{
  "index": "1961-A-4",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "4. Define a function \\( f \\) over the domain of positive integers as follows: \\( f(1) \\) \\( =1 \\), and for \\( n>1, f(n)=(-1)^{k} \\) where \\( k \\) is the total number of prime factors of \\( n \\). For example \\( f(9)=(-1)^{2}, f(20)=(-1)^{3} \\). Define \\( F(n) \\) as \\( \\Sigma f(d) \\) where the sum ranges over all positive integer divisors of \\( n \\). Prove that for every positive integer \\( n, F(n)=0 \\) or \\( F(n)=1 \\). For which integers \\( n \\) is \\( F(n) \\) \\( =1 ? \\quad \\)",
  "solution": "Solution. Suppose \\( m \\) and \\( n \\) are relatively prime positive integers. Then every divisor of \\( m n \\) is uniquely a product \\( d_{1} d_{2} \\) where \\( d_{1}\\left|m, d_{2}\\right| n \\), and conversely. Also \\( f\\left(d_{1} d_{2}\\right)=f\\left(d_{1}\\right) f\\left(d_{2}\\right) \\).\n\\[\n\\begin{aligned}\nF(m n) & =\\sum_{d \\mid m n} f(d)=\\sum_{d_{1} \\mid m} \\sum_{d_{2} \\mid n} f\\left(d_{1}\\right) f\\left(d_{2}\\right) \\\\\n& =F(m) F(n)\n\\end{aligned}\n\\]\n\nThus \\( F \\) is a multiplicative numerical function and it suffices to evaluate \\( F \\) on prime powers. Evidently for \\( p \\) a prime\n\\[\n\\begin{aligned}\nF\\left(p^{\\alpha}\\right) & =f(1)+f(p)+f\\left(p^{2}\\right)+\\cdots+f\\left(p^{\\alpha}\\right) \\\\\n& =1-1+1-\\cdots+(-1)^{\\alpha} \\\\\n& =\\left\\{\\begin{array}{ll}\n1 & \\text { if } \\alpha \\text { is even } \\\\\n0 & \\text { if } \\alpha \\text { is odd. }\n\\end{array}\\right.\n\\end{aligned}\n\\]\n\nLet \\( n \\) be any positive integer and suppose\n\\[\nn=p_{1}{ }^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\cdots p_{k}^{\\alpha_{k}}\n\\]\nis its canonical factorization into primes. Then \\( F(n)=F\\left(p_{1}{ }^{\\alpha_{1}}\\right) F\\left(p_{2}{ }^{\\alpha_{2}}\\right) \\) \\( \\cdots F\\left(p_{k}{ }^{\\alpha}\\right) \\) and we see that \\( F(n)=0 \\) if some prime appears with odd exponent in the prime factorization of \\( n \\), and \\( F(n)=1 \\) if all primes appear with even exponents. In other words,\n\\[\nF(n)=1\n\\]\nif \\( n \\) is a perfect square, and\n\\[\nF(n)=0\n\\]\nif \\( n \\) is not a perfect square.",
  "vars": [
    "f",
    "n",
    "k",
    "F",
    "d",
    "m",
    "p",
    "\\\\alpha",
    "d_1",
    "d_2",
    "p_1",
    "p_2",
    "p_k",
    "\\\\alpha_1",
    "\\\\alpha_2"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "primeparity",
        "n": "positint",
        "k": "primecount",
        "F": "divisorparity",
        "d": "divisor",
        "m": "coprimeone",
        "p": "primevar",
        "\\alpha": "exponentgeneral",
        "d_1": "divisorone",
        "d_2": "divisortwo",
        "p_1": "primeone",
        "p_2": "primetwo",
        "p_k": "primek",
        "\\alpha_1": "exponentone",
        "\\alpha_2": "exponenttwo"
      },
      "question": "4. Define a function \\( primeparity \\) over the domain of positive integers as follows: \\( primeparity(1) =1 \\), and for \\( positint>1, primeparity(positint)=(-1)^{primecount} \\) where \\( primecount \\) is the total number of prime factors of \\( positint \\). For example \\( primeparity(9)=(-1)^{2}, primeparity(20)=(-1)^{3} \\). Define \\( divisorparity(positint) \\) as \\( \\Sigma primeparity(divisor) \\) where the sum ranges over all positive integer divisors of \\( positint \\). Prove that for every positive integer \\( positint, divisorparity(positint)=0 \\) or \\( divisorparity(positint)=1 \\). For which integers \\( positint \\) is \\( divisorparity(positint) =1 ? \\quad \\)",
      "solution": "Solution. Suppose \\( coprimeone \\) and \\( positint \\) are relatively prime positive integers. Then every divisor of \\( coprimeone positint \\) is uniquely a product \\( divisorone divisortwo \\) where \\( divisorone\\left|coprimeone, divisortwo\\right| positint \\), and conversely. Also \\( primeparity\\left(divisorone divisortwo\\right)=primeparity\\left(divisorone\\right) primeparity\\left(divisortwo\\right) \\).\n\\[\n\\begin{aligned}\ndivisorparity(coprimeone positint) & =\\sum_{divisor \\mid coprimeone positint} primeparity(divisor)=\\sum_{divisorone \\mid coprimeone} \\sum_{divisortwo \\mid positint} primeparity\\left(divisorone\\right) primeparity\\left(divisortwo\\right) \\\\\n& =divisorparity(coprimeone) divisorparity(positint)\n\\end{aligned}\n\\]\n\nThus \\( divisorparity \\) is a multiplicative numerical function and it suffices to evaluate \\( divisorparity \\) on prime powers. Evidently for \\( primevar \\) a prime\n\\[\n\\begin{aligned}\ndivisorparity\\left(primevar^{exponentgeneral}\\right) & =primeparity(1)+primeparity(primevar)+primeparity\\left(primevar^{2}\\right)+\\cdots+primeparity\\left(primevar^{exponentgeneral}\\right) \\\\\n& =1-1+1-\\cdots+(-1)^{exponentgeneral} \\\\\n& =\\left\\{\\begin{array}{ll}\n1 & \\text { if } exponentgeneral \\text { is even } \\\\\n0 & \\text { if } exponentgeneral \\text { is odd. }\n\\end{array}\\right.