{
  "index": "1995-B-6",
  "type": "NT",
  "tag": [
    "NT",
    "COMB",
    "ANA"
  ],
  "difficulty": "",
  "question": "\\[\nS(\\alpha) = \\{ \\lfloor n\\alpha \\rfloor : n = 1,2,3,\\dots \\}.\n\\]\nProve that $\\{1,2,3,\\dots\\}$ cannot be expressed as the disjoint\nunion of three sets $S(\\alpha), S(\\beta)$ and $S(\\gamma)$. [As\nusual, $\\lfloor x \\rfloor$ is the greatest integer $\\leq x$.]\n\n\\end{itemize}\n\\end{document}",
  "solution": "Obviously $\\alpha, \\beta, \\gamma$ have to be greater than 1, and no\ntwo can both be rational, so without loss of generality assume that\n$\\alpha$ and $\\beta$ are irrational.\nLet $\\{x\\} = x - \\lfloor x \\rfloor$ denote the fractional part of $x$.\nThen $m \\in S(\\alpha)$ if and only if $f(m/\\alpha) \\in (1-1/\\alpha,1)\n\\cup \\{0\\}$. In particular, this means that $S(\\alpha) \\cap \\{1,\n\\dots, n\\}$ contains $\\lceil (n+1)/\\alpha \\rceil -1$ elements, and\nsimilarly. Hence for every integer $n$,\n\\[\nn = \\left\\lceil \\frac{n+1}\\alpha \\right\\rceil +\n \\left\\lceil \\frac{n+1}\\beta \\right\\rceil +\n \\left\\lceil \\frac{n+1}\\gamma \\right\\rceil -3.\n\\]\nDividing through by $n$ and taking the limit as $n \\to \\infty$ shows\nthat $1/\\alpha + 1/\\beta + 1/\\gamma = 1$. That in turn implies that\nfor all $n$,\n\\[\n\\left\\{ - \\frac{n+1}{\\alpha} \\right\\} +\n\\left\\{ - \\frac{n+1}{\\beta} \\right\\} +\n\\left\\{ - \\frac{n+1}{\\gamma} \\right\\} = 2.\n\\]\nOur desired contradiction is equivalent to showing that the left side actually\ntakes the value 1 for some $n$. Since the left side is\nan integer, it suffices to show that $\\{ -(n+1)/\\alpha\\} +\n\\{-(n+1)/\\beta\\} < 1$ for some $n$.\n\nA result in ergodic theory (the two-dimensional version of the Weil\nequidistribution theorem) states that if $1,r,s$ are linearly\nindependent over the rationals, then the set of points $(\\{nr\\},\n\\{ns\\}$ is dense (and in fact equidistributed) in the unit square. In\nparticular, our claim definitely holds unless $a/\\alpha + b/\\beta =\nc$ for some integers $a,b,c$.\n\nOn the other hand, suppose that such a relation does hold. Since\n$\\alpha$ and $\\beta$ are irrational, by the one-dimensional Weil\ntheorem, the set of points $(\\{-n/\\alpha\\}, \\{-n/\\beta\\}$ is dense in\nthe set of $(x,y)$ in the unit square such that $ax + by$ is an integer.\nIt is simple enough to show that this set meets the region $\\{(x,y)\n\\in [0,1]^{2}: x+y<1\\}$ unless $a+b$ is an integer, and that would\nimply that $1/\\alpha + 1/\\beta$, a quantity between 0 and 1, is an\ninteger. We have our desired contradiction.\n\n\\end{itemize}\n\\end{document}",
  "vars": [
    "n",
    "m",
    "x",
    "r",
    "s",
    "a",
    "b",
    "c"
  ],
  "params": [
    "S",
    "\\\\alpha",
    "\\\\beta",
    "\\\\gamma"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexer",
        "m": "member",
        "x": "placeholder",
        "r": "density",
        "s": "sequence",
        "a": "coeffone",
        "b": "coefftwo",
        "c": "coeffthr",
        "S": "setfunc",
        "\\alpha": "alphavar",
        "\\beta": "betavar",
        "\\gamma": "gammavar"
      },
