{
  "index": "1976-B-6",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "B-6. As usual, let \\( \\sigma(N) \\) denote the sum of all the (positive integral) divisors of \\( N \\). (Included among these divisors are \\( I \\) and \\( N \\) itself.) For example, if \\( p \\) is a prime, then \\( \\sigma(p)=p+1 \\). Motivated by the notion of a \"perfect\" number, a positive integer \\( N \\) is called \"quasiperfect\" if \\( \\sigma(N)=2 N+1 \\). Prove that every quasiperfect number is the square of an odd integer.",
  "solution": "B-6.\nLet \\( N=2^{\\alpha} p_{1}^{\\beta_{1}} p_{2}^{\\beta_{2}} \\cdots p_{k}^{\\beta_{k}} \\) where \\( \\alpha \\) and the \\( \\beta_{1} \\) are nonnegative integers and the \\( p_{1} \\) are distinct odd primes. Then\n\\[\n\\sigma(N)=\\sigma\\left(2^{\\alpha}\\right) \\sigma\\left(p_{1}^{\\beta_{1}}\\right) \\cdots \\sigma\\left(p_{k}^{\\beta_{k}}\\right) .\n\\]\n\nSince \\( \\sigma(N)=2 N+1 \\) is odd, it follows that \\( \\sigma\\left(p_{i}^{\\beta_{1}}\\right) \\) is odd, \\( 1 \\leqq i \\leqq k \\). But\n\\[\n\\sigma\\left(p_{i}^{\\beta_{i}}\\right)=1+p_{i}+p_{i}^{2}+\\cdots+p_{i}^{\\beta_{1}}\n\\]\nis odd if and only if \\( \\beta_{1} \\) is even; for if \\( \\beta_{1} \\) were odd, the right hand side would be the sum of an even number of odd numbers and hence even. It follows that the odd part of \\( N \\) must be a square, so that we may write\n\\[\nN=2^{\\alpha} M^{2}, \\alpha \\geqq 0 .\n\\]\nwhere \\( M \\) is odd. It remains to show that \\( \\alpha=0 \\).\nSince \\( N \\) is quasiperfect, \\( \\sigma(N)=2^{\\alpha+1} M^{2}+1 \\), while from (1) we deduce \\( \\sigma(N)=\\sigma\\left(2^{\\alpha}\\right) \\sigma\\left(M^{2}\\right)= \\) \\( \\left(2^{\\alpha+1}-1\\right) \\sigma\\left(M^{2}\\right) \\). Hence \\( 2^{\\alpha+1} M^{2}+1=\\left(2^{\\alpha+1}-1\\right) \\sigma\\left(M^{2}\\right) \\) so that\n\\[\nM^{2}+1 \\equiv 0\\left(\\bmod 2^{\\alpha+1}-1\\right)\n\\]\n\nIf \\( \\alpha>0,2^{\\alpha+1}-1 \\equiv 3(\\bmod 4) \\). Consequently \\( 2^{\\alpha+1}-1 \\) has a prime divisor \\( p \\equiv 3(\\bmod 4) \\). Equation (2) implies\n\\[\nM^{2}+1 \\equiv 0 \\quad(\\bmod p)\n\\]\n\nBut since -1 is a quadratic non-residue modulo \\( p \\) whenever \\( p \\equiv 3(\\bmod 4) \\), (3) is impossible. Thus. \\( \\alpha=0 \\).",
  "vars": [
    "N",
    "p",
    "p_1",
    "p_2",
    "p_k",
    "p_i",
    "k",
    "M",
    "i"
  ],
  "params": [
    "\\\\alpha",
    "\\\\beta_1",
    "\\\\beta_i",
    "\\\\sigma"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "N": "quasint",
        "p": "primevar",
        "p_1": "primeone",
        "p_2": "primetwo",
        "p_k": "primekth",
        "p_i": "primeith",
        "k": "primcount",
        "M": "oddroot",
        "i": "indexvar",
        "\\alpha": "alphaexpt",
        "\\beta_1": "betaoneex",
        "\\beta_i": "betaithex",
        "\\sigma": "divisorsum"
      },
      "question": "B-6. As usual, let \\( divisorsum(quasint) \\) denote the sum of all the (positive integral) divisors of \\( quasint \\). (Included among these divisors are \\( I \\) and \\( quasint \\) itself.) For example, if \\( primevar \\) is a prime, then \\( divisorsum(primevar)=primevar+1 \\). Motivated by the notion of a \"perfect\" number, a positive integer \\( quasint \\) is called \"quasiperfect\" if \\( divisorsum(quasint)=2 quasint+1 \\). Prove that every quasiperfect number is the square of an odd integer.",
