k − s̄ B̄_k.                           (9′)\n\nModulo p we have Ā_k≡ā, B̄_k≡b̄, so\n\n det M_k ≡ r̄ ā − s̄ b̄ = c ∈ 𝔽_p^×                    (10′)\n\nby the hypothesis (3).  Therefore det M_k is the non-zero constant c\nand is a unit of the polynomial ring K, so M_k is invertible over K.\nConsequently (8′) admits a unique solution (Δ̄_A,Δ̄_B) in K².  Choose\narbitrary lifts Δ_A,Δ_B ∈ Γ of those residue–class solutions and\ndefine A_{k+1},B_{k+1} by (5′).  Then (6′) and (7′) are satisfied, so\n𝒫_{k+1} holds.\n\nCompletion.  \nInduction produces A_ℓ,B_ℓ with 𝒫_ℓ.  Setting Â:=A_ℓ and 𝔅:=B_ℓ we\nobtain\n\n Â ≡ a, 𝔅 ≡ b (mod p)                (by (a)),  \n rÂ+s𝔅 ≡ 1 (mod p^ℓ)                (by (b)),  \n Â 𝔅 ≡ h (mod p^ℓ)                  (by (c)),\n\nwhich completes the proof.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.543635",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional setting.  \n   • The problem works in the multivariate polynomial ring Γ = ℤ[x₁,…,x_d] with d≥1,  \n     rather than the univariate ring of the original statement.  \n\n2. Non-commutative objects.  \n   • We pass from single polynomials to m×m polynomial *matrices*.  \n     All computations must therefore respect non-commutativity, and right/left multiplications\n     have to be distinguished throughout.\n\n3. Multiple simultaneous constraints.  \n   • We are asked to lift *three* identities at once: a Bézout combination,\n     equality of the two products AB and BA with a prescribed matrix H, and\n     commutativity of the pair (A,B).  \n   • Each extra constraint introduces an additional linear condition that has\n     to be preserved at every induction step.\n\n4. Surjectivity in a simple Artinian ring.  \n   • Ensuring solvability of the linear system for the corrections\n     (Δ_A,Δ_B) can no longer rely on elementary gcd arguments (which make sense only\n     in commutative rings).  One needs the structure theory of full matrix\n     rings, in particular their simplicity, to prove that a certain Λ̄-linear\n     map is surjective.\n\n5. Depth of induction.  \n   • The proof is an elaborate matrix-valued version of Hensel’s lemma,\n     carried out simultaneously for several conditions and inside a\n     non-commutative ring.  Each induction step requires working with three\n     coupled matrix equations, instead of one scalar equation in the original\n     Olympiad problem.\n\nThese layers of extra dimensions, non-commutativity, and interacting\nconstraints make the enhanced variant substantially harder than both the\ninitial question and the current kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}