{
  "index": "2010-B-1",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Is there an infinite sequence of real numbers $a_1, a_2, a_3, \\dots$ such that\n\\[\na_1^m + a_2^m + a_3^m + \\cdots = m\n\\]\nfor every positive integer $m$?",
  "solution": "\\textbf{First solution.}\nNo such sequence exists. If it did, then the Cauchy-Schwartz inequality would imply\n\\begin{align*}\n8 &= (a_1^2 + a_2^2 +  \\cdots)(a_1^4 + a_2^4 + \\cdots) \\\\\n&\\geq (a_1^3 + a_2^3 + \\cdots)^2 = 9,\n\\end{align*}\ncontradiction.\n\n\\textbf{Second solution.}\n(Communicated by Catalin Zara.)\nSuppose that such a sequence exists.\nIf $a_k^2 \\in [0,1]$ for all $k$, then $a_k^4 \\leq a_k^2$ for all $k$, and so\n\\[\n4 = a_1^4 + a_2^4 + \\cdots \\leq a_1^2 + a_2^2 + \\cdots = 2,\n\\]\ncontradiction. There thus exists a positive integer $k$ for which $a_k^2 \\geq 1$.\nHowever, in this case, for $m$ large, $a_k^{2m} > 2m$ and so\n$a_1^{2m} + a_2^{2m} + \\cdots \\neq 2m$.\n\n\\textbf{Third solution.}\nWe generalize the second solution to show that for any positive integer $k$, it is impossible for a sequence\n$a_1, a_2,\\dots$ of complex numbers to satisfy the given conditions in case\nthe series $a_1^k + a_2^k + \\cdots$ converges absolutely. This includes the original problem by taking\n$k=2$, in which case the series $a_1^2 + a_2^2 + \\cdots$ consists of nonnegative\nreal numbers and so converges absolutely if it converges at all.\n\nSince the sum $\\sum_{i=1}^\\infty |a_i|^k$ converges by hypothesis, we can find a positive integer $n$\nsuch that $\\sum_{i=n+1}^\\infty |a_i|^k < 1$. For each positive integer $d$, we then have\n\\[\n\\left|kd - \\sum_{i=1}^n a_i^{kd} \\right|\n\\leq \\sum_{i=n+1}^\\infty |a_i|^{kd} < 1.\n\\]\nWe thus cannot have $|a_1|,\\dots,|a_n| \\leq 1$, or else the sum $\\sum_{i=1}^n a_i^{kd}$ would be bounded\nin absolute value by $n$ independently of $d$. But if we put $r = \\max\\{|a_1|,\\dots,|a_n|\\} > 1$, we\nobtain another contradiction because for any $\\epsilon > 0$,\n\\[\n\\limsup_{d \\to \\infty} (r-\\epsilon)^{-kd} \\left| \\sum_{i=1}^n a_i^{kd} \\right| > 0.\n\\]\nFor instance, this follows from applying the root test to the rational function\n\\[\n\\sum_{i=1}^n \\frac{1}{1 - a_i^k z} = \\sum_{d=0}^\\infty \\left( \\sum_{i=1}^n a_i^{kd} \\right) z^d,\n\\]\nwhich has a pole within the circle $|z| \\leq r^{-1/k}$.\n(An elementary proof is also possible.)\n\n\\textbf{Fourth solution.}\n(Communicated by Noam Elkies.)\nSince $\\sum_k a_k^2 = 2$, for each positive integer $k$ we have $a_k^2 \\leq 2$ and so $a_k^4 \\leq 2 a_k^2$,\nwith equality only for $a_k^2 \\in \\{0,2\\}$. Thus to have $\\sum_k a_k^4 = 4$, there must be a single index\n$k$ for which $a_k^2 = 2$, and the other $a_k$ must all equal 0. But then $\\sum_k a_k^{2m} = 2^m \\neq 2m$\nfor any positive integer $m>2$.\n\n\\textbf{Remark.} Manjul Bhargava points out it is easy to construct sequences of complex numbers with the\ndesired property if we drop the condition of absolute convergence. Here is an inductive construction\n(of which several variants are possible).\nFor $n=1,2,\\dots$ and $z \\in \\CC$, define the finite sequence\n\\[\ns_{n,z} = \\left( \\frac{1}{z} e^{2 \\pi i j/n}: j = 0, \\dots, n-1 \\right).\n\\]\nThis sequence has the property that for any positive integer $j$, the sum of the $j$-th powers\nof the terms of $s_{n,z}$ equals $1/z^{j}$ if $j$ is divisible by $n$ and 0 otherwise.\nMoreover, any partial sum of $j$-th powers is bounded in absolute value by $n/|z|^j$.\n\nThe desired sequence will be constructed as follows. Suppose that we have a finite sequence\nwhich has the correct sum of $j$-th powers for $j=1,\\dots,m$. (For instance, for $m=1$,\nwe may start with the singleton sequence 1.) We may then extend it to a new sequence which has\nthe correct sum of $j$-th powers for $j=1,\\dots,m+1$, by appending $k$ copies of $s_{m+1,z}$\nfor suitable choices of a positive integer $k$ and a complex number $z$ with $|z| < m^{-2}$.\nThis last restriction ensures that the resulting infinite sequence $a_1,a_2,\\dots$ is such that\nfor each positive integer $m$, the series $a_1^m + a_2^m + \\cdots$ is convergent (though not absolutely\nconvergent). Its partial sums include a subsequence equal to the constant value $m$, so the sum of the series\nmust equal $m$ as desired.",
