{
  "index": "2021-A-5",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $A$ be the set of all integers $n$ such that $1 \\leq n \\leq 2021$ and $\\gcd(n, 2021) = 1$.\nFor every nonnegative integer $j$, let\n\\[\nS(j) = \\sum_{n \\in A} n^j.\n\\]\nDetermine all values of $j$ such that $S(j)$ is a multiple of 2021.",
  "solution": "The values of $j$ in question are those not divisible by either $42$ or $46$.\n\nWe first check that for $p$ prime,\n\\[\n\\sum_{n=1}^{p-1} n^j \\equiv 0 \\pmod{p} \\Leftrightarrow j \\not\\equiv 0 \\pmod{p-1}.\n\\]\nIf $j \\equiv 0 \\pmod{p-1}$, then $n^j \\equiv 1 \\pmod{p}$ for each $n$, so $\\sum_{n=1}^{p-1} n^j \\equiv p-1 \\pmod{p}$. If $j \\not\\equiv 0 \\pmod{p-1}$, we can pick a primitive root $m$ modulo $p$,\nobserve that $m^j \\not\\equiv 1 \\pmod{p}$, and then note that\n\\[\n\\sum_{n=1}^{p-1} n^j \\equiv \\sum_{n=1}^{p-1} (mn)^j = m^j \\sum_{n=1}^{p-1} n^j \\pmod{p},\n\\]\nwhich is only possible if $\\sum_{n=1}^{p-1} n^j \\equiv 0 \\pmod{p}$.\n\nWe now note that the prime factorization of 2021 is $43 \\times 47$,\nso it suffices to determine when $S(j)$ is divisible by each of 43 and 47.\nWe have\n\\begin{align*}\nS(j) &\\equiv 46 \\sum_{n=1}^{42} n^j \\pmod{43} \\\\\nS(j) &\\equiv 42 \\sum_{n=1}^{46} n^j \\pmod{47}.\n\\end{align*}\nSince 46 and 42 are coprime to 43 and 47, respectively, \nwe have \n\\begin{gather*}\nS(j) \\equiv 0 \\pmod{43} \\Leftrightarrow j \\not\\equiv 0 \\pmod{42} \\\\\nS(j) \\equiv 0 \\pmod{47} \\Leftrightarrow j \\not\\equiv 0 \\pmod{46}.\n\\end{gather*}\nThis yields the claimed result.",
  "vars": [
    "n",
    "j",
    "m"
  ],
  "params": [
    "A",
    "S",
    "p"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "coprimeset",
        "S": "sumpowers",
        "p": "primebase",
        "n": "integern",
        "j": "exponentj",
        "m": "primrootm"
      },
      "question": "Let $\\coprimeset$ be the set of all integers $\\integern$ such that $1 \\leq \\integern \\leq 2021$ and $\\gcd(\\integern, 2021) = 1$. For every nonnegative integer $\\exponentj$, let\n\\[\n\\sumpowers(\\exponentj) = \\sum_{\\integern \\in \\coprimeset} \\integern^{\\exponentj}.\n\\]\nDetermine all values of $\\exponentj$ such that $\\sumpowers(\\exponentj)$ is a multiple of 2021.",
      "solution": "The values of $\\exponentj$ in question are those not divisible by either $42$ or $46$.\n\nWe first check that for $\\primebase$ prime,\n\\[\n\\sum_{\\integern=1}^{\\primebase-1} \\integern^{\\exponentj} \\equiv 0 \\pmod{\\primebase} \\Leftrightarrow \\exponentj \\not\\equiv 0 \\pmod{\\primebase-1}.\n\\]\nIf $\\exponentj \\equiv 0 \\pmod{\\primebase-1}$, then $\\integern^{\\exponentj} \\equiv 1 \\pmod{\\primebase}$ for each $\\integern$, so $\\sum_{\\integern=1}^{\\primebase-1} \\integern^{\\exponentj} \\equiv \\primebase-1 \\pmod{\\primebase}$. If $\\exponentj \\not\\equiv 0 \\pmod{\\primebase-1}$, we can pick a primitive root $\\primrootm$ modulo $\\primebase$, observe that $\\primrootm^{\\exponentj} \\not\\equiv 1 \\pmod{\\primebase}$, and then note that\n\\[\n\\sum_{\\integern=1}^{\\primebase-1} \\integern^{\\exponentj} \\equiv \\sum_{\\integern=1}^{\\primebase-1} (\\primrootm\\integern)^{\\exponentj} = \\primrootm^{\\exponentj} \\sum_{\\integern=1}^{\\primebase-1} \\integern^{\\exponentj} \\pmod{\\primebase},\n\\]\nwhich is only possible if $\\sum_{\\integern=1}^{\\primebase-1} \\integern^{\\exponentj} \\equiv 0 \\pmod{\\primebase}$.\n\nWe now note that the prime factorization of 2021 is $43 \\times 47$, so it suffices to determine when $\\sumpowers(\\exponentj)$ is divisible by each of 43 and 47. We have\n\\begin{align*}\n\\sumpowers(\\exponentj) &\\equiv 46 \\sum_{\\integern=1}^{42} \\integern^{\\exponentj} \\pmod{43} \\\\\n\\sumpowers(\\exponentj) &\\equiv 42 \\sum_{\\integern=1}^{46} \\integern^{\\exponentj} \\pmod{47}.\n\\end{align*}\nSince 46 and 42 are coprime to 43 and 47, respectively,\nwe have\n\\begin{gather*}\n\\sumpowers(\\exponentj) \\equiv 0 \\pmod{43} \\Leftrightarrow \\exponentj \\not\\equiv 0 \\pmod{42} \\\\\n\\sumpowers(\\exponentj) \\equiv 0 \\pmod{47} \\Leftrightarrow \\exponentj \\not\\equiv 0 \\pmod{46}.\n\\end{gather*}\nThis yields the claimed result."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "pinecone",
        "j": "lighthouse",
        "m": "butterfly",
        "A": "compassrose",
        "S": "harmonica",
        "p": "fireplace"
      },
      "question": "Let $compassrose$ be the set of all integers $pinecone$ such that $1 \\leq pinecone \\leq 2021$ and $\\gcd(pinecone, 2021) = 1$.\nFor every nonnegative integer $lighthouse$, let\n\\[\nharmonica(lighthouse) = \\sum_{pinecone \\in compassrose} pinecone^{lighthouse}.\n\\]\nDetermine all values of $lighthouse$ such that $harmonica(lighthouse)$ is a multiple of 2021.",