      "solution": "B-6.\nLet \\( quasint=2^{alphaexpt}\\,primeone^{betaoneex}\\,primetwo^{\\beta_{2}} \\cdots primekth^{\\beta_{k}} \\) where \\( alphaexpt \\) and \\( betaoneex \\) are nonnegative integers and the \\( primeone \\) are distinct odd primes. Then\n\\[\ndivisorsum(quasint)=divisorsum\\left(2^{alphaexpt}\\right)\\,divisorsum\\left(primeone^{betaoneex}\\right)\\cdots divisorsum\\left(primekth^{\\beta_{k}}\\right).\n\\]\nSince \\( divisorsum(quasint)=2\\,quasint+1 \\) is odd, it follows that \\( divisorsum\\left(primeith^{betaithex}\\right) \\) is odd, \\( 1\\leqq indexvar\\leqq primcount \\). But\n\\[\ndivisorsum\\left(primeith^{betaithex}\\right)=1+primeith+primeith^{2}+\\cdots+primeith^{betaithex}\n\\]\nis odd if and only if \\( betaithex \\) is even; for if \\( betaithex \\) were odd, the right-hand side would be the sum of an even number of odd summands and hence even. Therefore the odd part of \\( quasint \\) must be a square, so we may write\n\\[\nquasint=2^{alphaexpt}\\,oddroot^{2},\\qquad alphaexpt\\geqq 0,\n\\]\nwith \\( oddroot \\) odd. It remains to prove that \\( alphaexpt=0 \\).\n\nBecause \\( quasint \\) is quasiperfect, \\( divisorsum(quasint)=2^{alphaexpt+1}oddroot^{2}+1 \\); on the other hand, from (1) we obtain\n\\[\ndivisorsum(quasint)=divisorsum\\left(2^{alphaexpt}\\right)\\,divisorsum\\left(oddroot^{2}\\right)=\\left(2^{alphaexpt+1}-1\\right)divisorsum\\left(oddroot^{2}\\right).\n\\]\nHence\n\\[\n2^{alphaexpt+1}oddroot^{2}+1=\\left(2^{alphaexpt+1}-1\\right)divisorsum\\left(oddroot^{2}\\right),\n\\]\nso that\n\\[\noddroot^{2}+1\\equiv 0\\pmod{2^{alphaexpt+1}-1}.\n\\]\nIf \\( alphaexpt>0 \\) then \\( 2^{alphaexpt+1}-1\\equiv 3\\pmod{4} \\). Consequently \\( 2^{alphaexpt+1}-1 \\) possesses a prime divisor \\( primevar\\equiv 3\\pmod{4} \\). From the congruence above we would have\n\\[\noddroot^{2}+1\\equiv 0\\pmod{primevar},\n\\]\nbut \\(-1\\) is a quadratic non-residue modulo any prime congruent to \\(3\\pmod 4\\), a contradiction. Therefore \\( alphaexpt=0 \\).\n\nThus every quasiperfect number is the square of an odd integer."
    },
    "descriptive_long_confusing": {
      "map": {
        "N": "marshmallow",
        "p": "lollipop",
        "p_1": "peppermint",
        "p_2": "butterscotch",
        "p_k": "gingerbread",
        "p_i": "cheesecake",
        "k": "shortcake",
        "M": "blueberry",
        "i": "tangerine",
        "\\alpha": "cinnamon",
        "\\beta_1": "chocolate",
        "\\beta_i": "strawberry",
        "\\sigma": "raspberry"
      },
      "question": "B-6. As usual, let \\( raspberry(marshmallow) \\) denote the sum of all the (positive integral) divisors of \\( marshmallow \\). (Included among these divisors are \\( I \\) and \\( marshmallow \\) itself.) For example, if \\( lollipop \\) is a prime, then \\( raspberry(lollipop)=lollipop+1 \\). Motivated by the notion of a \"perfect\" number, a positive integer \\( marshmallow \\) is called \"quasiperfect\" if \\( raspberry(marshmallow)=2 marshmallow+1 \\). Prove that every quasiperfect number is the square of an odd integer.",
      "solution": "B-6.\nLet \\( marshmallow=2^{cinnamon} peppermint^{chocolate} butterscotch^{\\beta_{2}} \\cdots gingerbread^{\\beta_{k}} \\) where \\( cinnamon \\) and the \\( chocolate \\) are nonnegative integers and the \\( peppermint \\) are distinct odd primes. Then\n\\[\nraspberry(marshmallow)=raspberry\\left(2^{cinnamon}\\right) raspberry\\left(peppermint^{chocolate}\\right) \\cdots raspberry\\left(gingerbread^{\\beta_{k}}\\right) .\n\\]\n\nSince \\( raspberry(marshmallow)=2 marshmallow+1 \\) is odd, it follows that \\( raspberry\\left(cheesecake^{chocolate}\\right) \\) is odd, \\( 1 \\leqq tangerine \\leqq shortcake \\). But\n\\[\nraspberry\\left(cheesecake^{strawberry}\\right)=1+cheesecake+cheesecake^{2}+\\cdots+cheesecake^{chocolate}\n\\]\nis odd if and only if \\( chocolate \\) is even; for if \\( chocolate \\) were odd, the right hand side would be the sum of an even number of odd numbers and hence even. It follows that the odd part of marshmallow must be a square, so that we may write\n\\[\nmarshmallow=2^{cinnamon} blueberry^{2}, cinnamon \\geqq 0 .\n\\]\nwhere blueberry is odd. It remains to show that \\( cinnamon=0 \\).