  "vars": [
    "a_1",
    "a_2",
    "a_3",
    "a_k",
    "a_i",
    "a_n",
    "m",
    "k",
    "n",
    "d",
    "r",
    "z",
    "j",
    "\\\\epsilon"
  ],
  "params": [],
  "sci_consts": [
    "e",
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "a_1": "firstseq",
        "a_2": "secondseq",
        "a_3": "thirdseq",
        "a_k": "kthseq",
        "a_i": "iseqelem",
        "a_n": "nseqelem",
        "m": "powerint",
        "k": "smallintk",
        "n": "smallintn",
        "d": "smallintd",
        "r": "radiusr",
        "z": "complexz",
        "j": "intindexj",
        "\\epsilon": "epsilonv"
      },
      "question": "Is there an infinite sequence of real numbers $firstseq, secondseq, thirdseq, \\dots$ such that\n\\[\nfirstseq^{powerint} + secondseq^{powerint} + thirdseq^{powerint} + \\cdots = powerint\n\\]\nfor every positive integer $powerint$?",
      "solution": "\\textbf{First solution.}\nNo such sequence exists. If it did, then the Cauchy\\textendash Schwarz inequality would imply\n\\begin{align*}\n8 &= (firstseq^2 + secondseq^2 +  \\cdots)(firstseq^4 + secondseq^4 + \\cdots) \\\\\n&\\geq (firstseq^3 + secondseq^3 + \\cdots)^2 = 9,\n\\end{align*}\ncontradiction.\n\n\\textbf{Second solution.}\n(Communicated by Catalin Zara.)\nSuppose that such a sequence exists.\nIf $kthseq^2 \\in [0,1]$ for all smallintk, then $kthseq^4 \\leq kthseq^2$ for all smallintk, and so\n\\[\n4 = firstseq^4 + secondseq^4 + \\cdots \\leq firstseq^2 + secondseq^2 + \\cdots = 2,\n\\]\ncontradiction. There thus exists a positive integer smallintk for which $kthseq^2 \\geq 1$.\nHowever, in this case, for powerint large, $kthseq^{2\\,powerint} > 2\\,powerint$ and so\n$firstseq^{2\\,powerint} + secondseq^{2\\,powerint} + \\cdots \\neq 2\\,powerint$.\n\n\\textbf{Third solution.}\nWe generalize the second solution to show that for any positive integer $smallintk$, it is impossible for a sequence\nfirstseq, secondseq,\\dots\\ of complex numbers to satisfy the given conditions in case\nthe series $firstseq^{smallintk} + secondseq^{smallintk} + \\cdots$ converges absolutely. This includes the original problem by taking\n$smallintk = 2$, in which case the series $firstseq^{2} + secondseq^{2} + \\cdots$ consists of nonnegative\nreal numbers and so converges absolutely if it converges at all.\n\nSince the sum $\\sum_{i=1}^{\\infty} |iseqelem|^{smallintk}$ converges by hypothesis, we can find a positive integer $smallintn$\nsuch that $\\sum_{i=smallintn+1}^{\\infty} |iseqelem|^{smallintk} < 1$. For each positive integer $smallintd$, we then have\n\\[\n\\left|\\,smallintk\\,smallintd - \\sum_{i=1}^{smallintn} iseqelem^{\\,{smallintk\\,smallintd}} \\right|\n\\leq \\sum_{i=smallintn+1}^{\\infty} |iseqelem|^{\\,{smallintk\\,smallintd}} < 1.\n\\]\nWe thus cannot have $|firstseq|,\\dots,|nseqelem| \\leq 1$, or else the sum $\\sum_{i=1}^{smallintn} iseqelem^{\\,{smallintk\\,smallintd}}$ would be bounded\nin absolute value by $smallintn$ independently of $smallintd$. But if we put $radiusr = \\max\\{|firstseq|,\\dots,|nseqelem|\\} > 1$, we\nobtain another contradiction because for any $epsilonv > 0$,\n\\[\n\\limsup_{smallintd \\to \\infty} (radiusr - epsilonv)^{-{smallintk\\,smallintd}} \\left| \\sum_{i=1}^{smallintn} iseqelem^{\\,{smallintk\\,smallintd}} \\right| > 0.\n\\]\nFor instance, this follows from applying the root test to the rational function\n\\[\n\\sum_{i=1}^{smallintn} \\frac{1}{1 - iseqelem^{smallintk} complexz} = \\sum_{smallintd=0}^{\\infty} \\left( \\sum_{i=1}^{smallintn} iseqelem^{\\,{smallintk\\,smallintd}} \\right) complexz^{smallintd},\n\\]\nwhich has a pole within the circle $|complexz| \\leq radiusr^{-1/smallintk}$.\n(An elementary proof is also possible.)\n\n\\textbf{Fourth solution.}\n(Communicated by Noam Elkies.)