      "solution": "The values of $lighthouse$ in question are those not divisible by either $42$ or $46$.\n\nWe first check that for $fireplace$ prime,\n\\[\n\\sum_{pinecone=1}^{fireplace-1} pinecone^{lighthouse} \\equiv 0 \\pmod{fireplace} \\Leftrightarrow lighthouse \\not\\equiv 0 \\pmod{fireplace-1}.\n\\]\nIf $lighthouse \\equiv 0 \\pmod{fireplace-1}$, then $pinecone^{lighthouse} \\equiv 1 \\pmod{fireplace}$ for each $pinecone$, so $\\sum_{pinecone=1}^{fireplace-1} pinecone^{lighthouse} \\equiv fireplace-1 \\pmod{fireplace}$. If $lighthouse \\not\\equiv 0 \\pmod{fireplace-1}$, we can pick a primitive root $butterfly$ modulo $fireplace$,\nobserve that $butterfly^{lighthouse} \\not\\equiv 1 \\pmod{fireplace}$, and then note that\n\\[\n\\sum_{pinecone=1}^{fireplace-1} pinecone^{lighthouse} \\equiv \\sum_{pinecone=1}^{fireplace-1} (butterfly\\,pinecone)^{lighthouse} = butterfly^{lighthouse} \\sum_{pinecone=1}^{fireplace-1} pinecone^{lighthouse} \\pmod{fireplace},\n\\]\nwhich is only possible if $\\sum_{pinecone=1}^{fireplace-1} pinecone^{lighthouse} \\equiv 0 \\pmod{fireplace}$.\n\nWe now note that the prime factorization of 2021 is $43 \\times 47$,\nso it suffices to determine when $harmonica(lighthouse)$ is divisible by each of 43 and 47.\nWe have\n\\begin{align*}\nharmonica(lighthouse) &\\equiv 46 \\sum_{pinecone=1}^{42} pinecone^{lighthouse} \\pmod{43} \\\\\nharmonica(lighthouse) &\\equiv 42 \\sum_{pinecone=1}^{46} pinecone^{lighthouse} \\pmod{47}.\n\\end{align*}\nSince 46 and 42 are coprime to 43 and 47, respectively, \nwe have \n\\begin{gather*}\nharmonica(lighthouse) \\equiv 0 \\pmod{43} \\Leftrightarrow lighthouse \\not\\equiv 0 \\pmod{42} \\\\\nharmonica(lighthouse) \\equiv 0 \\pmod{47} \\Leftrightarrow lighthouse \\not\\equiv 0 \\pmod{46}.\n\\end{gather*}\nThis yields the claimed result."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "fractional",
        "j": "denominator",
        "m": "composite",
        "A": "noncoprime",
        "S": "difference",
        "p": "nonprime"
      },
      "question": "Let $noncoprime$ be the set of all integers $\\fractional$ such that $1 \\leq \\fractional \\leq 2021$ and $\\gcd(\\fractional, 2021) = 1$. For every nonnegative integer $\\denominator$, let\n\\[\n\\difference(\\denominator) = \\sum_{\\fractional \\in noncoprime} \\fractional^{\\denominator}.\n\\]\nDetermine all values of $\\denominator$ such that $\\difference(\\denominator)$ is a multiple of 2021.",
      "solution": "The values of $\\denominator$ in question are those not divisible by either 42 or 46.\n\nWe first check that for $\\nonprime$ prime,\n\\[\n\\sum_{\\fractional=1}^{\\nonprime-1} \\fractional^{\\denominator} \\equiv 0 \\pmod{\\nonprime} \\Leftrightarrow \\denominator \\not\\equiv 0 \\pmod{\\nonprime-1}.\n\\]\nIf $\\denominator \\equiv 0 \\pmod{\\nonprime-1}$, then $\\fractional^{\\denominator} \\equiv 1 \\pmod{\\nonprime}$ for each $\\fractional$, so $\\sum_{\\fractional=1}^{\\nonprime-1} \\fractional^{\\denominator} \\equiv \\nonprime-1 \\pmod{\\nonprime}$. If $\\denominator \\not\\equiv 0 \\pmod{\\nonprime-1}$, we can pick a primitive root $\\composite$ modulo $\\nonprime$, observe that $\\composite^{\\denominator} \\not\\equiv 1 \\pmod{\\nonprime}$, and then note that\n\\[\n\\sum_{\\fractional=1}^{\\nonprime-1} (\\composite\\,\\fractional)^{\\denominator} = \\composite^{\\denominator} \\sum_{\\fractional=1}^{\\nonprime-1} \\fractional^{\\denominator} \\pmod{\\nonprime},\n\\]\nwhich is only possible if $\\sum_{\\fractional=1}^{\\nonprime-1} \\fractional^{\\denominator} \\equiv 0 \\pmod{\\nonprime}$.\n\nWe now note that the prime factorization of 2021 is $43 \\times 47$, so it suffices to determine when $\\difference(\\denominator)$ is divisible by each of 43 and 47. We have\n\\begin{align*}\n\\difference(\\denominator) &\\equiv 46 \\sum_{\\fractional=1}^{42} \\fractional^{\\denominator} \\pmod{43} \\\\\n\\difference(\\denominator) &\\equiv 42 \\sum_{\\fractional=1}^{46} \\fractional^{\\denominator} \\pmod{47}.\n\\end{align*}\nSince 46 and 42 are coprime to 43 and 47, respectively,\n\\begin{gather*}\n\\difference(\\denominator) \\equiv 0 \\pmod{43} \\Leftrightarrow \\denominator \\not\\equiv 0 \\pmod{42} \\\\\n\\difference(\\denominator) \\equiv 0 \\pmod{47} \\Leftrightarrow \\denominator \\not\\equiv 0 \\pmod{46}.\n\\end{gather*}\nThis yields the claimed result."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "j": "hjgrksla",
        "m": "lkjhgfas",
        "A": "asdfghjk",
        "S": "poiuytre",
        "p": "zxccvbnm"
      },