      "question": "\\[\nsetfunc(alphavar) = \\{ \\lfloor indexer\\,alphavar \\rfloor : indexer = 1,2,3,\\dots \\}.\n\\]\nProve that $\\{1,2,3,\\dots\\}$ cannot be expressed as the disjoint\nunion of three sets $setfunc(alphavar), setfunc(betavar)$ and $setfunc(gammavar)$. [As\nusual, $\\lfloor placeholder \\rfloor$ is the greatest integer $\\leq placeholder$.]",
      "solution": "Obviously $alphavar, betavar, gammavar$ have to be greater than 1, and no\ntwo can both be rational, so without loss of generality assume that\n$alphavar$ and $betavar$ are irrational.\nLet $\\{placeholder\\} = placeholder - \\lfloor placeholder \\rfloor$ denote the fractional part of $placeholder$.\nThen $member \\in setfunc(alphavar)$ if and only if $f(member/alphavar) \\in (1-1/alphavar,1)\n\\cup \\{0\\}$. In particular, this means that $setfunc(alphavar) \\cap \\{1,\n\\dots, indexer\\}$ contains $\\lceil (indexer+1)/alphavar \\rceil -1$ elements, and\nsimilarly. Hence for every integer $indexer$,\n\\[\nindexer = \\left\\lceil \\frac{indexer+1}{alphavar} \\right\\rceil +\n \\left\\lceil \\frac{indexer+1}{betavar} \\right\\rceil +\n \\left\\lceil \\frac{indexer+1}{gammavar} \\right\\rceil -3.\n\\]\nDividing through by $indexer$ and taking the limit as $indexer \\to \\infty$ shows\nthat $1/alphavar + 1/betavar + 1/gammavar = 1$. That in turn implies that\nfor all $indexer$,\n\\[\n\\left\\{ - \\frac{indexer+1}{alphavar} \\right\\} +\n\\left\\{ - \\frac{indexer+1}{betavar} \\right\\} +\n\\left\\{ - \\frac{indexer+1}{gammavar} \\right\\} = 2.\n\\]\nOur desired contradiction is equivalent to showing that the left side actually\ntakes the value 1 for some $indexer$. Since the left side is\nan integer, it suffices to show that $\\{ -(indexer+1)/alphavar\\} +\n\\{-(indexer+1)/betavar\\} < 1$ for some $indexer$.\n\nA result in ergodic theory (the two-dimensional version of the Weil\nequidistribution theorem) states that if $1,density,sequence$ are linearly\nindependent over the rationals, then the set of points $(\\{indexer\\,density\\},\n\\{indexer\\,sequence\\})$ is dense (and in fact equidistributed) in the unit square. In\nparticular, our claim definitely holds unless $coeffone/alphavar + coefftwo/betavar =\ncoeffthr$ for some integers $coeffone,coefftwo,coeffthr$.\n\nOn the other hand, suppose that such a relation does hold. Since\n$alphavar$ and $betavar$ are irrational, by the one-dimensional Weil\ntheorem, the set of points $(\\{-indexer/alphavar\\}, \\{-indexer/betavar\\})$ is dense in\nthe set of $(placeholder,y)$ in the unit square such that $coeffone\\,placeholder + coefftwo\\,y$ is an integer.\nIt is simple enough to show that this set meets the region $\\{(placeholder,y)\n\\in [0,1]^{2}: placeholder+y<1\\}$ unless $coeffone+coefftwo$ is an integer, and that would\nimply that $1/alphavar + 1/betavar$, a quantity between 0 and 1, is an\ninteger. We have our desired contradiction."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "hummingbird",
        "m": "pinecones",
        "x": "sandstorm",
        "r": "watershed",
        "s": "driftwood",
        "a": "moonlight",
        "b": "starlight",
        "c": "afterglow",
        "S": "windswept",
        "\\\\alpha": "evergreen",
        "\\\\beta": "lighthouse",
        "\\\\gamma": "raincloud"
      },