\nSince $\\sum_{smallintk} kthseq^2 = 2$, for each positive integer $smallintk$ we have $kthseq^2 \\leq 2$ and so $kthseq^4 \\leq 2\\,kthseq^2$,\nwith equality only for $kthseq^2 \\in \\{0,2\\}$. Thus to have $\\sum_{smallintk} kthseq^4 = 4$, there must be a single index\n$smallintk$ for which $kthseq^2 = 2$, and the other $kthseq$ must all equal 0. But then $\\sum_{smallintk} kthseq^{2\\,powerint} = 2^{powerint} \\neq 2\\,powerint$\nfor any positive integer $powerint > 2$.\n\n\\textbf{Remark.} Manjul Bhargava points out it is easy to construct sequences of complex numbers with the\ndesired property if we drop the condition of absolute convergence. Here is an inductive construction\n(of which several variants are possible).\nFor $smallintn = 1,2,\\dots$ and $complexz \\in \\CC$, define the finite sequence\n\\[\ns_{smallintn,complexz} = \\left( \\frac{1}{complexz} e^{2 \\pi i \\; intindexj/ smallintn}: intindexj = 0, \\dots, smallintn-1 \\right).\n\\]\nThis sequence has the property that for any positive integer intindexj, the sum of the intindexj-th powers\nof the terms of $s_{smallintn,complexz}$ equals $1/complexz^{intindexj}$ if intindexj is divisible by smallintn and 0 otherwise.\nMoreover, any partial sum of intindexj-th powers is bounded in absolute value by $smallintn/|complexz|^{intindexj}$.\n\nThe desired sequence will be constructed as follows. Suppose that we have a finite sequence\nwhich has the correct sum of intindexj-th powers for intindexj = 1,\\dots,powerint. (For instance, for powerint = 1,\nwe may start with the singleton sequence 1.) We may then extend it to a new sequence which has\nthe correct sum of intindexj-th powers for intindexj = 1,\\dots,powerint+1, by appending smallintk copies of $s_{powerint+1,complexz}$\nfor suitable choices of a positive integer smallintk and a complex number complexz with $|complexz| < powerint^{-2}$.\nThis last restriction ensures that the resulting infinite sequence firstseq, secondseq,\\dots\\ is such that\nfor each positive integer powerint, the series firstseq^{powerint} + secondseq^{powerint} + \\cdots is convergent (though not absolutely\nconvergent). Its partial sums include a subsequence equal to the constant value powerint, so the sum of the series\nmust equal powerint as desired."
    },
    "descriptive_long_confusing": {
      "map": {
        "a_1": "alabaster",
        "a_2": "blueberry",
        "a_3": "canaryway",
        "a_k": "doughnutt",
        "a_i": "elephanty",
        "a_n": "fernweary",
        "m": "marigolds",
        "k": "kangaroos",
        "n": "nightowls",
        "d": "daffodils",
        "r": "raspberry",
        "z": "zephyrus",
        "j": "jellyfish",
        "\\epsilon": "sugarcane"
      },
      "question": "Is there an infinite sequence of real numbers $alabaster, blueberry, canaryway, \\dots$ such that\n\\[\nalabaster^{marigolds} + blueberry^{marigolds} + canaryway^{marigolds} + \\cdots = marigolds\n\\]\nfor every positive integer $marigolds$?",
      "solution": "\\textbf{First solution.}\nNo such sequence exists. If it did, then the Cauchy–Schwarz inequality would imply\n\\begin{align*}\n8 &= (alabaster^2 + blueberry^2 +  \\cdots)(alabaster^4 + blueberry^4 + \\cdots) \\\\\n&\\geq (alabaster^3 + blueberry^3 + \\cdots)^2 = 9,\n\\end{align*}\ncontradiction.\n\n\\textbf{Second solution.}\n(Communicated by Catalin Zara.)\nSuppose that such a sequence exists.\nIf $doughnutt^2 \\in [0,1]$ for all $kangaroos$, then $doughnutt^4 \\leq doughnutt^2$ for all $kangaroos$, and so\n\\[\n4 = alabaster^4 + blueberry^4 + \\cdots \\leq alabaster^2 + blueberry^2 + \\cdots = 2,\n\\]\ncontradiction. There thus exists a positive integer $kangaroos$ for which $doughnutt^2 \\geq 1$.\nHowever, in this case, for $marigolds$ large, $doughnutt^{2 marigolds} > 2 marigolds$ and so\n$alabaster^{2 marigolds} + blueberry^{2 marigolds} + \\cdots \\neq 2 marigolds$.\n\n\\textbf{Third solution.