\n\\end{aligned}\n\\]\n\nLet \\( positint \\) be any positive integer and suppose\n\\[\npositint=primeone^{exponentone} primetwo^{exponenttwo} \\cdots primek^{exponentgeneral_{k}}\n\\]\nis its canonical factorization into primes. Then \\( divisorparity(positint)=divisorparity\\left(primeone^{exponentone}\\right) divisorparity\\left(primetwo^{exponenttwo}\\right) \\cdots divisorparity\\left(primek^{exponentgeneral}\\right) \\) and we see that \\( divisorparity(positint)=0 \\) if some prime appears with odd exponent in the prime factorization of \\( positint \\), and \\( divisorparity(positint)=1 \\) if all primes appear with even exponents. In other words,\n\\[\ndivisorparity(positint)=1\n\\]\nif \\( positint \\) is a perfect square, and\n\\[\ndivisorparity(positint)=0\n\\]\nif \\( positint \\) is not a perfect square."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "marigold",
        "n": "telescope",
        "k": "aardvarks",
        "F": "raincloud",
        "d": "oatmealie",
        "m": "suitcase",
        "p": "violinist",
        "\\alpha": "sandcastle",
        "d_1": "oatmealone",
        "d_2": "oatmealtwo",
        "p_1": "violinistone",
        "p_2": "violinisttwo",
        "p_k": "violinistmany",
        "\\alpha_1": "sandcastleone",
        "\\alpha_2": "sandcastletwo"
      },
      "question": "4. Define a function \\( marigold \\) over the domain of positive integers as follows: \\( marigold(1) =1 \\), and for \\( telescope>1, marigold(telescope)=(-1)^{aardvarks} \\) where \\( aardvarks \\) is the total number of prime factors of \\( telescope \\). For example \\( marigold(9)=(-1)^{2}, marigold(20)=(-1)^{3} \\). Define \\( raincloud(telescope) \\) as \\( \\Sigma marigold(oatmealie) \\) where the sum ranges over all positive integer divisors of \\( telescope \\). Prove that for every positive integer \\( telescope, raincloud(telescope)=0 \\) or \\( raincloud(telescope)=1 \\). For which integers \\( telescope \\) is \\( raincloud(telescope) =1 ? \\quad \\)",
      "solution": "Solution. Suppose \\( suitcase \\) and \\( telescope \\) are relatively prime positive integers. Then every divisor of \\( suitcase telescope \\) is uniquely a product \\( oatmealone oatmealtwo \\) where \\( oatmealone\\mid suitcase, oatmealtwo\\mid telescope \\), and conversely. Also \\( marigold(oatmealone oatmealtwo)=marigold(oatmealone) marigold(oatmealtwo) \\).\n\\[\n\\begin{aligned}\nraincloud(suitcase telescope) & =\\sum_{oatmealie \\mid suitcase telescope} marigold(oatmealie)=\\sum_{oatmealone \\mid suitcase} \\sum_{oatmealtwo \\mid telescope} marigold(oatmealone) marigold(oatmealtwo) \\\\\n& =raincloud(suitcase) raincloud(telescope)\n\\end{aligned}\n\\]\n\nThus \\( raincloud \\) is a multiplicative numerical function and it suffices to evaluate \\( raincloud \\) on prime powers. Evidently for \\( violinist \\) a prime\n\\[\n\\begin{aligned}\nraincloud\\left(violinist^{sandcastle}\\right) & =marigold(1)+marigold(violinist)+marigold\\left(violinist^{2}\\right)+\\cdots+marigold\\left(violinist^{sandcastle}\\right) \\\\\n& =1-1+1-\\cdots+(-1)^{sandcastle} \\\\\n& =\\left\\{\\begin{array}{ll}\n1 & \\text { if } sandcastle \\text { is even } \\\\\n0 & \\text { if } sandcastle \\text { is odd. }\n\\end{array}\\right.\n\\end{aligned}\n\\]\n\nLet \\( telescope \\) be any positive integer and suppose\n\\[\ntelescope=violinistone^{sandcastleone} violinisttwo^{sandcastletwo} \\cdots violinistmany^{\\alpha_{k}}\n\\]\nis its canonical factorization into primes. Then \\( raincloud(telescope)=raincloud\\left(violinistone^{sandcastleone}\\right) raincloud\\left(violinisttwo^{sandcastletwo}\\right) \\cdots raincloud\\left(violinistmany^{\\alpha}\\right) \\) and we see that \\( raincloud(telescope)=0 \\) if some prime appears with odd exponent in the prime factorization of \\( telescope \\), and \\( raincloud(telescope)=1 \\) if all primes appear with even exponents. In other words,\n\\[\nraincloud(telescope)=1\n\\]\nif \\( telescope \\) is a perfect square, and\n\\[\nraincloud(telescope)=0\n\\]\nif \\( telescope \\) is not a perfect square."
