{
  "index": "1980-A-4",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem A-4\n(a) Prove that there exist integers \\( a, b, c \\), not all zero and each of absolute value less than one million, such that\n\\[\n|a+b \\sqrt{2}+c \\sqrt{3}|<10^{-11}\n\\]\n(b) Let \\( a, b, c \\) be integers, not all zero and each of absolute value less than one million. Prove that\n\\[\n|a+b \\sqrt{2}+c \\sqrt{3}|>10^{-21}\n\\]",
  "solution": "A-4.\n(a). Let \\( S \\) be the set of the \\( 10^{18} \\) real numbers \\( r+s \\sqrt{2}+t \\sqrt{3} \\) with each of \\( r, s, t \\) in \\( \\left\\{0,1, \\ldots, 10^{6}-1\\right\\} \\) and let \\( d=(1+\\sqrt{2}+\\sqrt{3}) 10^{6} \\). Then each \\( x \\) in \\( S \\) is in the interval \\( 0 \\leqslant x<d \\). This interval is partitioned into \\( 10^{18}-1 \\) \"small\" intervals \\( (k-1) e \\leqslant x<k e \\) with \\( e=d /\\left(10^{18}-1\\right) \\) and \\( k \\) taking on the values \\( 1,2, \\ldots, 10^{18}-1 \\). By the pigeonhole principle, two of the \\( 10^{18} \\) numbers of \\( S \\) must be in the same small interval and their difference \\( a+b \\sqrt{2}+c \\sqrt{3} \\) gives the desired \\( a, b, c \\) since \\( c<10^{-11} \\).\n(b) Let \\( F_{1}=a+b \\sqrt{2}+c \\sqrt{3} \\) and \\( F_{2}, F_{3}, F_{4} \\) be the other numbers of the form \\( a \\pm b \\sqrt{2} \\pm c \\sqrt{3} \\). Using the irrationality of \\( \\sqrt{2} \\) and \\( \\sqrt{3} \\) and the fact that \\( a, b, c \\) are not all zero, one easily shows that no \\( F_{1} \\) is zero. (The demonstration of this was Problem A-1 of the 15th Competition, held on March 5, 1955. For the proof, see page 402 of The William Lowell Putnam Mathematical Competition Problems and Solutions: 1938-1964, published by the MAA, or see this Monthly, 62\n(1955) 561.) One also sees readily that the product \\( P=F_{1} F_{2} F_{3} F_{4} \\) is an integer. Hence \\( |P| \\geqslant 1 \\). Then \\( \\left|F_{1}\\right| \\geqslant 1 /\\left|F_{2} F_{3} F_{4}\\right|>10^{-21} \\) since \\( \\left|F_{1}\\right|<10^{7} \\) and thus \\( 1 /\\left|F_{1}\\right|>10^{-7} \\) for each \\( i \\).",
  "vars": [
    "a",
    "b",
    "c",
    "r",
    "s",
    "t",
    "x",
    "d",
    "e",
    "k",
    "F_1",
    "F_2",
    "F_3",
    "F_4",
    "P"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "alphaint",
        "b": "betaint",
        "c": "charlie",
        "r": "radical",
        "s": "segment",
        "t": "tangent",
        "x": "unknown",
        "d": "diameter",
        "e": "element",
        "k": "counter",
        "F_1": "factorone",
        "F_2": "factortwo",
        "F_3": "factorthree",
        "F_4": "factorfour",
        "P": "product"
      },
      "question": "Problem:\n<<<\nProblem A-4\n(a) Prove that there exist integers \\( alphaint, betaint, charlie \\), not all zero and each of absolute value less than one million, such that\n\\[\n|alphaint+ betaint \\sqrt{2}+ charlie \\sqrt{3}|<10^{-11}\n\\]\n(b) Let \\( alphaint, betaint, charlie \\) be integers, not all zero and each of absolute value less than one million. Prove that\n\\[\n|alphaint+ betaint \\sqrt{2}+ charlie \\sqrt{3}|>10^{-21}\n\\]\n>>>",
      "solution": "Solution:\n<<<\nA-4.\n(a). Let \\( S \\) be the set of the \\( 10^{18} \\) real numbers \\( radical+ segment \\sqrt{2}+ tangent \\sqrt{3} \\) with each of \\( radical, segment, tangent \\) in \\( \\left\\{0,1, \\ldots, 10^{6}-1\\right\\} \\) and let \\( diameter=(1+\\sqrt{2}+\\sqrt{3}) 10^{6} \\). Then each \\( unknown \\) in \\( S \\) is in the interval \\( 0 \\leqslant unknown<diameter \\). This interval is partitioned into \\( 10^{18}-1 \\) \"small\" intervals \\( (counter-1) element \\leqslant unknown<counter element \\) with \\( element=diameter /\\left(10^{18}-1\\right) \\) and \\( counter \\) taking on the values \\( 1,2, \\ldots, 10^{18}-1 \\). By the pigeonhole principle, two of the \\( 10^{18} \\) numbers of \\( S \\) must be in the same small interval and their difference \\( alphaint+ betaint \\sqrt{2}+ charlie \\sqrt{3} \\) gives the desired \\( alphaint, betaint, charlie \\) since \\( charlie<10^{-11} \\).