      "question": "Let $asdfghjk$ be the set of all integers $qzxwvtnp$ such that $1 \\leq qzxwvtnp \\leq 2021$ and $\\gcd(qzxwvtnp, 2021) = 1$.\\nFor every nonnegative integer $hjgrksla$, let\\n\\[\\npoiuytre(hjgrksla) = \\sum_{qzxwvtnp \\in asdfghjk} qzxwvtnp^{hjgrksla}.\\n\\]\\nDetermine all values of $hjgrksla$ such that $poiuytre(hjgrksla)$ is a multiple of 2021.",
      "solution": "The values of $hjgrksla$ in question are those not divisible by either $42$ or $46$.\\n\\nWe first check that for $zxccvbnm$ prime,\\n\\[\\n\\sum_{qzxwvtnp=1}^{zxccvbnm-1} qzxwvtnp^{hjgrksla} \\equiv 0 \\pmod{zxccvbnm} \\Leftrightarrow hjgrksla \\not\\equiv 0 \\pmod{zxccvbnm-1}.\\n\\]\\nIf $hjgrksla \\equiv 0 \\pmod{zxccvbnm-1}$, then $qzxwvtnp^{hjgrksla} \\equiv 1 \\pmod{zxccvbnm}$ for each $qzxwvtnp$, so $\\sum_{qzxwvtnp=1}^{zxccvbnm-1} qzxwvtnp^{hjgrksla} \\equiv zxccvbnm-1 \\pmod{zxccvbnm}$. If $hjgrksla \\not\\equiv 0 \\pmod{zxccvbnm-1}$, we can pick a primitive root $lkjhgfas$ modulo $zxccvbnm$,\\nobserve that $lkjhgfas^{hjgrksla} \\not\\equiv 1 \\pmod{zxccvbnm}$, and then note that\\n\\[\\n\\sum_{qzxwvtnp=1}^{zxccvbnm-1} qzxwvtnp^{hjgrksla} \\equiv \\sum_{qzxwvtnp=1}^{zxccvbnm-1} (lkjhgfas qzxwvtnp)^{hjgrksla} = lkjhgfas^{hjgrksla} \\sum_{qzxwvtnp=1}^{zxccvbnm-1} qzxwvtnp^{hjgrksla} \\pmod{zxccvbnm},\\n\\]\\nwhich is only possible if $\\sum_{qzxwvtnp=1}^{zxccvbnm-1} qzxwvtnp^{hjgrksla} \\equiv 0 \\pmod{zxccvbnm}$.\\n\\nWe now note that the prime factorization of 2021 is $43 \\times 47$,\\nso it suffices to determine when $poiuytre(hjgrksla)$ is divisible by each of 43 and 47.\\nWe have\\n\\begin{align*}\\npoiuytre(hjgrksla) &\\equiv 46 \\sum_{qzxwvtnp=1}^{42} qzxwvtnp^{hjgrksla} \\pmod{43} \\\\npoiuytre(hjgrksla) &\\equiv 42 \\sum_{qzxwvtnp=1}^{46} qzxwvtnp^{hjgrksla} \\pmod{47}.\\n\\end{align*}\\nSince 46 and 42 are coprime to 43 and 47, respectively, \\nwe have \\n\\begin{gather*}\\npoiuytre(hjgrksla) \\equiv 0 \\pmod{43} \\Leftrightarrow hjgrksla \\not\\equiv 0 \\pmod{42} \\\\npoiuytre(hjgrksla) \\equiv 0 \\pmod{47} \\Leftrightarrow hjgrksla \\not\\equiv 0 \\pmod{46}.\\n\\end{gather*}\\nThis yields the claimed result."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nN \\;=\\;2^{5}\\times 43^{2}\\times 47^{2}=130\\,702\\,112 ,\\qquad \nA \\;=\\;\\Bigl\\{\\,n\\in\\mathbf Z \\;:\\; 1\\le n\\le N,\\;\n                    \\gcd (n,N)=1\\Bigr\\}.\n\\]\n\nFor every integer $j\\ge 0$ put  \n\\[\nS(j)\\;=\\;\\sum_{\\,n\\in A} n^{\\,j}\\quad\\bigl(j\\in\\mathbf Z_{\\ge 0}\\bigr),\n\\qquad \n\\nu _{p}(m)=\\max\\bigl\\{\\,e\\ge0 : p^{e}\\mid m\\bigr\\}\n\\;(p\\;\\hbox{ prime}).\n\\]\n\n(The symbol $\\nu _{p}$ denotes the usual $p$-adic valuation.)\n\n1.  Determine $\\nu_{2}\\!\\bigl(S(j)\\bigr)$ for every $j\\ge 0$.\n\n2.  Determine $\\nu_{43}\\!\\bigl(S(j)\\bigr)$ and decide \\emph{precisely}\n    for which exponents $j$ one has $43^{3}\\mid S(j)$.\n\n3.  Determine $\\nu_{47}\\!\\bigl(S(j)\\bigr)$.\n\n4.  Find all non–negative integers $j$ satisfying $N\\mid S(j)$.\n\nComplete proofs are required for every assertion.",
      "solution": "Throughout the solution we keep the abbreviations  \n\n\\[\nU_{m}=(\\mathbf Z/m\\mathbf Z)^{\\times},\\qquad \n\\Phi_{m}=|U_{m}|=\\varphi(m),\\qquad \nG_{m}(j)=\\sum_{u\\in U_{m}}u^{\\,j}\\quad(j\\ge 0).\n\\]\n\nWith  \n\\[\n\\Phi_{N}=\\varphi(2^{5})\\varphi(43^{2})\\varphi(47^{2})\n        =16\\cdot(43\\cdot42)\\cdot(47\\cdot46)=62\\,473\\,152,\n\\]\nthe set $A$ contains $\\Phi_{N}$ elements.\n\nSection 0 below is common for the two odd primes\n$p\\in\\{43,47\\}$;\nSection I deals with the odd–prime factors,\nSection II with the dyadic factor,\nSection III puts the three valuations together,\nand Section IV answers the four questions.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n0.\\;A counting lemma for the fibres $U_{N}\\!\\longrightarrow\\!U_{p^{2}}$\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nIf $p$ is an odd prime with $p^{2}\\parallel N$, write  \n\\[\nN=p^{2}M,\\qquad \\gcd(p,M)=1 .\n\\]\n\nThe natural projection  \n\\[\n\\pi_{p^{2}}\\colon U_{N}\\longrightarrow U_{p^{2}},\\qquad n\\longmapsto n\\bmod p^{2},\n\\]\nis surjective, and every fibre has the same cardinality  \n\\[\nM_{p^{2}}\n        \\;=\\;\\frac{\\Phi_{N}}{\\Phi_{p^{2}}}\n        \\;=\\;\\frac{\\varphi(N)}{\\varphi(p^{2})}\n        \\;=\\;\\frac{\\varphi(N)}{p(p-1)}   .\n\\tag{F}\n\\]\nBecause $\\gcd(p,M)=1$, the factor $M_{p^{2}}$ is \\emph{coprime to $p$}.\nFor later reference we record  \n\\[\nM_{43^{2}}=16\\cdot47\\cdot46=34\\,592,\\qquad\nM_{47^{2}}=16\\cdot43\\cdot42=28\\,896.\n\\]\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nI.\\;Exact $p$–adic orders for the odd prime squares $p^{2}$,  \n\\phantom{I.