}\nWe generalize the second solution to show that for any positive integer $kangaroos$, it is impossible for a sequence\n$alabaster, blueberry,\\dots$ of complex numbers to satisfy the given conditions in case\nthe series $alabaster^{kangaroos} + blueberry^{kangaroos} + \\cdots$ converges absolutely. This includes the original problem by taking\n$kangaroos=2$, in which case the series $alabaster^2 + blueberry^2 + \\cdots$ consists of nonnegative\nreal numbers and so converges absolutely if it converges at all.\n\nSince the sum $\\sum_{i=1}^\\infty |elephanty|^{kangaroos}$ converges by hypothesis, we can find a positive integer $nightowls$\nsuch that $\\sum_{i=nightowls+1}^\\infty |elephanty|^{kangaroos} < 1$. For each positive integer $daffodils$, we then have\n\\[\n\\left|kangaroos\\, daffodils - \\sum_{i=1}^{nightowls} elephanty^{\\, kangaroos\\, daffodils} \\right|\n\\leq \\sum_{i=nightowls+1}^\\infty |elephanty|^{\\, kangaroos\\, daffodils} < 1.\n\\]\nWe thus cannot have $|alabaster|,\\dots,|fernweary| \\leq 1$, or else the sum $\\sum_{i=1}^{nightowls} elephanty^{\\, kangaroos\\, daffodils}$ would be bounded\nin absolute value by $nightowls$ independently of $daffodils$. But if we put $raspberry = \\max\\{|alabaster|,\\dots,|fernweary|\\} > 1$, we\nobtain another contradiction because for any $sugarcane > 0$,\n\\[\n\\limsup_{daffodils \\to \\infty} (raspberry-sugarcane)^{-\\, kangaroos\\, daffodils} \\left| \\sum_{i=1}^{nightowls} elephanty^{\\, kangaroos\\, daffodils} \\right| > 0.\n\\]\nFor instance, this follows from applying the root test to the rational function\n\\[\n\\sum_{i=1}^{nightowls} \\frac{1}{1 - elephanty^{\\, kangaroos} zephyrus} = \\sum_{daffodils=0}^\\infty \\left( \\sum_{i=1}^{nightowls} elephanty^{\\, kangaroos\\, daffodils} \\right) zephyrus^{daffodils},\n\\]\nwhich has a pole within the circle $|zephyrus| \\leq raspberry^{-1/kangaroos}$.\n(An elementary proof is also possible.)\n\n\\textbf{Fourth solution.}\n(Communicated by Noam Elkies.)\nSince $\\sum_{kangaroos} doughnutt^2 = 2$, for each positive integer $kangaroos$ we have $doughnutt^2 \\leq 2$ and so $doughnutt^4 \\leq 2 doughnutt^2$,\nwith equality only for $doughnutt^2 \\in \\{0,2\\}$. Thus to have $\\sum_{kangaroos} doughnutt^4 = 4$, there must be a single index\n$kangaroos$ for which $doughnutt^2 = 2$, and the other $doughnutt$ must all equal 0. But then $\\sum_{kangaroos} doughnutt^{2 marigolds} = 2^{marigolds} \\neq 2 marigolds$\nfor any positive integer $marigolds>2$.\n\n\\textbf{Remark.} Manjul Bhargava points out it is easy to construct sequences of complex numbers with the\ndesired property if we drop the condition of absolute convergence. Here is an inductive construction\n(of which several variants are possible).\nFor $nightowls=1,2,\\dots$ and $zephyrus \\in \\CC$, define the finite sequence\n\\[\ns_{nightowls,zephyrus} = \\left( \\frac{1}{\\zephyrus} e^{2 \\pi i jellyfish/nightowls}: jellyfish = 0, \\dots, nightowls-1 \\right).\n\\]\nThis sequence has the property that for any positive integer $jellyfish$, the sum of the $jellyfish$-th powers\nof the terms of $s_{nightowls,zephyrus}$ equals $1/\\zephyrus^{jellyfish}$ if $jellyfish$ is divisible by $nightowls$ and 0 otherwise.\nMoreover, any partial sum of $jellyfish$-th powers is bounded in absolute value by $nightowls/|\\zephyrus|^{jellyfish}$.\n\nThe desired sequence will be constructed as follows. Suppose that we have a finite sequence\nwhich has the correct sum of $jellyfish$-th powers for $jellyfish=1,\\dots,marigolds$. (For instance, for $marigolds=1$,\nwe may start with the singleton sequence 1.) We may then extend it to a new sequence which has\nthe correct sum of $jellyfish$-th powers for $jellyfish=1,\\dots,marigolds+1$, by appending $kangaroos$ copies of $s_{marigolds+1,\\zephyrus}$\nfor suitable choices of a positive integer $kangaroos$ and a complex number $\\zephyrus$ with $|\\zephyrus| < marigolds^{-2}$.\nThis last restriction ensures that the resulting infinite sequence $alabaster, blueberry,\\dots$ is such that\nfor each positive integer $marigolds$, the series $alabaster^{marigolds} + blueberry^{marigolds} + \\cdots$ is convergent (though not absolutely\nconvergent). Its partial sums include a subsequence equal to the constant value $marigolds$, so the sum of the series\nmust equal $marigolds$ as desired."