\nSince marshmallow is quasiperfect, \\( raspberry(marshmallow)=2^{cinnamon+1} blueberry^{2}+1 \\), while from (1) we deduce \\( raspberry(marshmallow)=raspberry\\left(2^{cinnamon}\\right) raspberry\\left(blueberry^{2}\\right)= \\left(2^{cinnamon+1}-1\\right) raspberry\\left(blueberry^{2}\\right) \\). Hence \\( 2^{cinnamon+1} blueberry^{2}+1=\\left(2^{cinnamon+1}-1\\right) raspberry\\left(blueberry^{2}\\right) \\) so that\n\\[\nblueberry^{2}+1 \\equiv 0\\left(\\bmod 2^{cinnamon+1}-1\\right)\n\\]\n\nIf \\( cinnamon>0,2^{cinnamon+1}-1 \\equiv 3(\\bmod 4) \\). Consequently \\( 2^{cinnamon+1}-1 \\) has a prime divisor \\( lollipop \\equiv 3(\\bmod 4) \\). Equation (2) implies\n\\[\nblueberry^{2}+1 \\equiv 0 \\quad(\\bmod lollipop)\n\\]\n\nBut since -1 is a quadratic non-residue modulo \\( lollipop \\) whenever \\( lollipop \\equiv 3(\\bmod 4) \\), (3) is impossible. Thus. \\( cinnamon=0 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "N": "infinitesimal",
        "p": "compositefactor",
        "p_1": "compositeone",
        "p_2": "compositetwo",
        "p_k": "compositekay",
        "p_i": "compositeindex",
        "k": "solitude",
        "M": "evenroot",
        "\\alpha": "zeropower",
        "\\beta_1": "unevenexponent",
        "\\beta_i": "unevenindex",
        "\\sigma": "divergence"
      },
      "question": "B-6. As usual, let \\( divergence(infinitesimal) \\) denote the sum of all the (positive integral) divisors of \\( infinitesimal \\). (Included among these divisors are \\( I \\) and \\( infinitesimal \\) itself.) For example, if \\( compositefactor \\) is a prime, then \\( divergence(compositefactor)=compositefactor+1 \\). Motivated by the notion of a \"perfect\" number, a positive integer \\( infinitesimal \\) is called \"quasiperfect\" if \\( divergence(infinitesimal)=2 infinitesimal+1 \\). Prove that every quasiperfect number is the square of an odd integer.",
      "solution": "B-6.\nLet \\( infinitesimal=2^{zeropower} compositeone^{unevenexponent} compositetwo^{\\beta_{2}} \\cdots compositekay^{\\beta_{k}} \\) where \\( zeropower \\) and the \\( unevenexponent \\) are nonnegative integers and the \\( compositeone \\) are distinct odd primes. Then\n\\[\ndivergence(infinitesimal)=divergence\\left(2^{zeropower}\\right) divergence\\left(compositeone^{unevenexponent}\\right) \\cdots divergence\\left(compositekay^{\\beta_{k}}\\right) .\n\\]\n\nSince \\( divergence(infinitesimal)=2 infinitesimal+1 \\) is odd, it follows that \\( divergence\\left(compositeindex^{\\beta_{1}}\\right) \\) is odd, \\( 1 \\leqq i \\leqq solitude \\). But\n\\[\ndivergence\\left(compositeindex^{\\beta_{i}}\\right)=1+compositeindex+compositeindex^{2}+\\cdots+compositeindex^{\\beta_{1}}\n\\]\nis odd if and only if \\( \\beta_{1} \\) is even; for if \\( \\beta_{1} \\) were odd, the right hand side would be the sum of an even number of odd numbers and hence even. It follows that the odd part of \\( infinitesimal \\) must be a square, so that we may write\n\\[\ninfinitesimal=2^{zeropower} evenroot^{2}, zeropower \\geqq 0 .\n\\]\nwhere \\( evenroot \\) is odd. It remains to show that \\( zeropower=0 \\).\nSince \\( infinitesimal \\) is quasiperfect, \\( divergence(infinitesimal)=2^{zeropower+1} evenroot^{2}+1 \\), while from (1) we deduce \\( divergence(infinitesimal)=divergence\\left(2^{zeropower}\\right) divergence\\left(evenroot^{2}\\right)= \\) \\( \\left(2^{zeropower+1}-1\\right) divergence\\left(evenroot^{2}\\right) \\). Hence \\( 2^{zeropower+1} evenroot^{2}+1=\\left(2^{zeropower+1}-1\\right) divergence\\left(evenroot^{2}\\right) \\) so that\n\\[\nevenroot^{2}+1 \\equiv 0\\left(\\bmod 2^{zeropower+1}-1\\right)\n\\]\n\nIf \\( zeropower>0,2^{zeropower+1}-1 \\equiv 3(\\bmod 4) \\). Consequently \\( 2^{zeropower+1}-1 \\) has a prime divisor \\( compositefactor \\equiv 3(\\bmod 4) \\). Equation (2) implies\n\\[\nevenroot^{2}+1 \\equiv 0 \\quad(\\bmod compositefactor)\n\\]\n\nBut since -1 is a quadratic non-residue modulo \\( compositefactor \\) whenever \\( compositefactor \\equiv 3(\\bmod 4) \\), (3) is impossible. Thus. \\( zeropower=0 \\)."