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "nonfunction",
        "n": "noninteger",
        "k": "noncount",
        "F": "nonmapping",
        "d": "multiple",
        "m": "noncoprime",
        "p": "composite",
        "\\\\alpha": "inverseexp",
        "d_1": "multipleone",
        "d_2": "multipletwo",
        "p_1": "compositeone",
        "p_2": "compositetwo",
        "p_k": "compositeend",
        "\\\\alpha_1": "inverseexpone",
        "\\\\alpha_2": "inverseexptwo"
      },
      "question": "4. Define a function \\( nonfunction \\) over the domain of positive integers as follows: \\( nonfunction(1) \\) \\( =1 \\), and for \\( noninteger>1, nonfunction(noninteger)=(-1)^{noncount} \\) where \\( noncount \\) is the total number of prime factors of \\( noninteger \\). For example \\( nonfunction(9)=(-1)^{2}, nonfunction(20)=(-1)^{3} \\). Define \\( nonmapping(noninteger) \\) as \\( \\Sigma nonfunction(multiple) \\) where the sum ranges over all positive integer divisors of \\( noninteger \\). Prove that for every positive integer \\( noninteger, nonmapping(noninteger)=0 \\) or \\( nonmapping(noninteger)=1 \\). For which integers \\( noninteger \\) is \\( nonmapping(noninteger) \\) \\( =1 ? \\quad \\)",
      "solution": "Solution. Suppose \\( noncoprime \\) and \\( noninteger \\) are relatively prime positive integers. Then every divisor of \\( noncoprime noninteger \\) is uniquely a product \\( multipleone multipletwo \\) where \\( multipleone\\left|noncoprime, multipletwo\\right| noninteger \\), and conversely. Also \\( nonfunction\\left(multipleone multipletwo\\right)=nonfunction\\left(multipleone\\right) nonfunction\\left(multipletwo\\right) \\).\n\\[\n\\begin{aligned}\nnonmapping(noncoprime noninteger) & =\\sum_{multiple \\mid noncoprime noninteger} nonfunction(multiple)=\\sum_{multipleone \\mid noncoprime} \\sum_{multipletwo \\mid noninteger} nonfunction\\left(multipleone\\right) nonfunction\\left(multipletwo\\right) \\\\ & =nonmapping(noncoprime) nonmapping(noninteger)\n\\end{aligned}\n\\]\n\nThus \\( nonmapping \\) is a multiplicative numerical function and it suffices to evaluate \\( nonmapping \\) on prime powers. Evidently for \\( composite \\) a prime\n\\[\n\\begin{aligned}\nnonmapping\\left(composite^{inverseexp}\\right) & =nonfunction(1)+nonfunction(composite)+nonfunction\\left(composite^{2}\\right)+\\cdots+nonfunction\\left(composite^{inverseexp}\\right) \\\\ & =1-1+1-\\cdots+(-1)^{inverseexp} \\\\ & =\\left\\{\\begin{array}{ll}\n1 & \\text { if } inverseexp \\text { is even } \\\\\n0 & \\text { if } inverseexp \\text { is odd. }\n\\end{array}\\right.\n\\end{aligned}\n\\]\n\nLet \\( noninteger \\) be any positive integer and suppose\n\\[\nnoninteger=compositeone{ }^{inverseexpone} compositetwo^{inverseexptwo} \\cdots compositeend^{inverseexp}\n\\]\nis its canonical factorization into primes. Then \\( nonmapping(noninteger)=nonmapping\\left(compositeone{ }^{inverseexpone}\\right) nonmapping\\left(compositetwo{ }^{inverseexptwo}\\right) \\cdots nonmapping\\left(compositeend{ }^{inverseexp}\\right) \\) and we see that \\( nonmapping(noninteger)=0 \\) if some prime appears with odd exponent in the prime factorization of \\( noninteger \\), and \\( nonmapping(noninteger)=1 \\) if all primes appear with even exponents. In other words,\n\\[\nnonmapping(noninteger)=1\n\\]\nif \\( noninteger \\) is a perfect square, and\n\\[\nnonmapping(noninteger)=0\n\\]\nif \\( noninteger \\) is not a perfect square."
    },
    "garbled_string": {
      "map": {
        "f": "qzxwvtnp",
        "n": "hjgrksla",
        "k": "ufldevpo",
        "F": "mtrelaxs",
        "d": "vbycrofe",
        "m": "sajkugbh",
        "p": "zqtrmnev",
        "\\alpha": "lopqstuv",
        "d_1": "quxsnmva",
        "d_2": "yirbkeoz",
        "p_1": "rfvbzxqu",
        "p_2": "hodnlysp",
        "p_k": "gicafmwe",
        "\\alpha_1": "aktewomz",
        "\\alpha_2": "lvdrusni"
      },
      "question": "4. Define a function \\( qzxwvtnp \\) over the domain of positive integers as follows: \\( qzxwvtnp(1)=1 \\), and for \\( hjgrksla>1, qzxwvtnp(hjgrksla)=(-1)^{ufldevpo} \\) where \\( ufldevpo \\) is the total number of prime factors of \\( hjgrksla \\). For example \\( qzxwvtnp(9)=(-1)^{2}, qzxwvtnp(20)=(-1)^{3} \\). Define \\( mtrelaxs(hjgrksla) \\) as \\( \\Sigma qzxwvtnp(vbycrofe) \\) where the sum ranges over all positive integer divisors of \\( hjgrksla \\). Prove that for every positive integer \\( hjgrksla, mtrelaxs(hjgrksla)=0 \\) or \\( mtrelaxs(hjgrksla)=1 \\). For which integers \\( hjgrksla \\) is \\( mtrelaxs(hjgrksla)=1 ? \\quad \\",
      "solution": "Solution. Suppose \\( sajkugbh \\) and \\( hjgrksla \\) are relatively prime positive integers. Then every divisor of \\( sajkugbh hjgrksla \\) is uniquely a product \\( quxsnmva yirbkeoz \\) where \\( quxsnmva\\left|sajkugbh, yirbkeoz\\right| hjgrksla \\), and conversely. Also \\( qzxwvtnp\\left(quxsnmva yirbkeoz\\right)=qzxwvtnp\\left(quxsnmva\\right) qzxwvtnp\\left(yirbkeoz\\right) \\).\n\\[\n\\begin{aligned}\nmtrelaxs(sajkugbh\\, hjgrksla) & =\\sum_{vbycrofe \\mid sajkugbh\\, hjgrksla} qzxwvtnp(vbycrofe)=\\sum_{quxsnmva \\mid sajkugbh} \\sum_{yirbkeoz \\mid hjgrksla} qzxwvtnp\\left(quxsnmva\\right) qzxwvtnp\\left(yirbkeoz\\right) \\\\\n& =mtrelaxs(sajkugbh)\\, mtrelaxs(hjgrksla)\n\\end{aligned}\n\\]\n\nThus \\( mtrelaxs \\) is a multiplicative numerical function and it suffices to evaluate \\( mtrelaxs \\) on prime powers. Evidently for \\( zqtrmnev \\) a prime\n\\[\n\\begin{aligned}\nmtrelaxs\\left(zqtrmnev^{lopqstuv}\\right) & =qzxwvtnp(1)+qzxwvtnp(zqtrmnev)+qzxwvtnp\\left(zqtrmnev^{2}\\right)+\\cdots+qzxwvtnp\\left(zqtrmnev^{lopqstuv}\\right) \\\\\n& =1-1+1-\\cdots+(-1)^{lopqstuv} \\\\\n& =\\left\\{\\begin{array}{ll}\n1 & \\text { if } lopqstuv \\text { is even } \\\\\n0 & \\text { if } lopqstuv \\text { is odd. }\n\\end{array}\\right.\n\\end{aligned}\n\\]\n\nLet \\( hjgrksla \\) be any positive integer and suppose\n\\[\nhjgrksla=rfvbzxqu^{aktewomz}\\, hodnlysp^{lvdrusni}\\, \\cdots\\, gicafmwe^{lopqstuv_{ufldevpo}}\n\\]\nis its canonical factorization into primes. Then \\( mtrelaxs(hjgrksla)=mtrelaxs\\left(rfvbzxqu^{aktewomz}\\right) mtrelaxs\\left(hodnlysp^{lvdrusni}\\right) \\cdots mtrelaxs\\left(gicafmwe^{lopqstuv}\\right) \\) and we see that \\( mtrelaxs(hjgrksla)=0 \\) if some prime appears with odd exponent in the prime factorization of \\( hjgrksla \\), and \\( mtrelaxs(hjgrksla)=1 \\) if all primes appear with even exponents. In other words,\n\\[\nmtrelaxs(hjgrksla)=1\n\\]\nif \\( hjgrksla \\) is a perfect square, and\n\\[\nmtrelaxs(hjgrksla)=0\n\\]\nif \\( hjgrksla \\) is not a perfect square."