      "question": "\\[\nwindswept(evergreen) = \\{ \\lfloor hummingbird evergreen \\rfloor : hummingbird = 1,2,3,\\dots \\}.\n\\]\nProve that $\\{1,2,3,\\dots\\}$ cannot be expressed as the disjoint\nunion of three sets $windswept(evergreen), windswept(lighthouse)$ and $windswept(raincloud)$. [As\nusual, $\\lfloor sandstorm \\rfloor$ is the greatest integer $\\leq sandstorm$.]",
      "solution": "Obviously evergreen, lighthouse, raincloud have to be greater than 1, and no\ntwo can both be rational, so without loss of generality assume that\nevergreen and lighthouse are irrational.\nLet $\\{sandstorm\\} = sandstorm - \\lfloor sandstorm \\rfloor$ denote the fractional part of sandstorm.\nThen $pinecones \\in windswept(evergreen)$ if and only if $f(pinecones/evergreen) \\in (1-1/evergreen,1)\n\\cup \\{0\\}$. In particular, this means that $windswept(evergreen) \\cap \\{1,\n\\dots, hummingbird\\}$ contains $\\lceil (hummingbird+1)/evergreen \\rceil -1$ elements, and\nsimilarly. Hence for every integer $hummingbird$,\n\\[\nhummingbird = \\left\\lceil \\frac{hummingbird+1}{evergreen} \\right\\rceil +\n \\left\\lceil \\frac{hummingbird+1}{lighthouse} \\right\\rceil +\n \\left\\lceil \\frac{hummingbird+1}{raincloud} \\right\\rceil -3.\n\\]\nDividing through by $hummingbird$ and taking the limit as $hummingbird \\to \\infty$ shows\nthat $1/evergreen + 1/lighthouse + 1/raincloud = 1$. That in turn implies that\nfor all $hummingbird$,\n\\[\n\\left\\{ - \\frac{hummingbird+1}{evergreen} \\right\\} +\n\\left\\{ - \\frac{hummingbird+1}{lighthouse} \\right\\} +\n\\left\\{ - \\frac{hummingbird+1}{raincloud} \\right\\} = 2.\n\\]\nOur desired contradiction is equivalent to showing that the left side actually\ntakes the value 1 for some $hummingbird$. Since the left side is\nan integer, it suffices to show that $\\{ -(hummingbird+1)/evergreen\\} +\n\\{-(hummingbird+1)/lighthouse\\} < 1$ for some $hummingbird$.\n\nA result in ergodic theory (the two-dimensional version of the Weil\nequidistribution theorem) states that if $1,watershed,driftwood$ are linearly\nindependent over the rationals, then the set of points $(\\{hummingbird watershed\\},\n\\{hummingbird driftwood\\}$ is dense (and in fact equidistributed) in the unit square. In\nparticular, our claim definitely holds unless $moonlight/evergreen + starlight/lighthouse =\nafterglow$ for some integers $moonlight,starlight,afterglow$.\n\nOn the other hand, suppose that such a relation does hold. Since\nevergreen and lighthouse are irrational, by the one-dimensional Weil\ntheorem, the set of points $(\\{-hummingbird/evergreen\\}, \\{-hummingbird/lighthouse\\}$ is dense in\nthe set of $(sandstorm,y)$ in the unit square such that $moonlight sandstorm + starlight y$ is an integer.\nIt is simple enough to show that this set meets the region $\\{(sandstorm,y)\n\\in [0,1]^{2}: sandstorm+y<1\\}$ unless $moonlight+starlight$ is an integer, and that would\nimply that $1/evergreen + 1/lighthouse$, a quantity between 0 and 1, is an\ninteger. We have our desired contradiction."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "fractional",
        "m": "irrational",
        "x": "constant",
        "r": "imaginary",
        "s": "disorderly",
        "a": "variable",
        "b": "stableone",
        "c": "changing",