    },
    "garbled_string": {
      "map": {
        "N": "qzxwvtnp",
        "p": "hjgrksla",
        "p_1": "vmnqtwor",
        "p_2": "gklsavnm",
        "p_k": "lxzmrptq",
        "p_i": "dsqplxne",
        "k": "rqhsmvzd",
        "M": "zfwopnrt",
        "i": "shpdkqvc",
        "\\\\alpha": "gnqtfwzs",
        "\\\\beta_1": "ndlqxzpr",
        "\\\\beta_i": "qprldmsx",
        "\\\\sigma": "wjkrgzop"
      },
      "question": "B-6. As usual, let \\( wjkrgzop(qzxwvtnp) \\) denote the sum of all the (positive integral) divisors of \\( qzxwvtnp \\). (Included among these divisors are \\( I \\) and \\( qzxwvtnp \\) itself.) For example, if \\( hjgrksla \\) is a prime, then \\( wjkrgzop(hjgrksla)=hjgrksla+1 \\). Motivated by the notion of a \"perfect\" number, a positive integer \\( qzxwvtnp \\) is called \"quasiperfect\" if \\( wjkrgzop(qzxwvtnp)=2 qzxwvtnp+1 \\). Prove that every quasiperfect number is the square of an odd integer.",
      "solution": "B-6.\nLet \\( qzxwvtnp=2^{gnqtfwzs} vmnqtwor^{ndlqxzpr} gklsavnm^{\\beta_{2}} \\cdots lxzmrptq^{\\beta_{k}} \\) where \\( gnqtfwzs \\) and the \\( ndlqxzpr \\) are nonnegative integers and the \\( vmnqtwor \\) are distinct odd primes. Then\n\\[\nwjkrgzop(qzxwvtnp)=wjkrgzop\\left(2^{gnqtfwzs}\\right) wjkrgzop\\left(vmnqtwor^{ndlqxzpr}\\right) \\cdots wjkrgzop\\left(lxzmrptq^{\\beta_{k}}\\right) .\n\\]\n\nSince \\( wjkrgzop(qzxwvtnp)=2 qzxwvtnp+1 \\) is odd, it follows that \\( wjkrgzop\\left(dsqplxne^{ndlqxzpr}\\right) \\) is odd, \\( 1 \\leqq shpdkqvc \\leqq rqhsmvzd \\). But\n\\[\nwjkrgzop\\left(dsqplxne^{qprldmsx}\\right)=1+dsqplxne+dsqplxne^{2}+\\cdots+dsqplxne^{ndlqxzpr}\n\\]\nis odd if and only if \\( ndlqxzpr \\) is even; for if \\( ndlqxzpr \\) were odd, the right hand side would be the sum of an even number of odd numbers and hence even. It follows that the odd part of \\( qzxwvtnp \\) must be a square, so that we may write\n\\[\nqzxwvtnp=2^{gnqtfwzs} zfwopnrt^{2}, gnqtfwzs \\geqq 0 .\n\\]\nwhere \\( zfwopnrt \\) is odd. It remains to show that \\( gnqtfwzs=0 \\).\nSince \\( qzxwvtnp \\) is quasiperfect, \\( wjkrgzop(qzxwvtnp)=2^{gnqtfwzs+1} zfwopnrt^{2}+1 \\), while from (1) we deduce \\( wjkrgzop(qzxwvtnp)=wjkrgzop\\left(2^{gnqtfwzs}\\right) wjkrgzop\\left(zfwopnrt^{2}\\right)= \\) \\( \\left(2^{gnqtfwzs+1}-1\\right) wjkrgzop\\left(zfwopnrt^{2}\\right) \\). Hence \\( 2^{gnqtfwzs+1} zfwopnrt^{2}+1=\\left(2^{gnqtfwzs+1}-1\\right) wjkrgzop\\left(zfwopnrt^{2}\\right) \\) so that\n\\[\nzfwopnrt^{2}+1 \\equiv 0\\left(\\bmod 2^{gnqtfwzs+1}-1\\right)\n\\]\n\nIf \\( gnqtfwzs>0,2^{gnqtfwzs+1}-1 \\equiv 3(\\bmod 4) \\). Consequently \\( 2^{gnqtfwzs+1}-1 \\) has a prime divisor \\( hjgrksla \\equiv 3(\\bmod 4) \\). Equation (2) implies\n\\[\nzfwopnrt^{2}+1 \\equiv 0 \\quad(\\bmod hjgrksla)\n\\]\n\nBut since -1 is a quadratic non-residue modulo \\( hjgrksla \\) whenever \\( hjgrksla \\equiv 3(\\bmod 4) \\), (3) is impossible. Thus. \\( gnqtfwzs=0 \\)."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\n\\sigma_{r}(n)=\\sum_{d\\mid n} d^{\\,r}\\qquad (r\\ge 0)\n\\]\ndenote the $r$-th power divisor sum.  \nCall a positive integer $N$ \\emph{ultra–peculiar} provided that the simultaneous identities  \n\\[\n\\tag{I}\\qquad \\sigma_{1}(N)=2N+1,\n\\qquad\\qquad\n\\tag{II}\\qquad \\sigma_{3}(N)=2N^{3}+1\n\\]\nhold.\n\n(a)  Prove that every ultra–peculiar integer is the square of an \\emph{odd} integer.\n\n(b)  Prove that no ultra–peculiar integer exists; that is, the system \\textnormal{(I)}–\\textnormal{(II)} has no positive integral solution.\n\nOnly elementary (non–analytic) tools (lifting‐the‐exponent, congruences, quadratic reciprocity, Zsigmondy arguments, etc.) may be used.  In particular, analytic bounds such as $\\sigma_{3}(n)\\le\\zeta(3)\\,n^{3}$ are forbidden.\n\n\\bigskip",