\n(b) Let \\( factorone =alphaint+ betaint \\sqrt{2}+ charlie \\sqrt{3} \\) and \\( factortwo, factorthree, factorfour \\) be the other numbers of the form \\( alphaint \\pm betaint \\sqrt{2} \\pm charlie \\sqrt{3} \\). Using the irrationality of \\( \\sqrt{2} \\) and \\( \\sqrt{3} \\) and the fact that \\( alphaint, betaint, charlie \\) are not all zero, one easily shows that no \\( factorone \\) is zero. (The demonstration of this was Problem A-1 of the 15th Competition, held on March 5, 1955. For the proof, see page 402 of The William Lowell Putnam Mathematical Competition Problems and Solutions: 1938-1964, published by the MAA, or see this Monthly, 62\n(1955) 561.) One also sees readily that the product \\( product =factorone factortwo factorthree factorfour \\) is an integer. Hence \\( |product| \\geqslant 1 \\). Then \\( \\left|factorone \\right| \\geqslant 1 /\\left|factortwo factorthree factorfour\\right|>10^{-21} \\) since \\( \\left|factorone \\right|<10^{7} \\) and thus \\( 1 /\\left|factorone \\right|>10^{-7} \\) for each \\( i \\).\n>>>"
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "pineapple",
        "b": "telescope",
        "c": "butterfly",
        "r": "sandwich",
        "s": "cylinder",
        "t": "harmonica",
        "x": "labyrinth",
        "d": "waterfall",
        "e": "cinnamon",
        "k": "microscope",
        "F_1": "umbrella",
        "F_2": "magnolia",
        "F_3": "chandelier",
        "F_4": "skateboard",
        "P": "marshmallow"
      },
      "question": "Problem A-4\n(a) Prove that there exist integers \\( pineapple, telescope, butterfly \\), not all zero and each of absolute value less than one million, such that\n\\[\n|pineapple+telescope \\sqrt{2}+butterfly \\sqrt{3}|<10^{-11}\n\\]\n(b) Let \\( pineapple, telescope, butterfly \\) be integers, not all zero and each of absolute value less than one million. Prove that\n\\[\n|pineapple+telescope \\sqrt{2}+butterfly \\sqrt{3}|>10^{-21}\n\\]",
      "solution": "A-4.\n(a). Let \\( S \\) be the set of the \\( 10^{18} \\) real numbers \\( sandwich+cylinder \\sqrt{2}+harmonica \\sqrt{3} \\) with each of \\( sandwich, cylinder, harmonica \\) in \\( \\{0,1, \\ldots, 10^{6}-1\\} \\) and let \\( waterfall=(1+\\sqrt{2}+\\sqrt{3}) 10^{6} \\). Then each \\( labyrinth \\) in \\( S \\) is in the interval \\( 0 \\leqslant labyrinth<waterfall \\). This interval is partitioned into \\( 10^{18}-1 \\) \"small\" intervals \\( (microscope-1) cinnamon \\leqslant labyrinth<microscope\\,cinnamon \\) with \\( cinnamon=waterfall /\\left(10^{18}-1\\right) \\) and \\( microscope \\) taking on the values \\( 1,2, \\ldots, 10^{18}-1 \\). By the pigeonhole principle, two of the \\( 10^{18} \\) numbers of \\( S \\) must be in the same small interval and their difference \\( pineapple+telescope \\sqrt{2}+butterfly \\sqrt{3} \\) gives the desired \\( pineapple, telescope, butterfly \\) since \\( butterfly<10^{-11} \\).\n\n(b) Let \\( umbrella=pineapple+telescope \\sqrt{2}+butterfly \\sqrt{3} \\) and \\( magnolia, chandelier, skateboard \\) be the other numbers of the form \\( pineapple \\pm telescope \\sqrt{2} \\pm butterfly \\sqrt{3} \\). Using the irrationality of \\( \\sqrt{2} \\) and \\( \\sqrt{3} \\) and the fact that \\( pineapple, telescope, butterfly \\) are not all zero, one easily shows that no \\( umbrella \\) is zero. (The demonstration of this was Problem A-1 of the 15th Competition, held on March 5, 1955. For the proof, see page 402 of The William Lowell Putnam Mathematical Competition Problems and Solutions: 1938-1964, published by the MAA, or see this Monthly, 62 (1955) 561.) One also sees readily that the product \\( marshmallow=umbrella magnolia chandelier skateboard \\) is an integer. Hence \\( |marshmallow| \\geqslant 1 \\). Then \\( \\left|umbrella\\right| \\geqslant 1 /\\left|magnolia chandelier skateboard\\right|>10^{-21} \\) since \\( \\left|umbrella\\right|<10^{7} \\) and thus \\( 1 /\\left|umbrella\\right|>10^{-7} \\) for each i."