    },
    "descriptive_long_misleading": {
      "map": {
        "a_1": "constantone",
        "a_2": "constanttwo",
        "a_3": "constantthree",
        "a_k": "constantindex",
        "a_i": "constantloop",
        "a_n": "constantlimit",
        "m": "fixedpower",
        "k": "fixedcount",
        "n": "fixedsize",
        "d": "fixedstep",
        "r": "minvalue",
        "z": "realnumber",
        "j": "outerindex",
        "\\epsilon": "hugetolerance"
      },
      "question": "Is there an infinite sequence of real numbers $constantone, constanttwo, constantthree, \\dots$ such that\n\\[\nconstantone^{fixedpower} + constanttwo^{fixedpower} + constantthree^{fixedpower} + \\cdots = fixedpower\n\\]\nfor every positive integer $fixedpower$?",
      "solution": "\\textbf{First solution.}\nNo such sequence exists. If it did, then the Cauchy--Schwarz inequality would imply\n\\begin{align*}\n8 &= (constantone^{2} + constanttwo^{2} +  \\cdots)(constantone^{4} + constanttwo^{4} + \\cdots) \\\\\n&\\geq (constantone^{3} + constanttwo^{3} + \\cdots)^2 = 9,\n\\end{align*}\ncontradiction.\n\n\\textbf{Second solution.}\n(Communicated by Catalin Zara.)\nSuppose that such a sequence exists.\nIf $constantindex^{2} \\in [0,1]$ for all $fixedcount$, then $constantindex^{4} \\leq constantindex^{2}$ for all $fixedcount$, and so\n\\[\n4 = constantone^{4} + constanttwo^{4} + \\cdots \\leq constantone^{2} + constanttwo^{2} + \\cdots = 2,\n\\]\ncontradiction. There thus exists a positive integer $fixedcount$ for which $constantindex^{2} \\geq 1$.\nHowever, in this case, for $fixedpower$ large, $constantindex^{2\\,fixedpower} > 2\\,fixedpower$ and so\n$constantone^{2\\,fixedpower} + constanttwo^{2\\,fixedpower} + \\cdots \\neq 2\\,fixedpower$.\n\n\\textbf{Third solution.}\nWe generalize the second solution to show that for any positive integer $fixedcount$, it is impossible for a sequence\n$constantone, constanttwo,\\dots$ of complex numbers to satisfy the given conditions in case\nthe series $constantone^{fixedcount} + constanttwo^{fixedcount} + \\cdots$ converges absolutely. This includes the original problem by taking\n$fixedcount=2$, in which case the series $constantone^{2} + constanttwo^{2} + \\cdots$ consists of nonnegative\nreal numbers and so converges absolutely if it converges at all.\n\nSince the sum $\\sum_{i=1}^\\infty |constantloop|^{fixedcount}$ converges by hypothesis, we can find a positive integer $fixedsize$\nsuch that $\\sum_{i=fixedsize+1}^\\infty |constantloop|^{fixedcount} < 1$. For each positive integer $fixedstep$, we then have\n\\[\n\\left|fixedcount\\,fixedstep - \\sum_{i=1}^{fixedsize} constantloop^{fixedcount\\,fixedstep} \\right|\n\\leq \\sum_{i=fixedsize+1}^\\infty |constantloop|^{fixedcount\\,fixedstep} < 1.\n\\]\nWe thus cannot have $|constantone|,\\dots,|constantlimit| \\leq 1$, or else the sum $\\sum_{i=1}^{fixedsize} constantloop^{fixedcount\\,fixedstep}$ would be bounded\nin absolute value by $fixedsize$ independently of $fixedstep$. But if we put $minvalue = \\max\\{|constantone|,\\dots,|constantlimit|\\} > 1$, we\nobtain another contradiction because for any $hugetolerance > 0$,\n\\[\n\\limsup_{fixedstep \\to \\infty} (minvalue-hugetolerance)^{-fixedcount\\,fixedstep} \\left| \\sum_{i=1}^{fixedsize} constantloop^{fixedcount\\,fixedstep} \\right| > 0.\n\\]\nFor instance, this follows from applying the root test to the rational function\n\\[\n\\sum_{i=1}^{fixedsize} \\frac{1}{1 - constantloop^{fixedcount} realnumber} = \\sum_{fixedstep=0}^\\infty \\left( \\sum_{i=1}^{fixedsize} constantloop^{fixedcount\\,fixedstep} \\right) realnumber^{fixedstep},\n\\]\nwhich has a pole within the circle $|realnumber| \\leq minvalue^{-1/fixedcount}$.