    },
    "kernel_variant": {
      "question": "Let $K$ be a number field and let $\\mathcal O_{K}$ denote its ring of integers.\n\nFor any non–zero integral ideal  \n\n\\[\n\\mathfrak a=\\prod_{i=1}^{t}\\mathfrak p_{i}^{\\,e_{i}}\\qquad\n(e_{i}\\ge 1,\\; \\mathfrak p_{i}\\ \\text{prime}),\n\\]\n\ndefine  \n\n\\[\n\\Omega(\\mathfrak a)=e_{1}+e_{2}+\\dots+e_{t},\n\\qquad\nf(\\mathfrak a)=(-1)^{\\Omega(\\mathfrak a)}.\n\\]\n\nFor another non–zero integral ideal $\\mathfrak A$ put  \n\n\\[\nF(\\mathfrak A)=\\sum_{\\mathfrak a\\mid\\mathfrak A} f(\\mathfrak a),\\tag{$\\dagger$}\n\\]\n\nwhere the sum runs over all non–zero \\emph{integral} ideals $\\mathfrak a$ that divide $\\mathfrak A$.\n\n(a)  Prove that $F(\\mathfrak A)$ takes only the values $0$ and $1$.\n\n(b)  Show that $F(\\mathfrak A)=1$ if and only if $\\mathfrak A$ is the \\emph{square of an integral ideal};  \nequivalently, every exponent occurring in the prime–ideal factorisation of $\\mathfrak A$ is even.\n\n(c)  Specialise to principal ideals.  For an algebraic integer $0\\neq\\alpha\\in\\mathcal O_{K}$ write $(\\alpha)$ for the principal ideal it generates and set  \n\n\\[\n\\mathscr S=\\bigl\\{\\,(\\alpha)\\subset \\mathcal O_{K}:\\alpha\\neq 0,\\;F((\\alpha))=1\\,\\bigr\\}.\n\\]\n\n(i)  Prove that $F((\\alpha))=1$ if and only if the principal ideal $(\\alpha)$ can be written uniquely in the form $(\\alpha)=\\mathfrak B^{2}$ with $\\mathfrak B$ an integral ideal.\n\n(ii)  Show that such a decomposition exists precisely when $\\mathfrak B$ represents a $2$–torsion element of the ideal class group; that is  \n\n\\[\n\\mathscr S\n=\\bigl\\{\\mathfrak B^{2}:\\mathfrak B\\text{ integral and }[\\mathfrak B]^{2}=1\\bigr\\}\n=\\bigl\\{\\mathfrak B^{2}:[\\mathfrak B]\\in \\operatorname{Cl}_{K}[2]\\bigr\\}.\n\\]\n\n(iii)  Call two ideals $(\\alpha),(\\beta)\\in\\mathscr S$ \\emph{square–equivalent} if  \n$(\\alpha)=(\\beta)\\,(\\gamma)^{2}$ for some $\\gamma\\in K^{\\times}$, and put  \n\n\\[\n\\overline{\\mathscr S}:=\\mathscr S\\big/\\text{(square–equivalence)}.\n\\]\n\nShow that  \n\n\\[\n\\operatorname{Cl}_{K}[2]\\times\\overline{\\mathscr S}\\;\\longrightarrow\\;\\overline{\\mathscr S},\n\\qquad\n([\\mathfrak C],[(\\alpha)])\\longmapsto[(\\alpha)\\,\\mathfrak C^{2}],\n\\]\n\ndefines a well–defined, free and transitive group action.  Hence $\\overline{\\mathscr S}$ is a torsor for $\\operatorname{Cl}_{K}[2]$.\n\n(iv)  Describe, up to multiplication by a \\emph{square} in $K^{\\times}$, the algebraic integers $\\alpha$ with $F((\\alpha))=1$ in terms of the torsor $\\overline{\\mathscr S}$.",
      "solution": "Throughout we write valuations additively: for a (fractional) ideal $\\mathfrak I$ and prime ideal $\\mathfrak p$ let $v_{\\mathfrak p}(\\mathfrak I)$ be the exponent of $\\mathfrak p$ in the prime decomposition of $\\mathfrak I$.\n\nStep 1.  \\textbf{Complete multiplicativity of $f$.}  \nBecause $\\Omega(\\,\\cdot\\,)$ is additive on products of ideals we have  \n\\[\nf(\\mathfrak a\\mathfrak b)=(-1)^{\\Omega(\\mathfrak a\\mathfrak b)}\n=(-1)^{\\Omega(\\mathfrak a)+\\Omega(\\mathfrak b)}=f(\\mathfrak a)\\,f(\\mathfrak b)\n\\]\nfor \\emph{all} non–zero integral ideals $\\mathfrak a,\\mathfrak b$.\n\nStep 2.  \\textbf{Multiplicativity of $F$.}  \nLet $\\mathfrak a,\\mathfrak b$ be coprime integral ideals.  Every divisor of $\\mathfrak a\\mathfrak b$ is uniquely of the form $\\mathfrak d_{1}\\mathfrak d_{2}$ with  \n$\\mathfrak d_{1}\\mid\\mathfrak a,\\;\\mathfrak d_{2}\\mid\\mathfrak b$.  Using the complete multiplicativity of $f$ we obtain  \n\\[\nF(\\mathfrak a\\mathfrak b)=\n\\sum_{\\mathfrak d_{1}\\mid\\mathfrak a}\n\\sum_{\\mathfrak d_{2}\\mid\\mathfrak b}\nf(\\mathfrak d_{1}\\mathfrak d_{2})\n=\n\\sum_{\\mathfrak d_{1}\\mid\\mathfrak a}\n\\sum_{\\mathfrak d_{2}\\mid\\mathfrak b}\nf(\\mathfrak d_{1})\\,f(\\mathfrak d_{2})\n=F(\\mathfrak a)\\,F(\\mathfrak b).\\tag{1}\n\\]\nHence $F$ is a (Dirichlet) multiplicative function on the monoid of integral ideals.\n\nStep 3.  \\textbf{Evaluation on prime powers.}  \nFor a prime ideal $\\mathfrak p$ and $\\alpha\\ge 0$\n\\[\nF(\\mathfrak p^{\\alpha})=\\sum_{j=0}^{\\alpha}(-1)^{j}\n=\\begin{cases}\n1 & \\text{if $\\alpha$ is even},\\\\[2mm]\n0 & \\text{if $\\alpha$ is odd}.