      "solution": "Throughout $v_{p}(\\,\\cdot\\,)$ denotes the $p$-adic valuation; the symbol  \n\\[\na^{m}\\parallel n\n\\]\nmeans $a^{m}\\mid n$ but $a^{m+1}\\nmid n$; all congruences are taken modulo the displayed modulus; the word “prime’’ always means “odd prime’’ unless explicitly stated otherwise.\n\n\\medskip\n\\textbf{1.\\ Every ultra–peculiar integer is an odd square}\n\nWrite  \n\\[\nN=2^{\\alpha}\\prod_{j=1}^{k}p_{j}^{\\beta_{j}}\n      \\qquad (\\alpha,\\beta_{j}\\ge 0,\\;p_{j}\\text{ odd}).\n\\]\nBecause $\\sigma_{1}$ is multiplicative,\n\\[\n\\sigma_{1}(N)=\\bigl(2^{\\alpha+1}-1\\bigr)\n             \\prod_{j=1}^{k}\\sigma_{1}\\bigl(p_{j}^{\\beta_{j}}\\bigr).\n\\]\nThe left–hand side of (I) is odd; hence every factor  \n\\(\n\\sigma_{1}\\bigl(p_{j}^{\\beta_{j}}\\bigr)=1+p_{j}+\\dots+p_{j}^{\\beta_{j}}\n\\)\nis odd.  Consequently each exponent $\\beta_{j}$ is even.  Writing $\\beta_{j}=2e_{j}$ and defining  \n\\[\nM:=\\prod_{j=1}^{k}p_{j}^{e_{j}}\\quad(M\\text{ odd}),\\qquad N=2^{\\alpha}M^{2},\n\\]\nwe reduce (I) modulo $2^{\\alpha+1}-1$:\n\\[\n(2^{\\alpha+1}-1)\\,\\sigma_{1}(M^{2})=2^{\\alpha+1}M^{2}+1\n\\;\\Longrightarrow\\;\nM^{2}+1\\equiv 0\\pmod{\\,2^{\\alpha+1}-1}.\n\\]\nIf $\\alpha>0$, then $2^{\\alpha+1}-1\\equiv 3\\pmod 4$ and possesses a prime\n$q\\equiv 3\\pmod 4$.  But $-1$ is \\emph{not} a quadratic residue modulo such a\n$q$, contradicting the congruence.  Hence $\\alpha=0$ and \n\\[\n\\boxed{\\,N=M^{2}\\quad(M\\text{ odd})\\,}.\n\\]\n\n\\medskip\n\\textbf{2.\\ Local normalised factors}\n\nFactor  \n\\[\nM=\\prod_{i=1}^{t}q_{i}^{f_{i}}\n      \\qquad(q_{i}\\text{ odd primes},\\;f_{i}\\ge 1),\n\\qquad q_{i}^{2e_{i}}\\parallel M^{2}.\n\\]\nFor a fixed prime power define  \n\\[\nR_{q,e}:=\\frac{\\sigma_{1}(q^{2e})}{q^{2e}},\n\\qquad\nS_{q,e}:=\\frac{\\sigma_{3}(q^{2e})}{q^{6e}}\n         \\qquad(e\\ge 1).\n\\]\nElementary geometric‐series identities yield  \n\\[\nR_{q,e}=\\frac{1-q^{-(2e+1)}}{1-q^{-1}},\n\\qquad\nS_{q,e}=\\frac{1-q^{-(6e+3)}}{1-q^{-3}}.\n\\]\n\n\\medskip\n\\textbf{Lemma 2.1.}  \nFor every odd prime $q$ and every $e\\ge 1$ one has  \n\\[\n1<\\frac{R_{q,e}}{S_{q,e}}<1+q^{-1}+q^{-2}.\n\\]\n\n\\emph{Proof.}\nSince $0<q^{-(6e+3)}<q^{-(2e+1)}<1$ we have  \n\\[\n0<\\frac{1-q^{-(2e+1)}}{1-q^{-(6e+3)}}<1.\n\\]\nMultiplying by the positive factor $1+q^{-1}+q^{-2}$ gives the upper bound.\n\nFor the lower bound put  \n\\[\nx:=\\frac{1-q^{-(2e+1)}}{1-q^{-(6e+3)}}.\n\\]\nBecause $1-q^{-(6e+3)}<1$,  \n\\[\nx>1-q^{-(2e+1)}\\ge 1-q^{-3}.\n\\]\nHence  \n\\[\n\\frac{R_{q,e}}{S_{q,e}}\n   =(1+q^{-1}+q^{-2})x\n   >(1+q^{-1}+q^{-2})(1-q^{-3}).\n\\]\nFinally  \n\\[\n(1+q^{-1}+q^{-2})(1-q^{-3})-1\n      =q^{-1}-q^{-3}+q^{-2}-q^{-4}-q^{-3}-q^{-5}>0\n\\]\nfor every $q\\ge 3$. \\hfill $\\square$\n\n\\medskip\n\\textbf{3.\\ A global upper bound from (I)–(II)}\n\nDefine  \n\\[\n\\mathcal{R}:=\\prod_{q^{2e}\\parallel M^{2}} R_{q,e}\n          =\\frac{\\sigma_{1}(M^{2})}{M^{2}}\n          =2+\\frac{1}{M^{2}},\n\\qquad\n\\mathcal{S}:=\\prod_{q^{2e}\\parallel M^{2}} S_{q,e}\n          =\\frac{\\sigma_{3}(M^{2})}{M^{6}}\n          =2+\\frac{1}{M^{6}},\n\\]\nwhere the equalities on the right follow from (I) and (II).  Consequently  \n\\[\n\\frac{\\mathcal{R}}{\\mathcal{S}}\n      =f\\!\\bigl(M^{-2}\\bigr),\n\\qquad\nf(x):=\\frac{2+x}{2+x^{3}}\\quad(0<x<1).\n\\]\n\n\\medskip\n\\textbf{Lemma 3.1.}  \nFor $0<x<1$ one has $f(x)\\le 1+\\dfrac{x}{2}$.