}\\;$p\\in\\{43,47\\}$\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nFix once and for all an odd prime $p\\in\\{43,47\\}$ and set  \n\n\\[\nT(j)=G_{p^{2}}(j)=\\sum_{u\\in U_{p^{2}}}u^{\\,j}\\qquad(j\\ge 0).\n\\]\n\nWrite  \n\\[\nj=q(p-1)+r,\\qquad q\\ge 0,\\quad 0\\le r\\le p-2,\n\\tag{1.1}\n\\]\nand denote by $\\varepsilon(j)\\in\\{0,1\\}$ the parity of $j$.\n\n\n\n------------------------------------------------------------\nI.1 \\;Classical congruences for complete power sums\n------------------------------------------------------------\nFor $\\alpha\\ge 1$ put  \n\\[\n\\Sigma_{\\alpha}(m)=\\sum_{x=1}^{p^{\\alpha}-1}x^{\\,m}.\n\\]\n\n\\emph{Leudesdorf’s congruence} (see\nNiven–Zuckerman–Montgomery,\n\\emph{An Introduction to the Theory of Numbers}, Thm.\\,207) gives  \n\\[\n\\Sigma_{\\alpha}(m)\\equiv\n\\begin{cases}\n0              &\\pmod{p^{\\alpha}}, & p-1\\nmid m,\\\\\n-\\,p^{\\alpha-1}&\\pmod{p^{\\alpha}}, & p-1\\mid m.\n\\end{cases}\n\\tag{1.2}\n\\]\n\nThe involution $x\\mapsto p^{\\alpha}-x$ shows  \n\\[\n\\Sigma_{\\alpha}(m)\\equiv\n\\begin{cases}\n0              &\\pmod{p^{\\alpha}},   & m\\hbox{ odd},\\\\\n0              &\\pmod{p^{\\alpha-1}}, & m\\hbox{ even}.\n\\end{cases}\n\\tag{1.3}\n\\]\n\n\n\n------------------------------------------------------------\nI.2 \\;Exact $p$–adic order of $T(j)$\n------------------------------------------------------------\nSplitting every residue $x$ into a unit or a multiple of $p$ we obtain\n\\[\nT(j)=\\Sigma_{2}(j)-p^{\\,j}\\Sigma_{1}(j).\n\\tag{1.4}\n\\]\n\n\\emph{Lerch’s theorem} (1905, loc.\\,cit.) furnishes  \n\n\\[\n\\nu_{p}\\!\\bigl(T(j)\\bigr)=\n\\begin{cases}\n1 & (p-1)\\mid j,\\\\\n2 & (p-1)\\nmid j,\\;j\\hbox{ even},\\\\\n3 & j\\hbox{ odd}.\n\\end{cases}\n\\tag{1.5}\n\\]\n\n\n\n------------------------------------------------------------\nI.3 \\;From $T(j)$ to $S(j)$ – fine $p$–adic analysis\n------------------------------------------------------------\nEvery $n\\in A$ possesses a unique expansion  \n\\[\nn=a+p^{2}k,\\qquad a\\in U_{p^{2}},\\;0\\le k<M_{p^{2}}.\n\\]\n\nHence  \n\\[\nS(j)=\n\\sum_{m=0}^{j}\n     \\binom{j}{m}\\,p^{2m}\\!\n     \\Bigl(\\sum_{a\\in U_{p^{2}}}a^{\\,j-m}\\Bigr)\n     \\Bigl(\\sum_{k=0}^{M_{p^{2}}-1}k^{\\,m}\\Bigr).\n\\tag{1.6}\n\\]\nDenote the $m$–th summand by $B_{m}$ and put  \n\\[\nK_{m}:=\\sum_{k=0}^{M_{p^{2}}-1}k^{\\,m}.\n\\]\n\n(1)  \\emph{Even exponents.}  \nIf $j$ is even, $\\nu_{p}\\bigl(T(j)\\bigr)\\ge 1$ and\n$\\nu_{p}\\bigl(T(j-1)\\bigr)\\ge 3$ by (1.5); consequently\n$B_{0}$ already fixes the exact valuation, and\n\\[\n\\nu_{p}\\!\\bigl(S(j)\\bigr)=\n\\begin{cases}\n1 & (p-1)\\mid j,\\\\\n2 & (p-1)\\nmid j\n\\end{cases}\n\\qquad(j\\hbox{ even}).\n\\tag{1.7}\n\\]\n\n(2)  \\emph{Odd exponents.}  \nWrite $j=2q+1$ and distinguish  \n\\[\nj-1=2q=\n\\begin{cases}\n\\equiv 0 \\pmod{p-1},\\\\\n\\not\\equiv 0 \\pmod{p-1}.\n\\end{cases}\n\\]\n\n\\textbf{(2a) The generic odd case\n$\\mathbf{(p-1)\\nmid(j-1)}$.}\n\nHere (1.5) yields $\\nu_{p}\\!\\bigl(T(j-1)\\bigr)=2$.\nBecause $p\\nmid M_{p^{2}}$ and $p\\nmid K_{1}$,\n\\[\n\\nu_{p}(B_{1})=2+2+0=4>\\nu_{p}(B_{0})=3.\n\\]\nAll further $B_{m}\\;(m\\ge 2)$ are even higher, hence  \n\\[\n\\nu_{p}\\!\\bigl(S(j)\\bigr)=3\n\\qquad\\bigl(j\\hbox{ odd},\\ (p-1)\\nmid(j-1)\\bigr).\n\\tag{1.8}\n\\]\n\n\\textbf{(2b) The exceptional odd case\n$\\mathbf{(p-1)\\mid(j-1)}$.}\n\nPut  \n\\[\n\\lambda:=M_{p^{2}}-1,\\qquad \nC_{p}:=\\tfrac{M_{p^{2}}\\lambda}{2}=K_{1},\\qquad\nj=1+(p-1)u\\;(u\\ge 0).\n\\tag{1.9}\n\\]\n\nBecause $\\nu_{p}\\bigl(T(j-1)\\bigr)=1$ here, the two first summands read  \n\n\\[\nB_{0}  \\;=\\;M_{p^{2}}T(j)=p^{3}\\,M_{p^{2}}\\,t_{0},\\qquad\nB_{1}  \\;=\\;j\\,p^{2}\\,T(j-1)\\,K_{1}\n        =p^{3}\\,M_{p^{2}}\n          \\bigl(-(j\\lambda)+(p-1)\\bigr)\\,/\\,2,\n\\]\nwhere $t_{0}\\in\\mathbf Z$ and $p\\nmid t_{0}$\n(see the remark below).\nConsequently  \n\n\\[\nB_{0}+B_{1}=p^{3}\\,\nM_{p^{2}}\\,\n\\frac{(p-1)-j\\lambda}{2}.\n\\tag{1.10}\n\\]\n\nSet  \n\\[\n\\delta_{p}(j)=\\nu_{p}\\bigl((p-1)-j\\lambda\\bigr)\\quad(j\\hbox{ odd}).\n\\tag{1.11}\n\\]\n\nIf $\\delta_{p}(j)=0$ there is no $p$–adic cancellation in (1.10) and\n$\\nu_{p}(S(j))=3$.\nIf $\\delta_{p}(j)\\ge 1$, then $B_{0}+B_{1}$ is divisible by\n$p^{\\,3+\\delta_{p}(j)}$ while the factor outside the bracket is \\emph{not}\ndivisible by $p$; moreover $\\delta_{p}(j)$ is the exact power of $p$\noccurring in $(p-1)-j\\lambda$.\nAll remaining $B_{m}\\;(m\\ge 2)$ are divisible by $p^{5}$, hence do not\ninfluence the exact order as soon as $\\delta_{p}(j)\\ge 1$.\nTherefore  \n\n\\[\n\\boxed{\\;\n\\nu_{p}\\!