\n\\end{cases}\\tag{2}\n\\]\n\nStep 4.  \\textbf{Proof of (a).}  \nWriting $\\mathfrak A=\\prod_{i=1}^{k}\\mathfrak p_{i}^{\\,\\alpha_{i}}$ with pairwise distinct $\\mathfrak p_{i}$ and applying (1)–(2),\n\\[\nF(\\mathfrak A)=\\prod_{i=1}^{k}F(\\mathfrak p_{i}^{\\,\\alpha_{i}})\\in\\{0,1\\}.\n\\]\n\nStep 5.  \\textbf{Proof of (b).}  \nBy (2) we have $F(\\mathfrak p^{\\alpha})=1$ precisely when $\\alpha$ is even.  Therefore  \n\\[\nF(\\mathfrak A)=1\n\\Longleftrightarrow\n\\alpha_{i}\\text{ even for all }i\n\\Longleftrightarrow\n\\mathfrak A\n=\\Bigl(\\prod_{i=1}^{k}\\mathfrak p_{i}^{\\,\\alpha_{i}/2}\\Bigr)^{2}\n=\\mathfrak B^{2}\n\\]\nwith $\\mathfrak B$ integral; conversely $F(\\mathfrak B^{2})=1$ for every integral $\\mathfrak B$.\n\nStep 6.  \\textbf{Principal ideals — parts (c)(i) and (ii).}\n\n(i)  For an algebraic integer $\\alpha\\neq 0$ write $(\\alpha)=\\prod_{i}\\mathfrak p_{i}^{\\,v_{i}(\\alpha)}$.  Then  \n\\[\nF((\\alpha))=1\n\\Longleftrightarrow\nv_{i}(\\alpha)\\text{ even }\\forall i\n\\Longleftrightarrow\n(\\alpha)=\\mathfrak B^{2}\\quad(\\text{integral }\\mathfrak B,\\text{ unique}).\\tag{3}\n\\]\n\n(ii)  If $(\\alpha)=\\mathfrak B^{2}$ then $[(\\alpha)]=[\\mathfrak B]^{2}=1$, so $[\\mathfrak B]\\in\\operatorname{Cl}_{K}[2]$.  Conversely, if $\\mathfrak B$ is an integral ideal with $[\\mathfrak B]^{2}=1$, then $\\mathfrak B^{2}$ is principal, and (3) shows that every member of $\\mathscr S$ arises in this way.  Hence  \n\\[\n\\mathscr S\n=\\bigl\\{\\mathfrak B^{2}:[\\mathfrak B]\\in\\operatorname{Cl}_{K}[2]\\bigr\\}.\\tag{4}\n\\]\n\nStep 7.  \\textbf{Torsor structure — part (c)(iii).}\n\nWe keep the notation  \n$(\\alpha)\\approx(\\beta)$ if $(\\alpha)=(\\beta)\\,(\\gamma)^{2}$ for some $\\gamma\\in K^{\\times}$, and set $\\overline{\\mathscr S}=\\mathscr S/\\!\\approx$.\n\nDefine  \n\\[\n[\\mathfrak C]\\circ[(\\alpha)]:=[(\\alpha)\\,\\mathfrak C^{2}],\\tag{5}\n\\]\nwhere $[\\mathfrak C]\\in\\operatorname{Cl}_{K}[2]$ and $[(\\alpha)]\\in\\overline{\\mathscr S}$.\n\n\\emph{Well–definedness.}  Precisely as in the original argument, (5) is independent of the chosen representatives for $[\\mathfrak C]$ and $[(\\alpha)]$.\n\n\\emph{Freeness (corrected argument).}  \nAssume  \n\\[\n[\\mathfrak C]\\circ[(\\alpha)]=[(\\alpha)].\n\\]\nChoose $\\alpha$ such that $(\\alpha)=\\mathfrak B^{2}$ with $\\mathfrak B$ integral (possible by (4)).  Then  \n\\[\n(\\alpha)\\mathfrak C^{2}=(\\alpha)\\,(\\gamma)^{2}\\quad\\text{for some }\\gamma\\in K^{\\times},\n\\]\nwhence $\\mathfrak C^{2}=(\\gamma)^{2}$.  For every prime ideal $\\mathfrak p$ we therefore have  \n\\[\n2\\,v_{\\mathfrak p}(\\mathfrak C)=2\\,v_{\\mathfrak p}((\\gamma)) \\;\\Longrightarrow\\; \nv_{\\mathfrak p}(\\mathfrak C)=v_{\\mathfrak p}((\\gamma)).\n\\]\nHence $\\mathfrak C=(\\gamma\\varepsilon)$ with $\\varepsilon\\in\\mathcal O_{K}^{\\times}$ a unit.  Consequently $\\mathfrak C$ is principal and $[\\mathfrak C]=1$ in $\\operatorname{Cl}_{K}[2]$.  No integrality assumption on any quotient ideal is needed, so the action is free.\n\n\\emph{Transitivity.}  \nFor $[(\\alpha)],[(\\beta)]\\in\\overline{\\mathscr S}$ choose integral $\\mathfrak B_{1},\\mathfrak B_{2}$ with $(\\alpha)=\\mathfrak B_{1}^{2},\\;(\\beta)=\\mathfrak B_{2}^{2}$.  Set $[\\mathfrak C]:=[\\mathfrak B_{1}^{-1}\\mathfrak B_{2}]$.  Since $[\\mathfrak B_{i}]^{2}=1$ we have $[\\mathfrak C]\\in\\operatorname{Cl}_{K}[2]$, and  \n\\[\n[\\mathfrak C]\\circ[(\\alpha)]=[(\\alpha)\\,\\mathfrak C^{2}]\n=[\\mathfrak B_{1}^{2}(\\mathfrak B_{1}^{-1}\\mathfrak B_{2})^{2}]\n=[\\mathfrak B_{2}^{2}]=[(\\beta)].\n\\]\nThus the action is transitive.\n\nTherefore $\\overline{\\mathscr S}$ is a principal homogeneous space (torsor) for $\\operatorname{Cl}_{K}[2]$.\n\nStep 8.  \\textbf{Description of the integers $\\alpha$ — part (c)(iv).}\n\nThe map $\\alpha\\longmapsto[(\\alpha)]$ induces a bijection  \n\\[\n\\bigl\\{\\alpha\\in\\mathcal O_{K}\\setminus\\{0\\}:F((\\alpha))=1\\bigr\\}\\big/\\!(K^{\\times})^{2}\n\\;\\xrightarrow{\\;\\sim\\;}\\;\n\\overline{\\mathscr S},\n\\]\nbecause multiplying $\\alpha$ by a square does not change its class in $\\overline{\\mathscr S}$.  