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "endvalue",
        "b": "penultima",
        "c": "finality",
        "r": "imaginary",
        "s": "gargantua",
        "t": "spacelike",
        "x": "knownvalue",
        "d": "closeness",
        "e": "gigantic",
        "k": "continuous",
        "F_1": "voidvalue",
        "F_2": "abyssvalue",
        "F_3": "vacuumval",
        "F_4": "nothingvl",
        "P": "quotient"
      },
      "question": "Problem A-4\n(a) Prove that there exist integers \\( endvalue, penultima, finality \\), not all zero and each of absolute value less than one million, such that\n\\[\n|endvalue+penultima \\sqrt{2}+finality \\sqrt{3}|<10^{-11}\n\\]\n(b) Let \\( endvalue, penultima, finality \\) be integers, not all zero and each of absolute value less than one million. Prove that\n\\[\n|endvalue+penultima \\sqrt{2}+finality \\sqrt{3}|>10^{-21}\n\\]",
      "solution": "A-4.\n(a). Let \\( S \\) be the set of the \\( 10^{18} \\) real numbers \\( imaginary+gargantua \\sqrt{2}+spacelike \\sqrt{3} \\) with each of \\( imaginary, gargantua, spacelike \\) in \\( \\left\\{0,1, \\ldots, 10^{6}-1\\right\\} \\) and let \\( closeness=(1+\\sqrt{2}+\\sqrt{3}) 10^{6} \\). Then each \\( knownvalue \\) in \\( S \\) is in the interval \\( 0 \\leqslant knownvalue<closeness \\). This interval is partitioned into \\( 10^{18}-1 \\) \"small\" intervals \\( (continuous-1) gigantic \\leqslant knownvalue<continuous gigantic \\) with \\( gigantic=closeness /\\left(10^{18}-1\\right) \\) and \\( continuous \\) taking on the values \\( 1,2, \\ldots, 10^{18}-1 \\). By the pigeonhole principle, two of the \\( 10^{18} \\) numbers of \\( S \\) must be in the same small interval and their difference \\( endvalue+penultima \\sqrt{2}+finality \\sqrt{3} \\) gives the desired \\( endvalue, penultima, finality \\) since \\( finality<10^{-11} \\).\n\n(b) Let \\( voidvalue=endvalue+penultima \\sqrt{2}+finality \\sqrt{3} \\) and \\( abyssvalue, vacuumval, nothingvl \\) be the other numbers of the form \\( endvalue \\pm penultima \\sqrt{2} \\pm finality \\sqrt{3} \\). Using the irrationality of \\( \\sqrt{2} \\) and \\( \\sqrt{3} \\) and the fact that \\( endvalue, penultima, finality \\) are not all zero, one easily shows that no \\( voidvalue \\) is zero. (The demonstration of this was Problem A-1 of the 15th Competition, held on March 5, 1955. For the proof, see page 402 of The William Lowell Putnam Mathematical Competition Problems and Solutions: 1938-1964, published by the MAA, or see this Monthly, 62\n(1955) 561.) One also sees readily that the product \\( quotient=voidvalue abyssvalue vacuumval nothingvl \\) is an integer. Hence \\( |quotient| \\geqslant 1 \\). Then \\( \\left|voidvalue\\right| \\geqslant 1 /\\left|abyssvalue vacuumval nothingvl\\right|>10^{-21} \\) since \\( \\left|voidvalue\\right|<10^{7} \\) and thus \\( 1 /\\left|voidvalue\\right|>10^{-7} \\) for each \\( i \\)."
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "mfjdlwre",
        "r": "bezqumns",
        "s": "vxcpldqo",
        "t": "nkjghswe",
        "x": "ofjzardc",
        "d": "ykpshuav",
        "e": "lbqtrnza",
        "k": "fzodrtmx",
        "F_1": "pqlswenk",
        "F_2": "xnydroqt",
        "F_3": "wbtkjzmp",
        "F_4": "hesvcmar",
        "P": "qmdryxks"
      },
      "question": "Problem A-4\n(a) Prove that there exist integers \\( qzxwvtnp, hjgrksla, mfjdlwre \\), not all zero and each of absolute value less than one million, such that\n\\[\n|qzxwvtnp+hjgrksla \\sqrt{2}+mfjdlwre \\sqrt{3}|<10^{-11}\n\\]\n(b) Let \\( qzxwvtnp, hjgrksla, mfjdlwre \\) be integers, not all zero and each of absolute value less than one million. Prove that\n\\[\n|qzxwvtnp+hjgrksla \\sqrt{2}+mfjdlwre \\sqrt{3}|>10^{-21}\n\\]",