\n(An elementary proof is also possible.)\n\n\\textbf{Fourth solution.}\n(Communicated by Noam Elkies.)\nSince $\\sum_{fixedcount} constantindex^{2} = 2$, for each positive integer $fixedcount$ we have $constantindex^{2} \\leq 2$ and so $constantindex^{4} \\leq 2\\,constantindex^{2}$,\nwith equality only for $constantindex^{2} \\in \\{0,2\\}$. Thus to have $\\sum_{fixedcount} constantindex^{4} = 4$, there must be a single index\n$fixedcount$ for which $constantindex^{2} = 2$, and the other $constantindex$ must all equal 0. But then $\\sum_{fixedcount} constantindex^{2\\,fixedpower} = 2^{fixedpower} \\neq 2\\,fixedpower$\nfor any positive integer $fixedpower>2$.\n\n\\textbf{Remark.} Manjul Bhargava points out it is easy to construct sequences of complex numbers with the\ndesired property if we drop the condition of absolute convergence. Here is an inductive construction\n(of which several variants are possible).\nFor $fixedsize=1,2,\\dots$ and $realnumber \\in \\CC$, define the finite sequence\n\\[\ns_{fixedsize,realnumber} = \\left( \\frac{1}{realnumber} e^{2 \\pi i \\, outerindex/fixedsize}: outerindex = 0, \\dots, fixedsize-1 \\right).\n\\]\nThis sequence has the property that for any positive integer $outerindex$, the sum of the $outerindex$-th powers\nof the terms of $s_{fixedsize,realnumber}$ equals $1/realnumber^{outerindex}$ if $outerindex$ is divisible by $fixedsize$ and 0 otherwise.\nMoreover, any partial sum of $outerindex$-th powers is bounded in absolute value by $fixedsize/|realnumber|^{outerindex}$.\n\nThe desired sequence will be constructed as follows. Suppose that we have a finite sequence\nwhich has the correct sum of $outerindex$-th powers for $outerindex=1,\\dots,fixedpower$. (For instance, for $fixedpower=1$, we may start with the singleton sequence 1.) We may then extend it to a new sequence which has\nthe correct sum of $outerindex$-th powers for $outerindex=1,\\dots,fixedpower+1$, by appending $fixedcount$ copies of $s_{fixedpower+1,realnumber}$\nfor suitable choices of a positive integer $fixedcount$ and a complex number $realnumber$ with $|realnumber| < fixedpower^{-2}$.\nThis last restriction ensures that the resulting infinite sequence $constantone,constanttwo,\\dots$ is such that\nfor each positive integer $fixedpower$, the series $constantone^{fixedpower} + constanttwo^{fixedpower} + \\cdots$ is convergent (though not absolutely\nconvergent). Its partial sums include a subsequence equal to the constant value $fixedpower$, so the sum of the series\nmust equal $fixedpower$ as desired."
    },
    "garbled_string": {
      "map": {
        "a_1": "wjkvbnqz",
        "a_2": "xbqlrtmc",
        "a_3": "pzmdsvfk",
        "a_k": "gnthwczr",
        "a_i": "vycfgjla",
        "a_n": "sdrkhpqm",
        "m": "lqtsbovw",
        "k": "hzrwdnce",
        "n": "qjpvmsor",
        "d": "btgkclhf",
        "r": "ysnwdxvu",
        "z": "kjflqzrb",
        "j": "vtgnscrh",
        "\\epsilon": "tqhrvmed"
      },
      "question": "Is there an infinite sequence of real numbers $wjkvbnqz, xbqlrtmc, pzmdsvfk, \\dots$ such that\n\\[\nwjkvbnqz^{lqtsbovw} + xbqlrtmc^{lqtsbovw} + pzmdsvfk^{lqtsbovw} + \\cdots = lqtsbovw\n\\]\nfor every positive integer $lqtsbovw$?",