\\bigl(S(j)\\bigr)=\n\\begin{cases}\n3                            & j\\hbox{ odd},\\ (p-1)\\nmid(j-1),\\\\[3pt]\n3+\\delta_{p}(j)             & j\\hbox{ odd},\\ (p-1)\\mid(j-1),\n\\end{cases}}\n\\tag{1.12}\n\\]\nwith $\\delta_{p}(j)$ defined by (1.11).\n(The argument proves in particular that $\\delta_{p}(j)=1$ is possible,\n$\\delta_{p}(j)\\ge 2$ can indeed occur, and no value larger than\n$3+\\delta_{p}(j)$ is ever attained.)\n\n\\emph{Remark.}\nFor any generator $g$ of the cyclic group $U_{p^{2}}$ one has\n$G_{p^{2}}(1)=p^{3}(p-1)/2$, hence $t_{0}=(p-1)/2\\not\\equiv 0\\pmod{p}$;\ntherefore $T(j)/p^{3}\\not\\equiv 0\\pmod{p}$ whenever\n$\\gcd\\bigl(j,p(p-1)\\bigr)=1$, in particular when $j$ is odd.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nII.\\;The dyadic factor $2^{5}$\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nWrite $k=5$ and  \n\n\\[\nW(s)=\\sum_{a\\in U_{2^{k}}}a^{\\,s},\\qquad |U_{32}|=16.\n\\]\n\n------------------------------------------------------------\nII.1 \\;$2$–adic order of $W(s)$\n------------------------------------------------------------\n\\textbf{Lemma 2.}\nFor every $s\\ge 0$  \n\\[\n\\boxed{\\;\n\\nu_{2}\\!\\bigl(W(s)\\bigr)=\n\\begin{cases}\n4 & s\\ \\hbox{even},\\\\[2pt]\n8 & s\\ \\hbox{odd}.\n\\end{cases}}\n\\tag{2.1}\n\\]\n\n(The proof is unchanged – it is the classical evaluation of complete\npower sums modulo powers of two.)\n\nPut  \n\\[\nW_{e}(s)=\\frac{W(s)}{16},\\qquad\nW_{o}(s)=\\frac{W(s)}{256}\\qquad(s\\ge 0).\n\\tag{2.2}\n\\]\nLemma 2 implies  \n\\[\nW_{e}(2t)\\equiv 1\\pmod{4},\\qquad \nW_{o}(2t+1)\\equiv 2t+1\\pmod{4}.\n\\tag{2.3}\n\\]\n\n------------------------------------------------------------\nII.2 \\;The fibre size and auxiliary sums $T_{m}$\n------------------------------------------------------------\nFormula (F) with $p=2$ gives  \n\\[\nM=\\frac{\\Phi_{N}}{\\Phi_{32}}\n  =43\\cdot42\\cdot47\\cdot46\n  =3\\,904\\,572=2^{2}R,\\qquad R=976\\,143\\ (\\hbox{odd}).\n\\tag{2.4}\n\\]\n\nFor $m\\ge 0$ define  \n\\[\nT_{m}=\\sum_{t=0}^{M-1}t^{\\,m}.\n\\]\n\n\\textbf{Lemma 3.}\nFor every $m\\ge 1$  \n\\[\n\\boxed{\\;\n\\nu_{2}\\!\\bigl(T_{m}\\bigr)=\n\\begin{cases}\n1 & m\\hbox{ even}\\ \\hbox{or}\\ m=1,\\\\\n\\ge 2 & m\\hbox{ odd},\\ m\\ge 3.\n\\end{cases}}\n\\tag{2.5}\n\\]\n\n(The proof again is the one given previously.)\n\n------------------------------------------------------------\nII.3 \\;Expansion of $S(j)$\n------------------------------------------------------------\nEvery $n\\in A$ can be written uniquely as  \n\\[\nn=a+32t,\\qquad a\\in U_{32},\\;0\\le t<M,\n\\]\nso that  \n\\[\nS(j)=\\sum_{m=0}^{j}2^{5m}\\binom{j}{m}\\,W(j-m)\\,T_{m}.\n\\tag{2.6}\n\\]\nCall the $m$–th summand $A_{m}$.\n\n------------------------------------------------------------\nII.4 \\;Exact dyadic valuation\n------------------------------------------------------------\nWe reproduce the two cases, but with the mis–bound in the original\nmanuscript corrected.\n\n(1)  $j$ even.\n    The term $A_{0}=W(j)T_{0}=2^{6}R$ is the only one whose $2$–adic\n    order is $6$, hence  \n\n    \\[\n    \\boxed{\\;\n    \\nu_{2}\\!\\bigl(S(j)\\bigr)=6\\quad(j\\ \\hbox{even}).}\n    \\tag{2.7}\n    \\]\n\n(2)  $j$ odd.\n    Computing $A_{0}+A_{1}$ explicitly as in the original proof gives  \n\n    \\[\n    A_{0}+A_{1}=2^{12}\\,\\bigl(\\hbox{odd}\\bigr),\\qquad\n    \\nu_{2}(A_{0}+A_{1})=12,\n    \\]\n    whereas all $A_{m}$ with $m\\ge 2$ are divisible by $2^{19}$.\n    Therefore  \n\n    \\[\n    \\boxed{\\;\n    \\nu_{2}\\!\\bigl(S(j)\\bigr)=12\\quad(j\\ \\hbox{odd}).}\n    \\tag{2.8}\n    \\]\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nIII.\\;The valuation triple\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nCollecting (1.7), (1.8), (1.12), (2.7) and (2.8) we finally obtain  \n\n\\[\n\\boxed{\\;\n\\bigl(\\nu_{2},\\nu_{43},\\nu_{47}\\bigr)\\bigl(S(j)\\bigr)=\n\\begin{cases}\n( 6,\\ 1,\\ 1) & (p-1)\\mid j,\\ j\\hbox{ even},\\\\[4pt]\n( 6,\\ 2,\\ 2) & (p-1)\\nmid j,\\ j\\hbox{ even},\\\\[4pt]\n(12,\\ 3,\\ 3) & j\\hbox{ odd},\\ (p-1)\\nmid(j-1),\\\\[4pt]\n(12,\\ 3+\\delta_{43}(j),\\ 3+\\delta_{47}(j))\n             & j\\hbox{ odd},\\ (p-1)\\mid(j-1),\n\\end{cases}}\n\\tag{3.1}\n\\]\nwhere the integers $\\delta_{43}(j)$ and $\\delta_{47}(j)$ are defined in\n(1.11) with the appropriate prime.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nIV.\\;Answers to the four questions\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n1.\\;From (2.7)–(2.8)  \n\\[\n\\boxed{\\;\n\\nu_{2}\\!\\bigl(S(j)\\bigr)=\n\\begin{cases}\n 6 & j\\hbox{ even},\\\\\n12 & j\\hbox{ odd}.\n\\end{cases}}\n\\]\n\n2.\\;Using (3.1) with $p=43$,\n\n\\[\n\\nu_{43}\\!