        "S": "sequence",
        "\\alpha": "smallvalue",
        "\\beta": "tinyvalue",
        "\\gamma": "minuscule"
      },
      "question": "\\[\nsequence(smallvalue) = \\{ \\lfloor fractional smallvalue \\rfloor : fractional = 1,2,3,\\dots \\}.\n\\]\nProve that $\\{1,2,3,\\dots\\}$ cannot be expressed as the disjoint\nunion of three sets $sequence(smallvalue), sequence(tinyvalue)$ and $sequence(minuscule)$. [As\nusual, $\\lfloor constant \\rfloor$ is the greatest integer $\\leq constant$.]\n\n\\end{itemize}\n\\end{document}",
      "solution": "\\<<<\nObviously $smallvalue, tinyvalue, minuscule$ have to be greater than 1, and no\ntwo can both be rational, so without loss of generality assume that\n$smallvalue$ and $tinyvalue$ are irrational.\nLet $\\{constant\\} = constant - \\lfloor constant \\rfloor$ denote the fractional part of $constant$.\nThen $irrational \\in sequence(smallvalue)$ if and only if $f(irrational/smallvalue) \\in (1-1/smallvalue,1)\n\\cup \\{0\\}$. In particular, this means that $sequence(smallvalue) \\cap \\{1,\n\\dots, fractional\\}$ contains $\\lceil (fractional+1)/smallvalue \\rceil -1$ elements, and\nsimilarly. Hence for every integer $fractional$,\n\\[\nfractional = \\left\\lceil \\frac{fractional+1}{smallvalue} \\right\\rceil +\n \\left\\lceil \\frac{fractional+1}{tinyvalue} \\right\\rceil +\n \\left\\lceil \\frac{fractional+1}{minuscule} \\right\\rceil -3.\n\\]\nDividing through by $fractional$ and taking the limit as $fractional \\to \\infty$ shows\nthat $1/smallvalue + 1/tinyvalue + 1/minuscule = 1$. That in turn implies that\nfor all $fractional$,\n\\[\n\\left\\{ - \\frac{fractional+1}{smallvalue} \\right\\} +\n\\left\\{ - \\frac{fractional+1}{tinyvalue} \\right\\} +\n\\left\\{ - \\frac{fractional+1}{minuscule} \\right\\} = 2.\n\\]\nOur desired contradiction is equivalent to showing that the left side actually\ntakes the value 1 for some $fractional$. Since the left side is\nan integer, it suffices to show that $\\{ -(fractional+1)/smallvalue\\} +\n\\{-(fractional+1)/tinyvalue\\} < 1$ for some $fractional$.\n\nA result in ergodic theory (the two-dimensional version of the Weil\nequidistribution theorem) states that if $1,imaginary,disorderly$ are linearly\nindependent over the rationals, then the set of points $(\\{fractionalimaginary\\},\n\\{fractionaldisorderly\\}$ is dense (and in fact equidistributed) in the unit square. In\nparticular, our claim definitely holds unless $variable/smallvalue + stableone/tinyvalue =\nchanging$ for some integers $variable,stableone,changing$.\n\nOn the other hand, suppose that such a relation does hold. Since\n$smallvalue$ and $tinyvalue$ are irrational, by the one-dimensional Weil\ntheorem, the set of points $(\\{-fractional/smallvalue\\}, \\{-fractional/tinyvalue\\}$ is dense in\nthe set of $(constant,y)$ in the unit square such that $variable constant + stableone y$ is an integer.\nIt is simple enough to show that this set meets the region $\\{(constant,y)\n\\in [0,1]^{2}: constant+y<1\\}$ unless $variable+stableone$ is an integer, and that would\nimply that $1/smallvalue + 1/tinyvalue$, a quantity between 0 and 1, is an\ninteger. We have our desired contradiction.\n\n\\end{itemize}\n\\end{document}\n>>>\n"