      "solution": "A-4.\n(a). Let \\( S \\) be the set of the \\( 10^{18} \\) real numbers \\( bezqumns+vxcpldqo \\sqrt{2}+nkjghswe \\sqrt{3} \\) with each of \\( bezqumns, vxcpldqo, nkjghswe \\) in \\( \\left\\{0,1, \\ldots, 10^{6}-1\\right\\} \\) and let \\( ykpshuav=(1+\\sqrt{2}+\\sqrt{3}) 10^{6} \\). Then each \\( ofjzardc \\) in \\( S \\) is in the interval \\( 0 \\leqslant ofjzardc<ykpshuav \\). This interval is partitioned into \\( 10^{18}-1 \\) \"small\" intervals \\( (fzodrtmx-1) lbqtrnza \\leqslant ofjzardc<fzodrtmx lbqtrnza \\) with \\( lbqtrnza=ykpshuav /\\left(10^{18}-1\\right) \\) and \\( fzodrtmx \\) taking on the values \\( 1,2, \\ldots, 10^{18}-1 \\). By the pigeonhole principle, two of the \\( 10^{18} \\) numbers of \\( S \\) must be in the same small interval and their difference \\( qzxwvtnp+hjgrksla \\sqrt{2}+mfjdlwre \\sqrt{3} \\) gives the desired \\( qzxwvtnp, hjgrksla, mfjdlwre \\) since \\( mfjdlwre<10^{-11} \\).\n\n(b) Let \\( pqlswenk=qzxwvtnp+hjgrksla \\sqrt{2}+mfjdlwre \\sqrt{3} \\) and \\( xnydroqt, wbtkjzmp, hesvcmar \\) be the other numbers of the form \\( qzxwvtnp \\pm hjgrksla \\sqrt{2} \\pm mfjdlwre \\sqrt{3} \\). Using the irrationality of \\( \\sqrt{2} \\) and \\( \\sqrt{3} \\) and the fact that \\( qzxwvtnp, hjgrksla, mfjdlwre \\) are not all zero, one easily shows that no \\( pqlswenk \\) is zero. (The demonstration of this was Problem A-1 of the 15th Competition, held on March 5, 1955. For the proof, see page 402 of The William Lowell Putnam Mathematical Competition Problems and Solutions: 1938-1964, published by the MAA, or see this Monthly, 62 (1955) 561.) One also sees readily that the product \\( qmdryxks=pqlswenk xnydroqt wbtkjzmp hesvcmar \\) is an integer. Hence \\( |qmdryxks| \\geqslant 1 \\). Then \\( \\left|pqlswenk\\right| \\geqslant 1 /\\left|xnydroqt wbtkjzmp hesvcmar\\right|>10^{-21} \\) since \\( \\left|pqlswenk\\right|<10^{7} \\) and thus \\( 1 /\\left|pqlswenk\\right|>10^{-7} \\) for each \\( i \\)."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nN \\;=\\; 271\\,828 \n\\quad\\bigl(= 2.71828 \\times 10^{5}\\bigr).\n\\]\n\n(a)\\;  Prove that there exist integers  \n\\[\na_{1},a_{2},\\dots ,a_{8}, \\qquad \\text{not all zero and each satisfying } |a_{i}|<N ,\n\\]\nsuch that  \n\n\\[\na_{1}+a_{2}+\\dots +a_{8}\\equiv 0 \\pmod{17}\n\\]\nand  \n\n\\[\n\\Bigl|\\,a_{1}+a_{2}\\sqrt{2}+a_{3}\\sqrt{3}+a_{4}\\sqrt{5}+a_{5}\\sqrt{7}\n        +a_{6}\\sqrt{11}+a_{7}\\sqrt{13}+a_{8}\\sqrt{17}\\,\\Bigr|\n        \\;<\\;10^{-34}.\n\\]\n\n(b)\\;  Show that for every octuple  \n\\[\n(a_{1},\\dots ,a_{8})\\in\\mathbb Z^{8}\\setminus\\{\\mathbf 0\\}\n\\quad\\text{with}\\quad |a_{i}|<N\\;\\;\\text{for all }i\n\\]\none has  \n\\[\n\\Bigl|\\,a_{1}+a_{2}\\sqrt{2}+a_{3}\\sqrt{3}+a_{4}\\sqrt{5}+a_{5}\\sqrt{7}\n        +a_{6}\\sqrt{11}+a_{7}\\sqrt{13}+a_{8}\\sqrt{17}\\,\\Bigr|\n        \\;>\\;10^{-861}.\n\\]\n\n(The congruence condition is required only in part (a); part (b) holds whether or not the octuple satisfies it.)",