Since $\\overline{\\mathscr S}$ is a torsor for $\\operatorname{Cl}_{K}[2]$, the same is true for the set of non–zero $\\alpha$ with $F((\\alpha))=1$, considered up to squares in $K^{\\times}$.  Explicitly,\n\\[\n\\alpha_{2}\\equiv\\alpha_{1}\\pmod{(K^{\\times})^{2}}\n\\Longleftrightarrow\n(\\alpha_{2})=(\\alpha_{1})\\,\\mathfrak C^{2}\\quad\n\\text{for some }[\\mathfrak C]\\in\\operatorname{Cl}_{K}[2].\n\\]\n\n\\medskip\n\\emph{Summary.}  \n(1) $F((\\alpha))=1$ iff $(\\alpha)=\\mathfrak B^{2}$ with $\\mathfrak B$ integral and unique.  \n(2) Such squares correspond bijectively to the $2$–torsion subgroup $\\operatorname{Cl}_{K}[2]$.  \n(3) After quotienting by principal squares the resulting classes form a torsor for $\\operatorname{Cl}_{K}[2]$.  \n(4) Hence the algebraic integers $\\alpha$ with $F((\\alpha))=1$, modulo multiplication by squares in $K^{\\times}$, are parametrised by $\\operatorname{Cl}_{K}[2]$.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.528780",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional setting: the problem is lifted from ℤ to the full ideal lattice of an arbitrary number field, requiring fluency with Dedekind domains, prime-ideal factorisation and ideal arithmetic.\n\n2. Additional structures: instead of ordinary integers, the argument must navigate unique factorisation of ideals (which may fail for elements) and manipulate ideal classes.\n\n3. Deeper theory: part (c) forces the solver to connect the combinatorial identity with the algebraic notion of 2-torsion in the ideal class group, a topic well beyond elementary number theory.\n\n4. Multiplicativity now involves the Chinese-remainder property for ideals; proving (1) rigorously needs knowledge of comaximality in Dedekind domains.\n\n5. While the core idea (alternating sum over powers of a prime) is retained, executing it in this generality and interpreting the outcome demands several non-trivial steps—prime-ideal factorisation, structure of Cl_K, and properties of units—making the variant substantially harder than either the original problem or the given kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $K$ be a number field and let $\\mathcal O_{K}$ denote its ring of integers.\n\nFor any non–zero integral ideal  \n\n\\[\n\\mathfrak a=\\prod_{i=1}^{t}\\mathfrak p_{i}^{\\,e_{i}}\\qquad\n(e_{i}\\ge 1,\\; \\mathfrak p_{i}\\ \\text{prime}),\n\\]\n\ndefine  \n\n\\[\n\\Omega(\\mathfrak a)=e_{1}+e_{2}+\\dots+e_{t},\n\\qquad\nf(\\mathfrak a)=(-1)^{\\Omega(\\mathfrak a)}.\n\\]\n\nFor another non–zero integral ideal $\\mathfrak A$ put  \n\n\\[\nF(\\mathfrak A)=\\sum_{\\mathfrak a\\mid\\mathfrak A} f(\\mathfrak a),\\tag{$\\dagger$}\n\\]\n\nwhere the sum runs over all non–zero \\emph{integral} ideals $\\mathfrak a$ that divide $\\mathfrak A$.\n\n(a)  Prove that $F(\\mathfrak A)$ takes only the values $0$ and $1$.\n\n(b)  Show that $F(\\mathfrak A)=1$ if and only if $\\mathfrak A$ is the \\emph{square of an integral ideal};  \nequivalently, every exponent occurring in the prime–ideal factorisation of $\\mathfrak A$ is even.\n\n(c)  Specialise to principal ideals.  For an algebraic integer $0\\neq\\alpha\\in\\mathcal O_{K}$ write $(\\alpha)$ for the principal ideal it generates and set  \n\n\\[\n\\mathscr S=\\bigl\\{\\,(\\alpha)\\subset \\mathcal O_{K}:\\alpha\\neq 0,\\;F((\\alpha))=1\\,\\bigr\\}.\n\\]\n\n(i)  Prove that $F((\\alpha))=1$ if and only if the principal ideal $(\\alpha)$ can be written uniquely in the form $(\\alpha)=\\mathfrak B^{2}$ with $\\mathfrak B$ an integral ideal.\n\n(ii)  Show that such a decomposition exists precisely when $\\mathfrak B$ represents a $2$–torsion element of the ideal class group; that is  \n\n\\[\n\\mathscr S\n=\\bigl\\{\\mathfrak B^{2}:\\mathfrak B\\text{ integral and }[\\mathfrak B]^{2}=1\\bigr\\}\n=\\bigl\\{\\mathfrak B^{2}:[\\mathfrak B]\\in \\operatorname{Cl}_{K}[2]\\bigr\\}.\n\\]\n\n(iii)  Call two ideals $(\\alpha),(\\beta)\\in\\mathscr S$ \\emph{square–equivalent} if  \n$(\\alpha)=(\\beta)\\,(\\gamma)^{2}$ for some $\\gamma\\in K^{\\times}$, and put  \n\n\\[\n\\overline{\\mathscr S}:=\\mathscr S\\big/\\text{(square–equivalence)}.\n\\]\n\nShow that  \n\n\\[\n\\operatorname{Cl}_{K}[2]\\times\\overline{\\mathscr S}\\;\\longrightarrow\\;\\overline{\\mathscr S},\n\\qquad\n([\\mathfrak C],[(\\alpha)])\\longmapsto[(\\alpha)\\,\\mathfrak C^{2}],\n\\]\n\ndefines a well–defined, free and transitive group action.  Hence $\\overline{\\mathscr S}$ is a torsor for $\\operatorname{Cl}_{K}[2]$.\n\n(iv)  Describe, up to multiplication by a \\emph{square} in $K^{\\times}$, the algebraic integers $\\alpha$ with $F((\\alpha))=1$ in terms of the torsor $\\overline{\\mathscr S}$.",