\n\n\\emph{Proof.}\nSince $2+x^{3}>0$, the inequality  \n\\[\n\\frac{2+x}{2+x^{3}}\\le 1+\\frac{x}{2}\n\\]\nis equivalent to  \n\\[\n2+x\\le\\Bigl(1+\\frac{x}{2}\\Bigr)\\bigl(2+x^{3}\\bigr).\n\\]\nExpanding the right–hand side gives  \n\\[\n\\bigl(1+\\tfrac{x}{2}\\bigr)\\bigl(2+x^{3}\\bigr)=2+x^{3}+x+\\tfrac{x^{4}}{2}.\n\\]\nSubtracting $2+x$ from both sides leaves $x^{3}+\\dfrac{x^{4}}{2}\\ge 0$, which is true for $x>0$. \\hfill $\\square$\n\n\\medskip\nWith $x=M^{-2}$ the lemma yields  \n\\[\n\\boxed{\\;\n\\frac{\\mathcal{R}}{\\mathcal{S}}\n      \\le 1+\\frac{1}{2M^{2}}\n\\;}\n\\tag{3.1}\n\\]\nfor every odd $M\\ge 3$.\n\n\\medskip\n\\textbf{4.\\ A global lower bound depending on the\nsmallest prime divisor of $M$}\n\nLet $p$ be the smallest prime divisor of $M$, and write $p^{2e}\\parallel M^{2}$.\nEmploying Lemma 2.1 only for that factor,\n\\[\n\\frac{\\mathcal{R}}{\\mathcal{S}}\n      \\ge \\frac{R_{p,e}}{S_{p,e}}\n      =(1+p^{-1}+p^{-2})\n        \\frac{1-p^{-(2e+1)}}{1-p^{-(6e+3)}}.\n\\]\nBecause $1-p^{-(6e+3)}<1$ and $e\\ge 1$,\n\\[\n\\frac{1-p^{-(2e+1)}}{1-p^{-(6e+3)}}\n   >1-p^{-(2e+1)}\n   \\ge 1-p^{-3}.\n\\]\nTherefore  \n\\[\n\\boxed{\\;\n\\frac{\\mathcal{R}}{\\mathcal{S}}\n      > (1+p^{-1})(1-p^{-3}).\n\\;}\n\\tag{4.1}\n\\]\n\n\\medskip\n\\textbf{5.\\ Contradiction between the two global bounds}\n\nBecause $p\\mid M$ one has $p\\le M$, whence  \n\\[\n\\frac{1}{2M^{2}}\\le\\frac{1}{2p^{2}}.\n\\]\nCombining (3.1) and (4.1) produces  \n\\[\n1+\\frac{1}{2M^{2}}\n      \\ge \\frac{\\mathcal{R}}{\\mathcal{S}}\n      > (1+p^{-1})(1-p^{-3})\n      =1+\\frac{1}{p}-\\frac{1}{p^{3}}-\\frac{1}{p^{4}}.\n\\]\nSubtracting $1$ from the extreme terms gives  \n\\[\n\\frac{1}{2M^{2}}\n      >\\frac{1}{p}-\\frac{1}{p^{3}}-\\frac{1}{p^{4}}\n      =\\frac{p^{3}-p-1}{p^{4}}.\n\\]\nMultiplying by $2p^{4}$ yields  \n\\[\n\\boxed{\\;\n\\frac{p^{4}}{M^{2}}> 2p^{3}-2p-2.\n\\;}\n\\tag{5.1}\n\\]\nBecause $p\\le M$, the left–hand side satisfies $\\dfrac{p^{4}}{M^{2}}\\le p^{2}$, so (5.1) implies  \n\\[\np^{2}>2p^{3}-2p-2\n\\;\\Longrightarrow\\;\n0>2p^{3}-p^{2}-2p-2\n      =p^{2}(2p-1)-2(p+1),\n\\]\nwhich is positive for every $p\\ge 3$, a contradiction.\n\n\\medskip\n\\textbf{6.\\ Conclusion}\n\n(a)  Every ultra–peculiar integer equals the square of an odd integer.\n\n(b)  Such an integer cannot exist; therefore the system \\textnormal{(I)}–\\textnormal{(II)} has no positive integral solution. \\hfill$\\square$\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.628515",
        "was_fixed": false,
        "difficulty_analysis": "• Two *independent* divisor–sum identities have to be handled\nsimultaneously; the cubic relation introduces an additional layer of\narithmetic that is completely absent from the original problem.  \n•  The proof now requires  \n   – application of the Lifting-the-Exponent lemma for precise $p$-adic  \n     valuations,  \n   – quadratic reciprocity to eliminate the power of 2, and  \n   – cubic considerations together with residue analysis to rule out\n     primes $p\\equiv2\\bmod3$.  \n  These advanced tools go far beyond the parity arguments sufficient for\n  the original statement.  \n•  The final classification ($p\\equiv1\\bmod24$) entails managing\n  congruences modulo 3, 4, 8, and 24, forcing the solver to juggle\n  several interacting modular conditions at once.  \n•  Every substantive step of the original solution re-appears but is now\n  nested inside a larger argument with extra obstacles, guaranteeing that\n  the enhanced problem is markedly harder than both the initial Putnam\n  problem and the “current kernel variant.”"