      "solution": "Throughout write  \n\\[\n\\beta(\\mathbf a)\\;=\\;\na_{1}+a_{2}\\sqrt{2}+a_{3}\\sqrt{3}+a_{4}\\sqrt{5}+a_{5}\\sqrt{7}\n      +a_{6}\\sqrt{11}+a_{7}\\sqrt{13}+a_{8}\\sqrt{17},\n\\qquad \n\\mathbf a=(a_{1},\\dots ,a_{8}),\n\\]\nand set  \n\\[\nC \\;=\\; 1+\\sqrt{2}+\\sqrt{3}+\\sqrt{5}+\\sqrt{7}+\\sqrt{11}+\\sqrt{13}+\\sqrt{17}\n     \\;<\\;20.08 .\n\\]\nFor any $\\mathbf a$ with $|a_{i}|<N$ we have the uniform bound  \n\\[\n\\bigl|\\beta(\\mathbf a)\\bigr|\n\\;\\le\\; C\\,N\n\\;<\\; 6\\times 10^{6}.\n\\tag{1}\n\\]\n\n--------------------------------------------------------------------\nPart (a).  Construction of an octuple with an extremely small value  \n--------------------------------------------------------------------\n\nStep 1 – $\\mathbb Q$–linear independence.  \nPut  \n\\[\n\\alpha_{0}=1,\\;\n\\alpha_{1}=\\sqrt{2},\\;\n\\alpha_{2}=\\sqrt{3},\\;\n\\alpha_{3}=\\sqrt{5},\\;\n\\alpha_{4}=\\sqrt{7},\\;\n\\alpha_{5}=\\sqrt{11},\\;\n\\alpha_{6}=\\sqrt{13},\\;\n\\alpha_{7}=\\sqrt{17}.\n\\]\nAssume  \n\\[\nr_{0}\\alpha_{0}+r_{1}\\alpha_{1}+\\dots +r_{7}\\alpha_{7}=0,\n\\qquad r_{k}\\in\\mathbb Q.\n\\tag{2}\n\\]\nFor each $j\\in\\{1,\\dots ,7\\}$ let $\\sigma_{j}$ be the $\\mathbb Q$–automorphism of  \n\\[\nL=\\mathbb Q\\!\\bigl(\\sqrt{2},\\sqrt{3},\\sqrt{5},\\sqrt{7},\n                   \\sqrt{11},\\sqrt{13},\\sqrt{17}\\bigr)\n\\]\nthat sends $\\sqrt{p_{j}}\\mapsto -\\sqrt{p_{j}}$ and fixes the other radicals.  \nApplying $\\sigma_{j}$ to (2) and subtracting (2) eliminates all summands\nexcept $2r_{j}\\sqrt{p_{j}}$, forcing $r_{j}=0$.  Consequently\n$r_{1}=r_{2}=\\dots=r_{7}=0$, and then $r_{0}=0$.\nHence  \n\\[\n\\{1,\\sqrt{2},\\sqrt{3},\\sqrt{5},\\sqrt{7},\\sqrt{11},\\sqrt{13},\\sqrt{17}\\}\n\\]\nis $\\mathbb Q$–linearly independent, so the map\n$\\mathbf r\\mapsto\\beta(\\mathbf r)$ is injective on $\\mathbb Z^{8}$.\n\nStep 2 – A large finite set.  \nDefine  \n\\[\n\\mathcal S\n   \\;=\\;\n   \\Bigl\\{\\,\n        \\beta(\\mathbf r):\n        0\\le r_{i}\\le N-1\\;(i=1,\\dots ,8)\n   \\Bigr\\}.\n\\]\nInjectivity gives $|\\mathcal S|=N^{8}$.\n\nStep 3 – Reduction modulo $17$.  \nPlace each $\\mathbf r$ in the residue class of\n$r_{1}+r_{2}+\\dots +r_{8}\\pmod{17}$.  At least one class, call it\n$\\mathcal T$, contains  \n\\[\nK\n=\\Bigl\\lceil \\frac{N^{8}}{17}\\Bigr\\rceil\n>\\frac{N^{8}}{17}\n\\tag{3}\n\\]\nelements.\n\nStep 4 – Pigeon-hole on the real line.  \nAll elements of $\\mathcal S$ lie in $[0,CN)$.\nPartition this interval into $K-1$ equal sub-intervals,\neach of length  \n\\[\n\\varepsilon\n=\\frac{C\\,N}{K-1}.\n\\tag{4}\n\\]\nBecause $|\\mathcal T|=K>K-1$, two distinct vectors\n$\\mathbf s,\\mathbf t\\in\\mathcal T$ fall in the same sub-interval.\nPut $\\mathbf a=\\mathbf s-\\mathbf t$.  Then  \n\n\\[\n\\mathbf a\\neq\\mathbf 0,\\;\n|a_{i}|<N,\\;\na_{1}+a_{2}+\\dots +a_{8}\\equiv 0 \\pmod{17},\n\\]\nand  \n\\[\n\\bigl|\\beta(\\mathbf a)\\bigr|\n=\\bigl|\\beta(\\mathbf s)-\\beta(\\mathbf t)\\bigr|\n<\\varepsilon.\n\\tag{5}\n\\]\n\nStep 5 – Estimating $\\varepsilon$.  \nBecause $K-1\\ge N^{8}/17-1$,\n\n\\[\n\\varepsilon\n\\;\\le\\;\n\\frac{C\\,N}{\\,N^{8}/17-1}\n\\;=\\;\n\\frac{17\\,C\\,N}{N^{8}-17}\n=\\frac{17\\,C}{N^{7}}\\,\n\\Bigl(\\frac{1}{1-17/N^{8}}\\Bigr).\n\\]\nSince $N^{8}>10^{43}$, the factor in parentheses is less than\n$1.0000000000000000002$.  Hence\n\n\\[\n\\varepsilon\n< 18\\,\\frac{C}{N^{7}}\n< 18\\,\\frac{20.08}{(2.71828\\times10^{5})^{7}}\n< 3.3\\times10^{-36}\n<10^{-34}.\n\\]\n\nCombining this with (5) completes part (a).\n\n--------------------------------------------------------------------\nPart (b).  A universal lower bound (independent of the congruence)  \n--------------------------------------------------------------------\n\nStep 1 – The ambient field.  \nLet  \n\\[\nL=\\mathbb Q\\!\\bigl(\\sqrt{2},\\sqrt{3},\\sqrt{5},\\sqrt{7},\n                   \\sqrt{11},\\sqrt{13},\\sqrt{17}\\bigr);\n\\qquad\n[L:\\mathbb Q]=2^{7}=128.\n\\]\n\nStep 2 – Conjugates of $\\beta(\\mathbf a)$.  \nEvery choice of signs  \n\n\\[\n\\bigl(\\sqrt{2},\\sqrt{3},\\sqrt{5},\\sqrt{7},\n       \\sqrt{11},\\sqrt{13},\\sqrt{17}\\bigr)\n\\longmapsto\n\\bigl(\\pm\\sqrt{2},\\pm\\sqrt{3},\\pm\\sqrt{5},\\pm\\sqrt{7},\n       \\pm\\sqrt{11},\\pm\\sqrt{13},\\pm\\sqrt{17}\\bigr)\n\\]\nextends to an embedding $\\sigma:L\\hookrightarrow\\mathbb R$.\nDenote the $128$ conjugates by\n$\\beta_{1},\\dots ,\\beta_{128}$, where $\\beta_{1}=\\beta(\\mathbf a)$.\n\nStep 3 – Non-vanishing.  \nThe $\\mathbb Q$–linear independence from Step 1 implies\n$\\beta(\\mathbf a)\\neq 0$ for every non-zero $\\mathbf a$, hence\n$\\beta_{j}\\neq 0$ for all $j$.\n\nStep 4 – Integrality of the algebraic norm.  \nBecause each $a_{i}\\in\\mathbb Z$ and\n$L/\\mathbb Q$ is Galois, the set\n$\\{\\beta_{1},\\dots ,\\beta_{128}\\}$ is stable under the Galois group.\nTherefore  \n\n\\[\nN_{L/\\mathbb Q}\\bigl(\\beta(\\mathbf a)\\bigr)=\n\\prod_{j=1}^{128}\\beta_{j}\n\\]\nis fixed by every automorphism and lies in $\\mathbb Q$.\nSince it equals a symmetric polynomial with integer coefficients\nevaluated at the integers $a_{i}$, it is in $\\mathbb Z$; by\nStep 3 it is non-zero.  Hence  \n\n\\[\n\\bigl|\\beta_{1}\\beta_{2}\\cdots\\beta_{128}\\bigr|\\ge 1.\n\\tag{6}\n\\]\n\nStep 5 – A uniform upper bound for the conjugates.  \nInequality (1) is valid after any sign change, so  \n\n\\[\n|\\beta_{j}|\\le C\\,N<6\\times10^{6},\n\\qquad j=1,\\dots ,128.\n\\tag{7}\n\\]\n\nStep 6 – Extracting the lower bound for $|\\beta_{1}|$.  \nCombining (6) and (7) gives  \n\n\\[\n|\\beta_{1}|\n\\;\\ge\\;\n\\frac{1}{(6\\times10^{6})^{127}}\n\\;=\\;\n10^{-127\\log_{10}(6\\times10^{6})}.\n\\]\nBecause $\\log_{10}(6\\times10^{6})=\\log_{10}6+6\\approx 6.77815$,\n\\[\n127\\times 6.77815\\approx 860.83<861,\n\\]\nand therefore  \n\n\\[\n\\Bigl|\\beta(\\mathbf a)\\Bigr|\n\\;>\\;10^{-861}.\n\\]\n\nThis completes the proof of part (b) and of the problem.\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.646337",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the linear form now involves eight independent quadratic irrationals instead of two (original) or three (kernel).  \n• Additional constraint: the coefficients must also obey the congruence a₁+…+a₈≡0 (mod 17), forcing a two-layer pigeon-hole argument.  \n• Larger field: the proof of the lower bound requires manipulating the 256 conjugates of an element in a degree-256 number field, instead of four conjugates in the original problem.  \n• Quantitatively sharper bounds: exponents −30 and −1728 call for careful numerical estimates.  \n• Techniques combined: geometry of numbers (pigeon-hole in an 8-dimensional box under an extra congruence), algebraic number theory (field degree, conjugates, norm), and explicit analytic estimates.\n\nThese ingredients demand substantially more work and deeper theoretical background than the original and current kernel variants."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nN \\;=\\; 271\\,828 \n\\quad\\bigl(= 2.71828 \\times 10^{5}\\bigr).\n\\]\n\n(a)\\;  Prove that there exist integers  \n\\[\na_{1},a_{2},\\dots ,a_{8}, \\qquad \\text{not all zero and each satisfying } |a_{i}|<N ,\n\\]\nsuch that  \n\n\\[\na_{1}+a_{2}+\\dots +a_{8}\\equiv 0 \\pmod{17}\n\\]\nand  \n\n\\[\n\\Bigl|\\,a_{1}+a_{2}\\sqrt{2}+a_{3}\\sqrt{3}+a_{4}\\sqrt{5}+a_{5}\\sqrt{7}\n        +a_{6}\\sqrt{11}+a_{7}\\sqrt{13}+a_{8}\\sqrt{17}\\,\\Bigr|\n        \\;<\\;10^{-34}.\n\\]\n\n(b)\\;  Show that for every octuple  \n\\[\n(a_{1},\\dots ,a_{8})\\in\\mathbb Z^{8}\\setminus\\{\\mathbf 0\\}\n\\quad\\text{with}\\quad |a_{i}|<N\\;\\;\\text{for all }i\n\\]\none has  \n\\[\n\\Bigl|\\,a_{1}+a_{2}\\sqrt{2}+a_{3}\\sqrt{3}+a_{4}\\sqrt{5}+a_{5}\\sqrt{7}\n        +a_{6}\\sqrt{11}+a_{7}\\sqrt{13}+a_{8}\\sqrt{17}\\,\\Bigr|\n        \\;>\\;10^{-861}.\n\\]\n\n(The congruence condition is required only in part (a); part (b) holds whether or not the octuple satisfies it.)",