      "solution": "Throughout we write valuations additively: for a (fractional) ideal $\\mathfrak I$ and prime ideal $\\mathfrak p$ let $v_{\\mathfrak p}(\\mathfrak I)$ be the exponent of $\\mathfrak p$ in the prime decomposition of $\\mathfrak I$.\n\nStep 1.  \\textbf{Complete multiplicativity of $f$.}  \nBecause $\\Omega(\\,\\cdot\\,)$ is additive on products of ideals we have  \n\\[\nf(\\mathfrak a\\mathfrak b)=(-1)^{\\Omega(\\mathfrak a\\mathfrak b)}\n=(-1)^{\\Omega(\\mathfrak a)+\\Omega(\\mathfrak b)}=f(\\mathfrak a)\\,f(\\mathfrak b)\n\\]\nfor \\emph{all} non–zero integral ideals $\\mathfrak a,\\mathfrak b$.\n\nStep 2.  \\textbf{Multiplicativity of $F$.}  \nLet $\\mathfrak a,\\mathfrak b$ be coprime integral ideals.  Every divisor of $\\mathfrak a\\mathfrak b$ is uniquely of the form $\\mathfrak d_{1}\\mathfrak d_{2}$ with  \n$\\mathfrak d_{1}\\mid\\mathfrak a,\\;\\mathfrak d_{2}\\mid\\mathfrak b$.  Using the complete multiplicativity of $f$ we obtain  \n\\[\nF(\\mathfrak a\\mathfrak b)=\n\\sum_{\\mathfrak d_{1}\\mid\\mathfrak a}\n\\sum_{\\mathfrak d_{2}\\mid\\mathfrak b}\nf(\\mathfrak d_{1}\\mathfrak d_{2})\n=\n\\sum_{\\mathfrak d_{1}\\mid\\mathfrak a}\n\\sum_{\\mathfrak d_{2}\\mid\\mathfrak b}\nf(\\mathfrak d_{1})\\,f(\\mathfrak d_{2})\n=F(\\mathfrak a)\\,F(\\mathfrak b).\\tag{1}\n\\]\nHence $F$ is a (Dirichlet) multiplicative function on the monoid of integral ideals.\n\nStep 3.  \\textbf{Evaluation on prime powers.}  \nFor a prime ideal $\\mathfrak p$ and $\\alpha\\ge 0$\n\\[\nF(\\mathfrak p^{\\alpha})=\\sum_{j=0}^{\\alpha}(-1)^{j}\n=\\begin{cases}\n1 & \\text{if $\\alpha$ is even},\\\\[2mm]\n0 & \\text{if $\\alpha$ is odd}.\n\\end{cases}\\tag{2}\n\\]\n\nStep 4.  \\textbf{Proof of (a).}  \nWriting $\\mathfrak A=\\prod_{i=1}^{k}\\mathfrak p_{i}^{\\,\\alpha_{i}}$ with pairwise distinct $\\mathfrak p_{i}$ and applying (1)–(2),\n\\[\nF(\\mathfrak A)=\\prod_{i=1}^{k}F(\\mathfrak p_{i}^{\\,\\alpha_{i}})\\in\\{0,1\\}.\n\\]\n\nStep 5.  \\textbf{Proof of (b).}  \nBy (2) we have $F(\\mathfrak p^{\\alpha})=1$ precisely when $\\alpha$ is even.  Therefore  \n\\[\nF(\\mathfrak A)=1\n\\Longleftrightarrow\n\\alpha_{i}\\text{ even for all }i\n\\Longleftrightarrow\n\\mathfrak A\n=\\Bigl(\\prod_{i=1}^{k}\\mathfrak p_{i}^{\\,\\alpha_{i}/2}\\Bigr)^{2}\n=\\mathfrak B^{2}\n\\]\nwith $\\mathfrak B$ integral; conversely $F(\\mathfrak B^{2})=1$ for every integral $\\mathfrak B$.\n\nStep 6.  \\textbf{Principal ideals — parts (c)(i) and (ii).}\n\n(i)  For an algebraic integer $\\alpha\\neq 0$ write $(\\alpha)=\\prod_{i}\\mathfrak p_{i}^{\\,v_{i}(\\alpha)}$.  Then  \n\\[\nF((\\alpha))=1\n\\Longleftrightarrow\nv_{i}(\\alpha)\\text{ even }\\forall i\n\\Longleftrightarrow\n(\\alpha)=\\mathfrak B^{2}\\quad(\\text{integral }\\mathfrak B,\\text{ unique}).\\tag{3}\n\\]\n\n(ii)  If $(\\alpha)=\\mathfrak B^{2}$ then $[(\\alpha)]=[\\mathfrak B]^{2}=1$, so $[\\mathfrak B]\\in\\operatorname{Cl}_{K}[2]$.  Conversely, if $\\mathfrak B$ is an integral ideal with $[\\mathfrak B]^{2}=1$, then $\\mathfrak B^{2}$ is principal, and (3) shows that every member of $\\mathscr S$ arises in this way.  Hence  \n\\[\n\\mathscr S\n=\\bigl\\{\\mathfrak B^{2}:[\\mathfrak B]\\in\\operatorname{Cl}_{K}[2]\\bigr\\}.\\tag{4}\n\\]\n\nStep 7.  \\textbf{Torsor structure — part (c)(iii).}\n\nWe keep the notation  \n$(\\alpha)\\approx(\\beta)$ if $(\\alpha)=(\\beta)\\,(\\gamma)^{2}$ for some $\\gamma\\in K^{\\times}$, and set $\\overline{\\mathscr S}=\\mathscr S/\\!\\approx$.\n\nDefine  \n\\[\n[\\mathfrak C]\\circ[(\\alpha)]:=[(\\alpha)\\,\\mathfrak C^{2}],\\tag{5}\n\\]\nwhere $[\\mathfrak C]\\in\\operatorname{Cl}_{K}[2]$ and $[(\\alpha)]\\in\\overline{\\mathscr S}$.\n\n\\emph{Well–definedness.