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "m": "hjgrksla",
        "x": "vkjzmqer",
        "r": "opdynmcl",
        "s": "wbrtqfsa",
        "a": "tlmcxoph",
        "b": "zqwrvekl",
        "c": "yplhndgt",
        "S": "pnsjdwkt",
        "\\alpha": "azldkqwe",
        "\\beta": "kmdzpryw",
        "\\gamma": "cxvtrhgu"
      },
      "question": "\\[\npnsjdwkt(azldkqwe) = \\{ \\lfloor qzxwvtnp azldkqwe \\rfloor : qzxwvtnp = 1,2,3,\\dots \\}.\n\\]\nProve that $\\{1,2,3,\\dots\\}$ cannot be expressed as the disjoint\nunion of three sets $pnsjdwkt(azldkqwe), pnsjdwkt(kmdzpryw)$ and $pnsjdwkt(cxvtrhgu)$. [As\nusual, $\\lfloor vkjzmqer \\rfloor$ is the greatest integer $\\leq vkjzmqer$.]",
      "solution": "Obviously $azldkqwe, kmdzpryw, cxvtrhgu$ have to be greater than 1, and no\ntwo can both be rational, so without loss of generality assume that\n$azldkqwe$ and $kmdzpryw$ are irrational.\nLet $\\{vkjzmqer\\} = vkjzmqer - \\lfloor vkjzmqer \\rfloor$ denote the fractional part of $vkjzmqer$.\nThen $hjgrksla \\in pnsjdwkt(azldkqwe)$ if and only if $f(hjgrksla/azldkqwe) \\in (1-1/azldkqwe,1)\n\\cup \\{0\\}$. In particular, this means that $pnsjdwkt(azldkqwe) \\cap \\{1,\n\\dots, qzxwvtnp\\}$ contains $\\lceil (qzxwvtnp+1)/azldkqwe \\rceil -1$ elements, and\nsimilarly. Hence for every integer $qzxwvtnp$,\n\\[\nqzxwvtnp = \\left\\lceil \\frac{qzxwvtnp+1}{azldkqwe} \\right\\rceil +\n \\left\\lceil \\frac{qzxwvtnp+1}{kmdzpryw} \\right\\rceil +\n \\left\\lceil \\frac{qzxwvtnp+1}{cxvtrhgu} \\right\\rceil -3.\n\\]\nDividing through by $qzxwvtnp$ and taking the limit as $qzxwvtnp \\to \\infty$ shows\nthat $1/azldkqwe + 1/kmdzpryw + 1/cxvtrhgu = 1$. That in turn implies that\nfor all $qzxwvtnp$,\n\\[\n\\left\\{ - \\frac{qzxwvtnp+1}{azldkqwe} \\right\\} +\n\\left\\{ - \\frac{qzxwvtnp+1}{kmdzpryw} \\right\\} +\n\\left\\{ - \\frac{qzxwvtnp+1}{cxvtrhgu} \\right\\} = 2.\n\\]\nOur desired contradiction is equivalent to showing that the left side actually\ntakes the value 1 for some $qzxwvtnp$. Since the left side is\nan integer, it suffices to show that $\\{ -(qzxwvtnp+1)/azldkqwe\\} +\n\\{-(qzxwvtnp+1)/kmdzpryw\\} < 1$ for some $qzxwvtnp$.\n\nA result in ergodic theory (the two-dimensional version of the Weil\nequidistribution theorem) states that if $1,opdynmcl,wbrtqfsa$ are linearly\nindependent over the rationals, then the set of points $(\\{qzxwvtnp opdynmcl\\},\n\\{qzxwvtnp wbrtqfsa\\}$ is dense (and in fact equidistributed) in the unit square. In\nparticular, our claim definitely holds unless $tlmcxoph/azldkqwe + zqwrvekl/kmdzpryw =\nyplhndgt$ for some integers $tlmcxoph,zqwrvekl,yplhndgt$.\n\nOn the other hand, suppose that such a relation does hold. Since\n$azldkqwe$ and $kmdzpryw$ are irrational, by the one-dimensional Weil\ntheorem, the set of points $(\\{-qzxwvtnp/azldkqwe\\}, \\{-qzxwvtnp/kmdzpryw\\}$ is dense in\nthe set of $(vkjzmqer,y)$ in the unit square such that $tlmcxoph vkjzmqer + zqwrvekl y$ is an integer.\nIt is simple enough to show that this set meets the region $\\{(vkjzmqer,y)\n\\in [0,1]^{2}: vkjzmqer+y<1\\}$ unless $tlmcxoph+zqwrvekl$ is an integer, and that would\nimply that $1/azldkqwe + 1/kmdzpryw$, a quantity between 0 and 1, is an\ninteger. We have our desired contradiction.\n"
    },