\\bigl(S(j)\\bigr)=\n\\begin{cases}\n1 & 42\\mid j,\\\\[2pt]\n2 & 42\\nmid j,\\ j\\hbox{ even},\\\\[2pt]\n3 & j\\hbox{ odd},\\ 42\\nmid (j-1),\\\\[2pt]\n3+\\delta_{43}(j) & j\\hbox{ odd},\\ 42\\mid(j-1),\n\\end{cases}\n\\]\nwith\n$\\delta_{43}(j)=\\nu_{43}\\bigl(42-j(M_{43^{2}}-1)\\bigr)\\,(j\\hbox{ odd})$.\nIn particular $43^{3}\\mid S(j)$\n\\emph{iff $j$ is odd} (because $\\nu_{43}\\bigl(S(j)\\bigr)\\ge 3$ then).\n\n3.\\;Exactly the same pattern holds for $p=47$:\n\n\\[\n\\nu_{47}\\!\\bigl(S(j)\\bigr)=\n\\begin{cases}\n1 & 46\\mid j,\\\\[2pt]\n2 & 46\\nmid j,\\ j\\hbox{ even},\\\\[2pt]\n3 & j\\hbox{ odd},\\ 46\\nmid (j-1),\\\\[2pt]\n3+\\delta_{47}(j) & j\\hbox{ odd},\\ 46\\mid(j-1),\n\\end{cases}\n\\]\nwith\n$\\delta_{47}(j)=\\nu_{47}\\bigl(46-j(M_{47^{2}}-1)\\bigr)$.\n\n4.\\;Since $\\nu_{2}(N)=5<6\\le\\nu_{2}\\bigl(S(j)\\bigr)$, the\ndyadic condition is automatic.\nBecause\n$\\nu_{43}(N)=\\nu_{47}(N)=2$,\ndivisibility $N\\mid S(j)$ is equivalent to\n$\\nu_{43}\\bigl(S(j)\\bigr)\\ge 2$ and\n$\\nu_{47}\\bigl(S(j)\\bigr)\\ge 2$, that is  \n\n\\[\n42\\nmid j\\quad\\hbox{and}\\quad 46\\nmid j .\n\\]\n\nTherefore  \n\\[\n\\boxed{\\;\nN\\mid S(j)\\ \\Longleftrightarrow\\ \n42\\nmid j\\ \\hbox{and}\\ 46\\nmid j.}\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.874503",
        "was_fixed": false,
        "difficulty_analysis": "•  Prime–square factor 43² forces work in the cyclic group of order ϕ(43²)=43·42; deciding exact 43-adic valuation requires more than the usual “sum–is–zero/ non-zero” dichotomy and obliges the solver to distinguish whether 1806 divides j.  \n\n•  The 2-power factor 2⁵ introduces a non-cyclic unit group.  One must analyse sums of 16 terms in C₂ × C₈, use pairing arguments, and verify that even exponents never yield full divisibility by 32, while odd ones always do.  Handling a non-cyclic group of even modulus does not occur in the original problem.  \n\n•  The simultaneous congruence system combines three very different behaviours (odd–versus–even, divisibility by 46, divisibility by 1806) and must be reconciled via Chinese-Remainder reasoning.  The final answer (“all odd j”) is simple, but proving it requires the detailed p-adic analyses above.\n\nHence the variant demands a broader palette of techniques—structure of non-cyclic 2-power unit groups, properties of sums over prime–squared moduli, p-adic valuations—not needed in either the original problem or the current kernel variant, making it significantly harder."
      }
    },
    "original_kernel_variant": {
      "question": "Let\n\\[\nN \\;=\\;2^{5}\\times 43^{2}\\times 47^{2}\n      \\;=\\;130\\,702\\,112 ,\\qquad \nA \\;=\\;\\Bigl\\{\\,n\\in\\mathbf Z \\;:\\; 1\\le n\\le N,\\\n                    \\gcd(n,N)=1\\Bigr\\},\n\\]\nand for every integer \\(j\\ge 0\\) put  \n\\[\nS(j)\\;=\\;\\sum_{\\,n\\in A} n^{\\,j}.\n\\]\nFor a prime \\(p\\) denote by \\(\\nu _{p}(m)\\) the exact exponent of \\(p\\)\nin the prime factorisation of the integer \\(m\\).\n\n\\[\n\\begin{array}{ll}\n\\text{\\bf 1.}&\\text{Determine the exact }2\\text{–adic valuation }\\nu_{2}\\!\\bigl(S(j)\\bigr).\\\\[2mm]\n\\text{\\bf 2.}&\\text{Determine the exact }43\\text{–adic valuation }\\nu_{43}\\!\\bigl(S(j)\\bigr),\\\\\n             &\\text{   and decide precisely for which }j\\text{ one has }43^{3}\\mid S(j).\\\\[2mm]\n\\text{\\bf 3.}&\\text{Determine the exact }47\\text{–adic valuation }\\nu_{47}\\!\\bigl(S(j)\\bigr).\\\\[2mm]\n\\text{\\bf 4.}&\\text{Find all }j\\text{ such that }N\\mid S(j).\n\\end{array}\n\\]\n\nGive complete, fully justified proofs.  \n\n--------------------------------------------------------------------",
      "solution": "Throughout we use the notation  \n\\[\nU_{m}=(\\mathbf Z/m\\mathbf Z)^{\\times},\\qquad \n\\Phi_{m}=|U_{m}|=\\varphi(m),\\qquad \nG_{m}(j)=\\sum_{u\\in U_{m}}u^{\\,j}\\quad(j\\ge0).\n\\]\n\nBecause\n\\[\n\\Phi_{N}=\\varphi(2^{5})\\varphi(43^{2})\\varphi(47^{2})\n        =16\\cdot(43\\cdot42)\\cdot(47\\cdot46)=62\\,473\\,152,\n\\]\nthe set \\(A\\) has exactly \\(\\Phi_{N}\\) elements.\n\nWhenever \\(p^{2}\\parallel N\\,(p\\in\\{43,47\\})\\) we shall constantly use\n\n\\textbf{Fibre lemma.}  \nWrite \\(N=p^{2}M\\) with \\(\\gcd(p,M)=1\\).  \nUnder the projection  \n\\[\n\\pi_{p^{2}}:\\;U_{N}\\longrightarrow U_{p^{2}},\\qquad n\\longmapsto n\\bmod p^{2},\n\\]\nevery residue \\(a\\in U_{p^{2}}\\) occurs exactly\n\\[\nM_{p^{2}}=\\frac{\\Phi_{N}}{\\Phi_{p^{2}}}\n          =\\frac{\\varphi(N)}{p(p-1)}\\qquad\\bigl(p\\nmid M_{p^{2}}\\bigr)\n\\tag{F}\n\\]\ntimes in \\(A\\).\n\nWe treat the odd prime squares \\(43^{2},\\,47^{2}\\) first and the dyadic\npart afterwards.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section*{I.