}  Precisely as in the original argument, (5) is independent of the chosen representatives for $[\\mathfrak C]$ and $[(\\alpha)]$.\n\n\\emph{Freeness (corrected argument).}  \nAssume  \n\\[\n[\\mathfrak C]\\circ[(\\alpha)]=[(\\alpha)].\n\\]\nChoose $\\alpha$ such that $(\\alpha)=\\mathfrak B^{2}$ with $\\mathfrak B$ integral (possible by (4)).  Then  \n\\[\n(\\alpha)\\mathfrak C^{2}=(\\alpha)\\,(\\gamma)^{2}\\quad\\text{for some }\\gamma\\in K^{\\times},\n\\]\nwhence $\\mathfrak C^{2}=(\\gamma)^{2}$.  For every prime ideal $\\mathfrak p$ we therefore have  \n\\[\n2\\,v_{\\mathfrak p}(\\mathfrak C)=2\\,v_{\\mathfrak p}((\\gamma)) \\;\\Longrightarrow\\; \nv_{\\mathfrak p}(\\mathfrak C)=v_{\\mathfrak p}((\\gamma)).\n\\]\nHence $\\mathfrak C=(\\gamma\\varepsilon)$ with $\\varepsilon\\in\\mathcal O_{K}^{\\times}$ a unit.  Consequently $\\mathfrak C$ is principal and $[\\mathfrak C]=1$ in $\\operatorname{Cl}_{K}[2]$.  No integrality assumption on any quotient ideal is needed, so the action is free.\n\n\\emph{Transitivity.}  \nFor $[(\\alpha)],[(\\beta)]\\in\\overline{\\mathscr S}$ choose integral $\\mathfrak B_{1},\\mathfrak B_{2}$ with $(\\alpha)=\\mathfrak B_{1}^{2},\\;(\\beta)=\\mathfrak B_{2}^{2}$.  Set $[\\mathfrak C]:=[\\mathfrak B_{1}^{-1}\\mathfrak B_{2}]$.  Since $[\\mathfrak B_{i}]^{2}=1$ we have $[\\mathfrak C]\\in\\operatorname{Cl}_{K}[2]$, and  \n\\[\n[\\mathfrak C]\\circ[(\\alpha)]=[(\\alpha)\\,\\mathfrak C^{2}]\n=[\\mathfrak B_{1}^{2}(\\mathfrak B_{1}^{-1}\\mathfrak B_{2})^{2}]\n=[\\mathfrak B_{2}^{2}]=[(\\beta)].\n\\]\nThus the action is transitive.\n\nTherefore $\\overline{\\mathscr S}$ is a principal homogeneous space (torsor) for $\\operatorname{Cl}_{K}[2]$.\n\nStep 8.  \\textbf{Description of the integers $\\alpha$ — part (c)(iv).}\n\nThe map $\\alpha\\longmapsto[(\\alpha)]$ induces a bijection  \n\\[\n\\bigl\\{\\alpha\\in\\mathcal O_{K}\\setminus\\{0\\}:F((\\alpha))=1\\bigr\\}\\big/\\!(K^{\\times})^{2}\n\\;\\xrightarrow{\\;\\sim\\;}\\;\n\\overline{\\mathscr S},\n\\]\nbecause multiplying $\\alpha$ by a square does not change its class in $\\overline{\\mathscr S}$.  Since $\\overline{\\mathscr S}$ is a torsor for $\\operatorname{Cl}_{K}[2]$, the same is true for the set of non–zero $\\alpha$ with $F((\\alpha))=1$, considered up to squares in $K^{\\times}$.  Explicitly,\n\\[\n\\alpha_{2}\\equiv\\alpha_{1}\\pmod{(K^{\\times})^{2}}\n\\Longleftrightarrow\n(\\alpha_{2})=(\\alpha_{1})\\,\\mathfrak C^{2}\\quad\n\\text{for some }[\\mathfrak C]\\in\\operatorname{Cl}_{K}[2].\n\\]\n\n\\medskip\n\\emph{Summary.}  \n(1) $F((\\alpha))=1$ iff $(\\alpha)=\\mathfrak B^{2}$ with $\\mathfrak B$ integral and unique.  \n(2) Such squares correspond bijectively to the $2$–torsion subgroup $\\operatorname{Cl}_{K}[2]$.  \n(3) After quotienting by principal squares the resulting classes form a torsor for $\\operatorname{Cl}_{K}[2]$.  \n(4) Hence the algebraic integers $\\alpha$ with $F((\\alpha))=1$, modulo multiplication by squares in $K^{\\times}$, are parametrised by $\\operatorname{Cl}_{K}[2]$.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.441527",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional setting: the problem is lifted from ℤ to the full ideal lattice of an arbitrary number field, requiring fluency with Dedekind domains, prime-ideal factorisation and ideal arithmetic.\n\n2. Additional structures: instead of ordinary integers, the argument must navigate unique factorisation of ideals (which may fail for elements) and manipulate ideal classes.\n\n3. Deeper theory: part (c) forces the solver to connect the combinatorial identity with the algebraic notion of 2-torsion in the ideal class group, a topic well beyond elementary number theory.\n\n4. Multiplicativity now involves the Chinese-remainder property for ideals; proving (1) rigorously needs knowledge of comaximality in Dedekind domains.\n\n5. While the core idea (alternating sum over powers of a prime) is retained, executing it in this generality and interpreting the outcome demands several non-trivial steps—prime-ideal factorisation, structure of Cl_K, and properties of units—making the variant substantially harder than either the original problem or the given kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}