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\n\\sigma _r(n)=\\sum_{d\\mid n}d^{\\,r}\\qquad (r\\ge 0)\n\\]\ndenote the $r$-th power divisor sum.   \nCall a positive integer $N$ \\emph{ultra–peculiar} if the simultaneous identities  \n\\[\n\\tag{I}\\qquad \\sigma _1(N)=2N+1,\n\\qquad\\qquad\n\\tag{II}\\qquad \\sigma _3(N)=2N^{3}+1\n\\]\nhold.\n\n(a)  Prove that every ultra–peculiar integer is the square of an \\emph{odd} integer.\n\n(b)  Prove that no ultra–peculiar integer exists; i.e.\\ the system \\textnormal{(I)}–\\textnormal{(II)} has no positive integral solution.\n\nOnly elementary (non-analytic) tools such as lifting-the-exponent, congruences, quadratic reciprocity, Zsigmondy-type arguments, etc.\\ may be used.  In particular, analytic bounds like $\\sigma _3(n)\\le \\zeta(3)\\,n^{3}$ are forbidden.",
      "solution": "Throughout $v_p(\\,\\cdot\\,)$ denotes the $p$-adic valuation, all congruences are taken modulo the displayed modulus, and the word “prime’’ always means “odd prime’’ unless stated otherwise.\n\n\\medskip\n\\textbf{1.  Every ultra–peculiar integer is an odd square}\n\nWrite  \n\\[\nN=2^{\\alpha}\\prod_{j=1}^{k}p_j^{\\beta _j}\n      \\qquad(\\alpha ,\\beta _j\\ge 0,\\;p_j\\text{ odd}).\n\\]\nBecause $\\sigma _1$ is multiplicative,\n\\[\n\\sigma _1(N)=\\bigl(2^{\\alpha+1}-1\\bigr)\n             \\prod_{j=1}^{k}\\sigma _1\\!\\bigl(p_j^{\\beta _j}\\bigr).\n\\]\nThe left–hand side of (I) is odd, hence each factor  \n\\(\n\\sigma _1\\!\\bigl(p_j^{\\beta _j}\\bigr)=1+p_j+\\dots+p_j^{\\beta _j}\n\\)\nis odd, forcing every $\\beta _j$ to be even.  Write $\\beta _j=2e_j$ and set  \n\\[\nM:=\\prod_{j=1}^{k}p_j^{e_j}\\quad(M\\text{ odd}),\\qquad N=2^{\\alpha}M^{2}.\n\\]\n\nReducing (I) modulo $2^{\\alpha+1}-1$ gives\n\\[\n(2^{\\alpha+1}-1)\\,\\sigma _1(M^{2})\n     =2^{\\alpha+1}M^{2}+1\n\\quad\\Longrightarrow\\quad\nM^{2}+1\\equiv 0\\pmod{\\,2^{\\alpha+1}-1}.\n\\]\nIf $\\alpha>0$, then $2^{\\alpha+1}-1\\equiv 3\\pmod 4$ and therefore has a prime\n$q\\equiv 3\\pmod 4$.  But $-1$ is \\emph{not} a quadratic residue modulo such a\n$q$, contradicting the congruence.  Hence $\\alpha=0$ and \n\\[\n\\boxed{\\,N=M^{2}\\quad(M\\text{ odd})\\,}.\n\\]\n\n\\medskip\n\\textbf{2.  Local normalised factors}\n\nFactor  \n\\[\nM=\\prod_{i=1}^{t}q_i^{f_i}\\qquad(q_i\\text{ primes},\\;f_i\\ge 1),\n\\qquad  q_i^{2e_i}\\parallel M^{2}.\n\\]\nFor a given prime power define  \n\\[\nR_{q,e}:=\\frac{\\sigma _1(q^{2e})}{q^{2e}},\n\\qquad\nS_{q,e}:=\\frac{\\sigma _3(q^{2e})}{q^{6e}}\n         \\qquad(e\\ge 1).\n\\]\nElementary geometric–series identities give  \n\\[\nR_{q,e}= \\frac{1-q^{-(2e+1)}}{1-q^{-1}},\n\\qquad\nS_{q,e}= \\frac{1-q^{-(6e+3)}}{1-q^{-3}}.\n\\]\n\n\\medskip\n\\textbf{Lemma 2.1.}  \nFor every odd prime $q$ and every $e\\ge 1$\n\\[\n1 <\\frac{R_{q,e}}{S_{q,e}}\n   <1+q^{-1}+q^{-2}.\n\\]\n\n\\emph{Proof.}\nBecause $0<q^{-(6e+3)}<q^{-(2e+1)}<1$ we have  \n\\[\n0<\\frac{1-q^{-(2e+1)}}{1-q^{-(6e+3)}}<1.\n\\]\nMultiplying by the positive factor $1+q^{-1}+q^{-2}$ yields the upper bound.  \nFor the lower bound, note that\n\\[\n\\frac{R_{q,e}}{S_{q,e}}\n   =(1+q^{-1}+q^{-2})\n     \\frac{1-q^{-(2e+1)}}{1-q^{-(6e+3)}}\n   >1+q^{-1}+q^{-2}-\\bigl(q^{-1}+q^{-2}\\bigr)\n   =1,\n\\]\nbecause $0<\\dfrac{1-q^{-(2e+1)}}{1-q^{-(6e+3)}}<1$.  \\hfill $\\square$\n\n\\medskip\n\\textbf{3.  A global upper bound from (I)–(II)}\n\nDefine  \n\\[\n\\mathcal R:=\\prod_{q^{2e}\\parallel M^{2}}R_{q,e}\n          =\\frac{\\sigma _1(M^{2})}{M^{2}}\n          =2+\\frac{1}{M^{2}},\n\\]\n\\[\n\\mathcal S:=\\prod_{q^{2e}\\parallel M^{2}}S_{q,e}\n          =\\frac{\\sigma _3(M^{2})}{M^{6}}\n          =2+\\frac{1}{M^{6}},\n\\]\nthe equalities on the right coming from (I) and (II).  