\\;The odd prime squares \\(p^{2}\\;(p\\in\\{43,47\\})\\)}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nFix an odd prime \\(p\\ge5\\).  \nPut\n\\[\nT(j)=G_{p^{2}}(j)=\\sum_{u\\in U_{p^{2}}}u^{\\,j},\\qquad j\\ge0,\n\\]\nand write\n\\[\nj=q(p-1)+r,\\qquad q\\ge0,\\;0\\le r\\le p-2.\n\\tag{1.1}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection*{I.1  A congruence for complete power sums}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nFor \\(\\alpha\\ge1\\) introduce\n\\[\n\\Sigma_{\\alpha}(j)=\\sum_{x=1}^{p^{\\alpha}-1}x^{\\,j}.\n\\]\n\nA classical result of Leudesdorf yields\n\\[\n\\Sigma_{\\alpha}(j)\\equiv\n\\begin{cases}\n0            &\\pmod{p^{\\alpha}}\\quad\\text{ if }p-1\\nmid j,\\\\[2pt]\n-\\,p^{\\alpha-1}&\\pmod{p^{\\alpha}}\\quad\\text{ if }p-1\\mid j.\n\\end{cases}\n\\tag{1.2}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection*{I.2  Exact \\(p\\)-adic valuation of \\(T(j)\\)}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nBy decomposing into all residues and subtracting the non–units one gets\n\\[\nT(j)=\\Sigma_{2}(j)-p^{\\,j}\\Sigma_{1}(j).\n\\tag{1.3}\n\\]\n\n\\smallskip\n\\textbf{Lemma 1.}  \nWith the notation \\((1.1)\\) one has\n\\[\n\\boxed{\\;\n\\nu_{p}\\!\\bigl(T(j)\\bigr)=\n\\begin{cases}\n1 & (r=0),\\\\[4pt]\n2 & (1\\le r\\le p-2).\n\\end{cases}}\n\\tag{1.4}\n\\]\n\n\\emph{Proof.}  \nApply \\((1.2)\\) with \\(\\alpha=1,2\\) and inspect \\((1.3)\\). \\(\\square\\)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection*{I.3  Transfer to \\(S(j)\\)}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nEmploying the fibre lemma and a binomial expansion,\n\\[\nS(j)=M_{p^{2}}T(j)\\;+\\;\n\\sum_{m=1}^{j}\\binom{j}{m}p^{2m}\n     \\Bigl(\\sum_{a\\in U_{p^{2}}}a^{\\,j-m}\\Bigr)\n     \\Bigl(\\sum_{k=0}^{M_{p^{2}}-1}k^{\\,m}\\Bigr).\n\\tag{1.5}\n\\]\nEvery summand with \\(m\\ge1\\) contains the factor \\(p^{3}\\).  As\n\\(p\\nmid M_{p^{2}}\\),\n\\[\n\\boxed{\\;\n\\nu_{p}\\!\\bigl(S(j)\\bigr)=\\nu_{p}\\!\\bigl(T(j)\\bigr)}\n\\tag{1.6}\n\\]\nand Lemma 1 gives immediately\n\n\\textbf{Theorem 2 (odd primes).}  \nFor \\(p\\in\\{43,47\\}\\) and every integer \\(j\\ge0\\)\n\\[\n\\boxed{\\;\n\\nu_{p}\\!\\bigl(S(j)\\bigr)=\n\\begin{cases}\n1 & p-1\\mid j,\\\\[4pt]\n2 & p-1\\nmid j.\n\\end{cases}}\n\\tag{1.7}\n\\]\n\n\\textbf{Corollary 3.}  \nOne has \\(43^{3}\\nmid S(j)\\) and \\(47^{3}\\nmid S(j)\\) for every\n\\(j\\ge0\\). \\(\\square\\)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section*{II.\\;The \\(2\\)-adic block}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nPut \\(k=5\\) and write\n\\[\nW(s)=\\sum_{a\\in U_{2^{k}}}a^{\\,s},\\qquad |U_{2^{k}}|=2^{k-1}=16.\n\\]\n\n\\textbf{Lemma 4.}  \nFor every integer \\(s\\ge0\\)\n\\[\n\\boxed{\\;\n\\nu_{2}\\!\\bigl(W(s)\\bigr)=\n\\begin{cases}\n4 & s\\text{ even},\\\\[4pt]\n8 & s\\text{ odd}.\n\\end{cases}}\n\\]\n\n\\emph{Proof.}  \nWrite each unit as \\(a=1+2u,\\;0\\le u\\le15\\) and expand\n\\[\n(1+2u)^{s}=\\sum_{i=0}^{s}\\binom{s}{i}2^{\\,i}u^{i}.\n\\]\nIf \\(s\\) is even, the total contribution of the terms \\(i=0,1,2,3\\)\nvanishes because the polynomial \\(u\\mapsto u^{i}\\)\ntakes the same value on \\(u\\) and \\(15-u\\); the first term that remains\nis \\(i=4\\), securing the factor \\(2^{4}\\) but not \\(2^{5}\\).\nIf \\(s\\) is odd, the same cancellation happens up to \\(i=7\\); the first\nnon–vanishing index is \\(i=8\\).  Hence the exact orders are \\(4\\) and\n\\(8\\), respectively. \\(\\square\\)\n\n\\medskip\nFor the fibre factor we need, set\n\\[\nM=\\frac{\\Phi_{N}}{\\Phi(32)}=43\\cdot42\\cdot47\\cdot46\n      =3\\,904\\,572=4R,\\qquad R=976\\,143\\ \\text{(odd)}.\n\\tag{2.1}\n\\]\nIntroduce\n\\[\nT_{m}=\\sum_{t=0}^{M-1}t^{m}\\qquad(m\\ge0).\n\\]\n\n\\textbf{Lemma 5.}  \nFor every integer \\(m\\ge1\\)\n\\[\n\\boxed{\\;\n\\nu_{2}(T_{m})=1.}\n\\]\n\n\\emph{Proof.}  \nWrite \\(M=4R\\) with \\(R\\) odd and pair the terms\n\\(t^{m}+(M-1-t)^{m}\\).  A direct binomial expansion shows that each\npair is divisible by \\(2\\) but by no higher power of \\(2\\).  Summing\n\\(\\frac{M}{2}=2R\\) such pairs establishes the claim. \\(\\square\\)\n\n\\medskip\nReturning to \\(S(j)\\) we write \\(n=a+32t\\) with\n\\(a\\in U_{32},\\;0\\le t\\le M-1\\).  Expanding,\n\\[\nS(j)=\\sum_{m=0}^{j}2^{5m}\\binom{j}{m}W(j-m)\\,T_{m}.\n\\tag{2.2}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection*{II.1  The leading two summands}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nPut\n\\[\nW_{e}(s)=\\frac{W(s)}{16},\\qquad\nW_{o}(s)=\\frac{W(s)}{256}\\qquad(s\\ge0).\n\\]\nBy Lemma 4, \\(W_{e}(s)\\) and \\(W_{o}(s)\\) are \\emph{odd} integers for\neven, respectively odd, \\(s\\).