      "solution": "\\textbf{First solution.}\nNo such sequence exists. If it did, then the Cauchy--Schwartz inequality would imply\n\\begin{align*}\n8 &= (wjkvbnqz^2 + xbqlrtmc^2 +  \\cdots)(wjkvbnqz^4 + xbqlrtmc^4 + \\cdots) \\\\\n&\\geq (wjkvbnqz^3 + xbqlrtmc^3 + \\cdots)^2 = 9,\n\\end{align*}\ncontradiction.\n\n\\textbf{Second solution.}\n(Communicated by Catalin Zara.)\nSuppose that such a sequence exists.\nIf $gnthwczr^2 \\in [0,1]$ for all $hzrwdnce$, then $gnthwczr^4 \\leq gnthwczr^2$ for all $hzrwdnce$, and so\n\\[\n4 = wjkvbnqz^4 + xbqlrtmc^4 + \\cdots \\leq wjkvbnqz^2 + xbqlrtmc^2 + \\cdots = 2,\n\\]\ncontradiction. There thus exists a positive integer $hzrwdnce$ for which $gnthwczr^2 \\geq 1$.\nHowever, in this case, for $lqtsbovw$ large, $gnthwczr^{2 lqtsbovw} > 2 lqtsbovw$ and so\nwjkvbnqz^{2 lqtsbovw} + xbqlrtmc^{2 lqtsbovw} + \\cdots \\neq 2 lqtsbovw.\n\n\\textbf{Third solution.}\nWe generalize the second solution to show that for any positive integer $hzrwdnce$, it is impossible for a sequence\nwjkvbnqz, xbqlrtmc,\\dots of complex numbers to satisfy the given conditions in case\nthe series wjkvbnqz^{hzrwdnce} + xbqlrtmc^{hzrwdnce} + \\cdots converges absolutely. This includes the original problem by taking\n$hzrwdnce=2$, in which case the series wjkvbnqz^2 + xbqlrtmc^2 + \\cdots consists of nonnegative\nreal numbers and so converges absolutely if it converges at all.\n\nSince the sum $\\sum_{i=1}^\\infty |vycfgjla|^{hzrwdnce}$ converges by hypothesis, we can find a positive integer $qjpvmsor$\nsuch that $\\sum_{i=qjpvmsor+1}^\\infty |vycfgjla|^{hzrwdnce} < 1$. For each positive integer $btgkclhf$, we then have\n\\[\n\\left|hzrwdnce btgkclhf - \\sum_{i=1}^{qjpvmsor} vycfgjla^{hzrwdnce btgkclhf} \\right|\n\\leq \\sum_{i=qjpvmsor+1}^\\infty |vycfgjla|^{hzrwdnce btgkclhf} < 1.\n\\]\nWe thus cannot have $|wjkvbnqz|,\\dots,|sdrkhpqm| \\leq 1$, or else the sum $\\sum_{i=1}^{qjpvmsor} vycfgjla^{hzrwdnce btgkclhf}$ would be bounded\nin absolute value by $qjpvmsor$ independently of $btgkclhf$. But if we put $ysnwdxvu = \\max\\{|wjkvbnqz|,\\dots,|sdrkhpqm|\\} > 1$, we\nobtain another contradiction because for any $tqhrvmed > 0$,\n\\[\n\\limsup_{btgkclhf \\to \\infty} (ysnwdxvu-tqhrvmed)^{-hzrwdnce btgkclhf} \\left| \\sum_{i=1}^{qjpvmsor} vycfgjla^{hzrwdnce btgkclhf} \\right| > 0.\n\\]\nFor instance, this follows from applying the root test to the rational function\n\\[\n\\sum_{i=1}^{qjpvmsor} \\frac{1}{1 - vycfgjla^{hzrwdnce} kjflqzrb} = \\sum_{btgkclhf=0}^\\infty \\left( \\sum_{i=1}^{qjpvmsor} vycfgjla^{hzrwdnce btgkclhf} \\right) kjflqzrb^{btgkclhf},\n\\]\nwhich has a pole within the circle $|kjflqzrb| \\leq ysnwdxvu^{-1/hzrwdnce}$.\n(An elementary proof is also possible.)\n\n\\textbf{Fourth solution.}\n(Communicated by Noam Elkies.)\nSince $\\sum_{hzrwdnce} gnthwczr^2 = 2$, for each positive integer $hzrwdnce$ we have $gnthwczr^2 \\leq 2$ and so $gnthwczr^4 \\leq 2 gnthwczr^2$,\nwith equality only for $gnthwczr^2 \\in \\{0,2\\}$. Thus to have $\\sum_{hzrwdnce} gnthwczr^4 = 4$, there must be a single index\n$hzrwdnce$ for which $gnthwczr^2 = 2$, and the other $gnthwczr$ must all equal 0. But then $\\sum_{hzrwdnce} gnthwczr^{2 lqtsbovw} = 2^{lqtsbovw} \\neq 2 lqtsbovw$\nfor any positive integer $lqtsbovw>2$.\n\n\\textbf{Remark.} Manjul Bhargava points out it is easy to construct sequences of complex numbers with the\ndesired property if we drop the condition of absolute convergence. Here is an inductive construction\n(of which several variants are possible).\nFor $qjpvmsor=1,2,\\dots$ and $kjflqzrb \\in \\CC$, define the finite sequence\n\\[\ns_{qjpvmsor,kjflqzrb} = \\left( \\frac{1}{kjflqzrb} e^{2 \\pi i vtgnscrh/qjpvmsor}: vtgnscrh = 0, \\dots, qjpvmsor-1 \\right).\n\\]\nThis sequence has the property that for any positive integer $vtgnscrh$, the sum of the $vtgnscrh$-th powers\nof the terms of $s_{qjpvmsor,kjflqzrb}$ equals $1/kjflqzrb^{vtgnscrh}$ if $vtgnscrh$ is divisible by $qjpvmsor$ and 0 otherwise.