      "solution": "Throughout write  \n\\[\n\\beta(\\mathbf a)\\;=\\;\na_{1}+a_{2}\\sqrt{2}+a_{3}\\sqrt{3}+a_{4}\\sqrt{5}+a_{5}\\sqrt{7}\n      +a_{6}\\sqrt{11}+a_{7}\\sqrt{13}+a_{8}\\sqrt{17},\n\\qquad \n\\mathbf a=(a_{1},\\dots ,a_{8}),\n\\]\nand set  \n\\[\nC \\;=\\; 1+\\sqrt{2}+\\sqrt{3}+\\sqrt{5}+\\sqrt{7}+\\sqrt{11}+\\sqrt{13}+\\sqrt{17}\n     \\;<\\;20.08 .\n\\]\nFor any $\\mathbf a$ with $|a_{i}|<N$ we have the uniform bound  \n\\[\n\\bigl|\\beta(\\mathbf a)\\bigr|\n\\;\\le\\; C\\,N\n\\;<\\; 6\\times 10^{6}.\n\\tag{1}\n\\]\n\n--------------------------------------------------------------------\nPart (a).  Construction of an octuple with an extremely small value  \n--------------------------------------------------------------------\n\nStep 1 – $\\mathbb Q$–linear independence.  \nPut  \n\\[\n\\alpha_{0}=1,\\;\n\\alpha_{1}=\\sqrt{2},\\;\n\\alpha_{2}=\\sqrt{3},\\;\n\\alpha_{3}=\\sqrt{5},\\;\n\\alpha_{4}=\\sqrt{7},\\;\n\\alpha_{5}=\\sqrt{11},\\;\n\\alpha_{6}=\\sqrt{13},\\;\n\\alpha_{7}=\\sqrt{17}.\n\\]\nAssume  \n\\[\nr_{0}\\alpha_{0}+r_{1}\\alpha_{1}+\\dots +r_{7}\\alpha_{7}=0,\n\\qquad r_{k}\\in\\mathbb Q.\n\\tag{2}\n\\]\nFor each $j\\in\\{1,\\dots ,7\\}$ let $\\sigma_{j}$ be the $\\mathbb Q$–automorphism of  \n\\[\nL=\\mathbb Q\\!\\bigl(\\sqrt{2},\\sqrt{3},\\sqrt{5},\\sqrt{7},\n                   \\sqrt{11},\\sqrt{13},\\sqrt{17}\\bigr)\n\\]\nthat sends $\\sqrt{p_{j}}\\mapsto -\\sqrt{p_{j}}$ and fixes the other radicals.  \nApplying $\\sigma_{j}$ to (2) and subtracting (2) eliminates all summands\nexcept $2r_{j}\\sqrt{p_{j}}$, forcing $r_{j}=0$.  Consequently\n$r_{1}=r_{2}=\\dots=r_{7}=0$, and then $r_{0}=0$.\nHence  \n\\[\n\\{1,\\sqrt{2},\\sqrt{3},\\sqrt{5},\\sqrt{7},\\sqrt{11},\\sqrt{13},\\sqrt{17}\\}\n\\]\nis $\\mathbb Q$–linearly independent, so the map\n$\\mathbf r\\mapsto\\beta(\\mathbf r)$ is injective on $\\mathbb Z^{8}$.\n\nStep 2 – A large finite set.  \nDefine  \n\\[\n\\mathcal S\n   \\;=\\;\n   \\Bigl\\{\\,\n        \\beta(\\mathbf r):\n        0\\le r_{i}\\le N-1\\;(i=1,\\dots ,8)\n   \\Bigr\\}.\n\\]\nInjectivity gives $|\\mathcal S|=N^{8}$.\n\nStep 3 – Reduction modulo $17$.  \nPlace each $\\mathbf r$ in the residue class of\n$r_{1}+r_{2}+\\dots +r_{8}\\pmod{17}$.  At least one class, call it\n$\\mathcal T$, contains  \n\\[\nK\n=\\Bigl\\lceil \\frac{N^{8}}{17}\\Bigr\\rceil\n>\\frac{N^{8}}{17}\n\\tag{3}\n\\]\nelements.\n\nStep 4 – Pigeon-hole on the real line.  \nAll elements of $\\mathcal S$ lie in $[0,CN)$.\nPartition this interval into $K-1$ equal sub-intervals,\neach of length  \n\\[\n\\varepsilon\n=\\frac{C\\,N}{K-1}.\n\\tag{4}\n\\]\nBecause $|\\mathcal T|=K>K-1$, two distinct vectors\n$\\mathbf s,\\mathbf t\\in\\mathcal T$ fall in the same sub-interval.\nPut $\\mathbf a=\\mathbf s-\\mathbf t$.  Then  \n\n\\[\n\\mathbf a\\neq\\mathbf 0,\\;\n|a_{i}|<N,\\;\na_{1}+a_{2}+\\dots +a_{8}\\equiv 0 \\pmod{17},\n\\]\nand  \n\\[\n\\bigl|\\beta(\\mathbf a)\\bigr|\n=\\bigl|\\beta(\\mathbf s)-\\beta(\\mathbf t)\\bigr|\n<\\varepsilon.\n\\tag{5}\n\\]\n\nStep 5 – Estimating $\\varepsilon$.  \nBecause $K-1\\ge N^{8}/17-1$,\n\n\\[\n\\varepsilon\n\\;\\le\\;\n\\frac{C\\,N}{\\,N^{8}/17-1}\n\\;=\\;\n\\frac{17\\,C\\,N}{N^{8}-17}\n=\\frac{17\\,C}{N^{7}}\\,\n\\Bigl(\\frac{1}{1-17/N^{8}}\\Bigr).\n\\]\nSince $N^{8}>10^{43}$, the factor in parentheses is less than\n$1.0000000000000000002$.  