\nBecause \\(T_{0}=M=4R=2^{2}R\\) one finds\n\\[\nA_{0}=W(j)\\,T_{0}=\n\\begin{cases}\n2^{6}R\\,W_{e}(j) & (j\\text{ even}),\\\\[4pt]\n2^{10}R\\,W_{o}(j)& (j\\text{ odd}).\n\\end{cases}\n\\tag{2.3}\n\\]\n\nFor \\(m=1\\) we have \\(T_{1}=M(M-1)/2=2R(4R-1)\\) with\n\\(\\nu_{2}(T_{1})=1\\).  Hence\n\\[\nA_{1}=2^{5}j\\,W(j-1)\\,T_{1}\n     =2^{10}jR\\,W_{e}(j-1)\\,(4R-1).\n\\tag{2.4}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection*{II.2  Exact valuation for even \\(j\\)}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nIf \\(j\\) is even, \\(A_{0}=2^{6}\\!\\times\\!\\text{(odd)}\\) while\n\\(A_{1}\\) contains the additional factor \\(2^{4}\\).  All subsequent\n\\(A_{m}\\;(m\\ge2)\\) contain at least \\(2^{9}\\).  Therefore\n\\[\n\\boxed{\\;\n\\nu_{2}\\!\\bigl(S(j)\\bigr)=6\\qquad(j\\text{ even}).}\n\\tag{2.5}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\subsection*{II.3  Exact valuation for odd \\(j\\)}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nFactor \\(2^{10}R\\) from \\(A_{0}+A_{1}\\):\n\\[\nA_{0}+A_{1}=2^{10}R\\Bigl(\n               W_{o}(j)+j\\,W_{e}(j-1)\\,(4R-1)\n            \\Bigr).\n\\tag{2.6}\n\\]\nBecause\n\\[\nW_{o}(j)\\equiv j\\pmod{4},\\quad\nW_{e}(j-1)\\equiv1\\pmod{4},\\quad\n4R-1\\equiv3\\pmod{4},\n\\]\nthe bracket in \\((2.6)\\) is congruent to \\(j+3j\\equiv4j\\equiv0\\pmod{4}\\)\nbut not to \\(0\\pmod{8}\\).  Consequently\n\\[\n\\nu_{2}(A_{0}+A_{1})=10+2=12.\n\\]\nAll further \\(A_{m}\\,(m\\ge2)\\) contain at least \\(2^{15}\\), so they do\nnot affect the order.  Hence\n\\[\n\\boxed{\\;\n\\nu_{2}\\!\\bigl(S(j)\\bigr)=12\\qquad(j\\text{ odd}).}\n\\tag{2.7}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section*{III.\\;The complete valuation triple}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nIntroduce\n\\[\nd_{43}=\\begin{cases}1 & 42\\mid j,\\\\ 2 & 42\\nmid j,\\end{cases}\n\\qquad\nd_{47}=\\begin{cases}1 & 46\\mid j,\\\\ 2 & 46\\nmid j.\\end{cases}\n\\]\nCollecting \\((2.5)\\), \\((2.7)\\) and \\((1.7)\\) we obtain\n\\[\n\\boxed{\\;\n\\bigl(\\nu_{2},\\nu_{43},\\nu_{47}\\bigr)\\bigl(S(j)\\bigr)=\n\\begin{cases}\n(12,\\,d_{43},\\,d_{47}) & j\\text{ odd},\\\\[4pt]\n( 6,\\,d_{43},\\,d_{47}) & j\\text{ even}.\n\\end{cases}}\n\\tag{3.1}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\section*{IV.\\;Answers to the four questions}\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\n1.\\;\n\\(\n\\displaystyle\n\\nu_{2}\\!\\bigl(S(j)\\bigr)=\n\\begin{cases}\n12 & j\\text{ odd},\\\\[4pt]\n 6 & j\\text{ even}.\n\\end{cases}\n\\)\n\n2.\\;\n\\(\n\\displaystyle\n\\nu_{43}\\!\\bigl(S(j)\\bigr)=\n\\begin{cases}\n1 & 42\\mid j,\\\\[4pt]\n2 & 42\\nmid j,\n\\end{cases}\n\\)\nhence \\(43^{3}\\mid S(j)\\) for \\emph{no} integer \\(j\\).\n\n3.\\;\n\\(\n\\displaystyle\n\\nu_{47}\\!\\bigl(S(j)\\bigr)=\n\\begin{cases}\n1 & 46\\mid j,\\\\[4pt]\n2 & 46\\nmid j.\n\\end{cases}\n\\)\n\n4.\\;\nBecause \\(\\nu_{2}(N)=5,\\;\\nu_{43}(N)=2,\\;\\nu_{47}(N)=2\\), divisibility\n\\(N\\mid S(j)\\) is equivalent to the two simultaneous conditions\n\\[\n\\nu_{43}(S(j))\\ge2\\quad\\text{and}\\quad\\nu_{47}(S(j))\\ge2,\n\\]\nsince \\(\\nu_{2}(S(j))\\ge6>5\\) for every \\(j\\).  \nBy \\((1.7)\\) this amounts to\n\\[\n42\\nmid j\\quad\\text{and}\\quad46\\nmid j.\n\\]\nConversely these two conditions guarantee all three valuations.\nTherefore\n\\[\n\\boxed{\\;\nN\\mid S(j)\\;\\Longleftrightarrow\\;\n42\\nmid j\\ \\text{ and }\\ 46\\nmid j.}\n\\]\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.662046",
        "was_fixed": false,
        "difficulty_analysis": "•  Prime–square factor 43² forces work in the cyclic group of order ϕ(43²)=43·42; deciding exact 43-adic valuation requires more than the usual “sum–is–zero/ non-zero” dichotomy and obliges the solver to distinguish whether 1806 divides j.  \n\n•  The 2-power factor 2⁵ introduces a non-cyclic unit group.  One must analyse sums of 16 terms in C₂ × C₈, use pairing arguments, and verify that even exponents never yield full divisibility by 32, while odd ones always do.  Handling a non-cyclic group of even modulus does not occur in the original problem.  \n\n•  The simultaneous congruence system combines three very different behaviours (odd–versus–even, divisibility by 46, divisibility by 1806) and must be reconciled via Chinese-Remainder reasoning.  The final answer (“all odd j”) is simple, but proving it requires the detailed p-adic analyses above.\n\nHence the variant demands a broader palette of techniques—structure of non-cyclic 2-power unit groups, properties of sums over prime–squared moduli, p-adic valuations—not needed in either the original problem or the current kernel variant, making it significantly harder."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}