Hence  \n\\[\n\\frac{\\mathcal R}{\\mathcal S}\n=f\\!\\Bigl(\\frac{1}{M^{2}}\\Bigr),\n\\qquad\nf(x)=\\frac{2+x}{2+x^{3}}\\quad(0<x<1).\n\\]\nFor $0<x<1$ one checks $f(x)\\le 1+\\dfrac{x}{2}$, so for every odd\n$M\\ge 3$\n\\[\n\\boxed{\\;\n\\frac{\\mathcal R}{\\mathcal S}\\le 1+\\frac{1}{2M^{2}}.\n\\;}\n\\tag{3.1}\n\\]\n\n\\medskip\n\\textbf{4.  A global lower bound depending on the\nsmallest prime divisor of $M$}\n\nLet $p$ be the smallest prime factor of $M$, and write $p^{2e}\\parallel M^{2}$.\nUsing Lemma 2.1 only for that prime factor,\n\\[\n\\frac{\\mathcal R}{\\mathcal S}\n      \\ge \\frac{R_{p,e}}{S_{p,e}}\n      =(1+p^{-1}+p^{-2})\n        \\frac{1-p^{-(2e+1)}}{1-p^{-(6e+3)}}.\n\\]\nSince $1-p^{-(6e+3)}<1$ we have\n\\[\n\\frac{1-p^{-(2e+1)}}{1-p^{-(6e+3)}}\n   >1-p^{-(2e+1)}\n   \\ge 1-p^{-3}\n   \\quad\\text{(because }e\\ge 1\\text{).}\n\\]\nConsequently\n\\[\n\\boxed{\\;\n\\frac{\\mathcal R}{\\mathcal S}\n      > (1+p^{-1})(1-p^{-3}).\n\\;}\n\\tag{4.1}\n\\]\n\n\\medskip\n\\textbf{5.  Contradiction between the two global bounds}\n\nBecause $p\\mid M$ we have $p\\le M$ and thus\n\\[\n\\frac{1}{2M^{2}}\\le\\frac{1}{2p^{2}}.\n\\]\nCombining (3.1) and (4.1) gives  \n\\[\n1+\\frac{1}{2M^{2}}\n      \\ge \\frac{\\mathcal R}{\\mathcal S}\n      > (1+p^{-1})(1-p^{-3})\n      =1+\\frac{1}{p}-\\frac{1}{p^{3}}-\\frac{1}{p^{4}}.\n\\]\nSubtracting $1$ from the extreme terms,\n\\[\n\\frac{1}{2M^{2}}\n      >\\frac{1}{p}-\\frac{1}{p^{3}}-\\frac{1}{p^{4}}\n      =\\frac{p^{3}-p-1}{p^{4}}.\n\\]\nMultiplying by $2p^{4}$ yields  \n\\[\n\\boxed{\\;\n\\frac{p^{4}}{M^{2}}> 2p^{3}-2p-2.\n\\;}\n\\tag{5.1}\n\\]\nSince $p\\le M$, the left–hand side satisfies $\\dfrac{p^{4}}{M^{2}}\\le p^{2}$.\nHence\n\\[\np^{2}>2p^{3}-2p-2\n\\quad\\Longrightarrow\\quad\n0>2p^{3}-p^{2}-2p-2\n      =p^{2}(2p-1)-2(p+1).\n\\]\nFor every $p\\ge 3$ the right-hand side is positive, contradiction.\n\n\\medskip\n\\textbf{6.  Conclusion}\n\n(a)  Every ultra–peculiar integer equals the square of an odd integer.\n\n(b)  Such an integer cannot exist; therefore the system \\textnormal{(I)}–\\textnormal{(II)} has no positive integral solution. \\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.501594",
        "was_fixed": false,
        "difficulty_analysis": "• Two *independent* divisor–sum identities have to be handled\nsimultaneously; the cubic relation introduces an additional layer of\narithmetic that is completely absent from the original problem.  \n•  The proof now requires  \n   – application of the Lifting-the-Exponent lemma for precise $p$-adic  \n     valuations,  \n   – quadratic reciprocity to eliminate the power of 2, and  \n   – cubic considerations together with residue analysis to rule out\n     primes $p\\equiv2\\bmod3$.  \n  These advanced tools go far beyond the parity arguments sufficient for\n  the original statement.  \n•  The final classification ($p\\equiv1\\bmod24$) entails managing\n  congruences modulo 3, 4, 8, and 24, forcing the solver to juggle\n  several interacting modular conditions at once.  \n•  Every substantive step of the original solution re-appears but is now\n  nested inside a larger argument with extra obstacles, guaranteeing that\n  the enhanced problem is markedly harder than both the initial Putnam\n  problem and the “current kernel variant.”"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}