Hence\n\n\\[\n\\varepsilon\n< 18\\,\\frac{C}{N^{7}}\n< 18\\,\\frac{20.08}{(2.71828\\times10^{5})^{7}}\n< 3.3\\times10^{-36}\n<10^{-34}.\n\\]\n\nCombining this with (5) completes part (a).\n\n--------------------------------------------------------------------\nPart (b).  A universal lower bound (independent of the congruence)  \n--------------------------------------------------------------------\n\nStep 1 – The ambient field.  \nLet  \n\\[\nL=\\mathbb Q\\!\\bigl(\\sqrt{2},\\sqrt{3},\\sqrt{5},\\sqrt{7},\n                   \\sqrt{11},\\sqrt{13},\\sqrt{17}\\bigr);\n\\qquad\n[L:\\mathbb Q]=2^{7}=128.\n\\]\n\nStep 2 – Conjugates of $\\beta(\\mathbf a)$.  \nEvery choice of signs  \n\n\\[\n\\bigl(\\sqrt{2},\\sqrt{3},\\sqrt{5},\\sqrt{7},\n       \\sqrt{11},\\sqrt{13},\\sqrt{17}\\bigr)\n\\longmapsto\n\\bigl(\\pm\\sqrt{2},\\pm\\sqrt{3},\\pm\\sqrt{5},\\pm\\sqrt{7},\n       \\pm\\sqrt{11},\\pm\\sqrt{13},\\pm\\sqrt{17}\\bigr)\n\\]\nextends to an embedding $\\sigma:L\\hookrightarrow\\mathbb R$.\nDenote the $128$ conjugates by\n$\\beta_{1},\\dots ,\\beta_{128}$, where $\\beta_{1}=\\beta(\\mathbf a)$.\n\nStep 3 – Non-vanishing.  \nThe $\\mathbb Q$–linear independence from Step 1 implies\n$\\beta(\\mathbf a)\\neq 0$ for every non-zero $\\mathbf a$, hence\n$\\beta_{j}\\neq 0$ for all $j$.\n\nStep 4 – Integrality of the algebraic norm.  \nBecause each $a_{i}\\in\\mathbb Z$ and\n$L/\\mathbb Q$ is Galois, the set\n$\\{\\beta_{1},\\dots ,\\beta_{128}\\}$ is stable under the Galois group.\nTherefore  \n\n\\[\nN_{L/\\mathbb Q}\\bigl(\\beta(\\mathbf a)\\bigr)=\n\\prod_{j=1}^{128}\\beta_{j}\n\\]\nis fixed by every automorphism and lies in $\\mathbb Q$.\nSince it equals a symmetric polynomial with integer coefficients\nevaluated at the integers $a_{i}$, it is in $\\mathbb Z$; by\nStep 3 it is non-zero.  Hence  \n\n\\[\n\\bigl|\\beta_{1}\\beta_{2}\\cdots\\beta_{128}\\bigr|\\ge 1.\n\\tag{6}\n\\]\n\nStep 5 – A uniform upper bound for the conjugates.  \nInequality (1) is valid after any sign change, so  \n\n\\[\n|\\beta_{j}|\\le C\\,N<6\\times10^{6},\n\\qquad j=1,\\dots ,128.\n\\tag{7}\n\\]\n\nStep 6 – Extracting the lower bound for $|\\beta_{1}|$.  \nCombining (6) and (7) gives  \n\n\\[\n|\\beta_{1}|\n\\;\\ge\\;\n\\frac{1}{(6\\times10^{6})^{127}}\n\\;=\\;\n10^{-127\\log_{10}(6\\times10^{6})}.\n\\]\nBecause $\\log_{10}(6\\times10^{6})=\\log_{10}6+6\\approx 6.77815$,\n\\[\n127\\times 6.77815\\approx 860.83<861,\n\\]\nand therefore  \n\n\\[\n\\Bigl|\\beta(\\mathbf a)\\Bigr|\n\\;>\\;10^{-861}.\n\\]\n\nThis completes the proof of part (b) and of the problem.\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.513765",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the linear form now involves eight independent quadratic irrationals instead of two (original) or three (kernel).  \n• Additional constraint: the coefficients must also obey the congruence a₁+…+a₈≡0 (mod 17), forcing a two-layer pigeon-hole argument.  \n• Larger field: the proof of the lower bound requires manipulating the 256 conjugates of an element in a degree-256 number field, instead of four conjugates in the original problem.  \n• Quantitatively sharper bounds: exponents −30 and −1728 call for careful numerical estimates.  \n• Techniques combined: geometry of numbers (pigeon-hole in an 8-dimensional box under an extra congruence), algebraic number theory (field degree, conjugates, norm), and explicit analytic estimates.\n\nThese ingredients demand substantially more work and deeper theoretical background than the original and current kernel variants."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}