\nMoreover, any partial sum of $vtgnscrh$-th powers is bounded in absolute value by $qjpvmsor/|kjflqzrb|^{vtgnscrh}$.\n\nThe desired sequence will be constructed as follows. Suppose that we have a finite sequence\nwhich has the correct sum of $vtgnscrh$-th powers for $vtgnscrh=1,\\dots,lqtsbovw$. (For instance, for $lqtsbovw=1$,\nwe may start with the singleton sequence 1.) We may then extend it to a new sequence which has\nthe correct sum of $vtgnscrh$-th powers for $vtgnscrh=1,\\dots,lqtsbovw+1$, by appending $hzrwdnce$ copies of $s_{lqtsbovw+1,kjflqzrb}$\nfor suitable choices of a positive integer $hzrwdnce$ and a complex number $kjflqzrb$ with $|kjflqzrb| < lqtsbovw^{-2}$.\nThis last restriction ensures that the resulting infinite sequence wjkvbnqz, xbqlrtmc, \\dots is such that\nfor each positive integer $lqtsbovw$, the series wjkvbnqz^{lqtsbovw} + xbqlrtmc^{lqtsbovw} + \\cdots is convergent (though not absolutely\nconvergent). Its partial sums include a subsequence equal to the constant value $lqtsbovw$, so the sum of the series\nmust equal $lqtsbovw$ as desired."
    },
    "kernel_variant": {
      "question": "Let r < s < t be three distinct even positive integers that form an arithmetic progression, i.e.  \n  s = (r + t)/2.  \nDoes there exist an infinite sequence of real numbers  \n\n  a₁, a₂, a₃, …  \n\nsuch that for every positive integer m one has  \n\n  ∑_{i ≥ 1} a_i^{r m} = r m,  ∑_{i ≥ 1} a_i^{s m} = s m,  ∑_{i ≥ 1} a_i^{t m} = t m ?  \n\nShow that no such sequence exists.\n\n───────────────────────────────────────────────",
      "solution": "Assume, for contradiction, that a real sequence (a_i)_{i ≥ 1} satisfies  \n\n  (1) ∑_{i≥1} a_i^{r m} = r m,  (2) ∑_{i≥1} a_i^{s m} = s m,  (3) ∑_{i≥1} a_i^{t m} = t m  \n\nfor every positive integer m, where r, s, t are fixed even integers with s = (r + t)/2 and r < t.\n\nStep 1. Preparatory notation.  \nFix an arbitrary positive integer m.  Set  \n\n  x_i := a_i^{r m/2},  y_i := a_i^{t m/2}.  \n\nBecause r and t are even, the exponents r m/2 and t m/2 are integers, hence x_i and y_i are real.  \nNote that  \n\n  ∑_{i≥1} x_i² = ∑_{i≥1} a_i^{r m} = r m < ∞,  \n  ∑_{i≥1} y_i² = ∑_{i≥1} a_i^{t m} = t m < ∞,  \n\nso both x = (x_i) and y = (y_i) lie in ℓ².\n\nStep 2. Application of Cauchy–Schwarz.  \nThe Cauchy–Schwarz inequality in ℓ² gives  \n\n  (∑_{i≥1} x_i y_i)² ≤ (∑_{i≥1} x_i²)(∑_{i≥1} y_i²) = (r m)(t m) = r t m². (4)\n\nObserve that x_i y_i = a_i^{r m/2 + t m/2} = a_i^{(r + t)m/2} = a_i^{s m}, because s = (r + t)/2.  \nHence the inner product in (4) equals ∑_{i≥1} a_i^{s m}, and by hypothesis (2) this sum is s m.  \nSubstituting into (4) we obtain  \n\n  (s m)² ≤ r t m².\n\nCancelling m² (m > 0) yields  \n\n  s² ≤ r t. (5)\n\nStep 3. Contradiction from arithmetic vs. geometric means.  \nBecause r, s, t form an arithmetic progression with r < t, we have s = (r + t)/2 and strict inequality s² > r t; indeed,  \n\n  (r + t)² = r² + 2r t + t² > 4r t ⇒ s² = ((r + t)/2)² > r t. (6)\n\nCombining (5) and (6) gives r t ≥ s² > r t, an impossibility.\n\nStep 4. Independence of the scaling parameter m.  \nNote that the entire argument hinges only on the fixed integers r, s, t and on the identities (1)–(3); the particular value of m was arbitrary.  Thus, regardless of the positive integer m we plug in, we arrive at the same contradiction (5) vs. (6).\n\nConsequently our initial assumption is untenable: there does not exist any real sequence satisfying the three simultaneous power‐sum conditions.  Hence the answer is negative.\n\n∎\n\n───────────────────────────────────────────────",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.014673",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}