    "kernel_variant": {
      "question": "For each index κ ∈ {α,β,γ,δ} choose a real pair  \n\n  θκ > 1 and 0 < φκ < 1.\n\nDefine  \n  Tκ := {⌊n θκ + φκ⌋ : n = 1,2,3,…}.                (♠)\n\nCan the positive integers be written as a disjoint union of the four shifted Beatty sets,\n\n  ℕ = Tα ⊔ Tβ ⊔ Tγ ⊔ Tδ ?\n\nProve that such a decomposition is impossible, even if the four shifts φκ are chosen arbitrarily (they need not sum to an integer, coincide, etc.).",
      "solution": "(≈ 320 words)  \nAssume, toward a contradiction, that\n\n (0) ℕ = Tα ⊔ Tβ ⊔ Tγ ⊔ Tδ,  θκ > 1, 0<φκ<1.\n\nStep 1 – How many terms of Tκ do not exceed N.  \nPut  \n\n Aκ(N) := |Tκ ∩ {1,…,N}|,  N ≥ 1.\n\nMembership m ∈ Tκ means m = ⌊n θκ+φκ⌋ for some n, hence  \n\n n θκ+φκ < m+1 ≤ N+1 ⇔ n < (N+1−φκ)/θκ.\n\nThus  \n\n Aκ(N)=⌈(N+1−φκ)/θκ⌉−1\n    =⌊(N+1−φκ)/θκ⌋−εκ(N),\n\nwhere εκ(N)=1 precisely when (N+1−φκ)/θκ is an integer, and 0 otherwise.  \nBecause of (0),\n\n ∑κ Aκ(N)=N.                                                        (1)\n\nInserting the formula for Aκ(N) gives  \n\n ∑κ ⌊(N+1−φκ)/θκ⌋=N+Z(N),                                        (2)\n\nwith Z(N):=∑κ εκ(N) ∈ {0,1,2,3,4}.\n\nStep 2 – The reciprocal identity.  \nDivide (2) by N and let N→∞: lim ⌊x⌋/N=x/N, hence  \n\n (3) 1/α+1/β+1/γ+1/δ = 1.\n\nStep 3 – A rigid fractional-part sum.  \nWrite { x }=x−⌊x⌋.  From (2) and the trivial equality  \n\n ∑κ (N+1−φκ)/θκ = N+1−∑κ φκ,\n\nsubtracting yields  \n\n ∑κ {(N+1−φκ)/θκ} = 1−∑κ φκ − Z(N).                        (4)\n\nThe left side always lies in [0,4), so the right side must as well; consequently\n\n Z(N)=0 for every N.                                            (5)\n\nTherefore (N+1−φκ)/θκ is never an integer; in particular, any θκ with an even reduced denominator is excluded.\n\nStep 4 – An unavoidable visit to the triangle x+y<1.  \nDefine\n\n xN:={−(N+1−φα)/α}, yN:={−(N+1−φβ)/β}.\n\nBecause xN+1≡xN−1/α (mod 1) and the analogous relation for yN, the sequence (xN,yN) is the orbit of v0:= (−(1−φα)/α, −(1−φβ)/β) under the toral translation τ(x,y)=(x−1/α, y−1/β).\n\nLet Δ:= {(x,y)∈[0,1)² : x+y<1}.  If some index N satisfies (xN,yN)∈Δ, then\n\n {−(N+1−φα)/α}+{−(N+1−φβ)/β}<1,                              (6)\n\nand combining (4),(5) yields\n\n sum of all four fractional parts = 1−∑κ φκ < 3,\n\ncontradicting (4).  Hence no orbit point may enter Δ.\n\nYet τ is an ergodic translation unless 1,1/α,1/β are rationally dependent.  In the ergodic case the orbit is dense and certainly meets Δ, contradiction.\n\nIf a linear relation a/α+b/β=c (integers, coprime) exists, then, by (5), neither 1/α nor 1/β is integral; at least one is irrational, so τ restricted to the closed subgroup L={ax+by≡λ (mod 1)} (λ determined by v0) is again ergodic, forcing an intersection with Δ.  \n\nFinally, when both 1/α=q/p and 1/β=s/r are rational with odd denominators p,r≥3, the orbit is strictly periodic of length lcm(p,r).  Averaging xN+yN over one period gives 1; therefore some term is <1, placing the orbit in Δ – the same contradiction.\n\nEvery possible arithmetic configuration leads to a violation of (4), so assumption (0) is impossible. ∎",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.111323",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}