{
  "index": "1947-B-5",
  "type": "ALG",
  "tag": [
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "11. \\( a, b, c, d \\) are distinct integers such that\n\\[\n(x-a)(x-b)(x-c)(x-d)-4=0\n\\]\nhas an integral root \\( r \\). Show that \\( 4 r=a+b+c+d \\).",
  "solution": "Solution. Since \\( r \\) is a root,\n\\[\n(r-a)(r-b)(r-c)(r-d)=4\n\\]\nand since \\( a, b, c, d \\) are distinct integers, \\( r-a, r-b, r-c, r-d \\) are distinct integers whose product is 4 . But the only set of four distinct integers whose product is 4 is the set \\( \\{1,-1,2,-2\\} \\). Hence\n\\[\n(r-a)+(r-b)+(r-c)+(r-d)=1-1+2-2=0\n\\]\nso\n\\[\n4 r=a+b+c+d\n\\]",
  "vars": [
    "x",
    "r"
  ],
  "params": [
    "a",
    "b",
    "c",
    "d"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "unknownx",
        "r": "rootval",
        "a": "constanta",
        "b": "constantb",
        "c": "constantc",
        "d": "constantd"
      },
      "question": "11. \\( \\mathrm{constanta}, \\mathrm{constantb}, \\mathrm{constantc}, \\mathrm{constantd} \\) are distinct integers such that\n\\[\n(\\mathrm{unknownx}-\\mathrm{constanta})(\\mathrm{unknownx}-\\mathrm{constantb})(\\mathrm{unknownx}-\\mathrm{constantc})(\\mathrm{unknownx}-\\mathrm{constantd})-4=0\n\\]\nhas an integral root \\( \\mathrm{rootval} \\). Show that \\( 4 \\mathrm{rootval}=\\mathrm{constanta}+\\mathrm{constantb}+\\mathrm{constantc}+\\mathrm{constantd} \\).",
      "solution": "Solution. Since \\( \\mathrm{rootval} \\) is a root,\n\\[\n(\\mathrm{rootval}-\\mathrm{constanta})(\\mathrm{rootval}-\\mathrm{constantb})(\\mathrm{rootval}-\\mathrm{constantc})(\\mathrm{rootval}-\\mathrm{constantd})=4\n\\]\nand since \\( \\mathrm{constanta}, \\mathrm{constantb}, \\mathrm{constantc}, \\mathrm{constantd} \\) are distinct integers, \\( \\mathrm{rootval}-\\mathrm{constanta}, \\mathrm{rootval}-\\mathrm{constantb}, \\mathrm{rootval}-\\mathrm{constantc}, \\mathrm{rootval}-\\mathrm{constantd} \\) are distinct integers whose product is 4. But the only set of four distinct integers whose product is 4 is the set \\( \\{1,-1,2,-2\\} \\). Hence\n\\[\n(\\mathrm{rootval}-\\mathrm{constanta})+(\\mathrm{rootval}-\\mathrm{constantb})+(\\mathrm{rootval}-\\mathrm{constantc})+(\\mathrm{rootval}-\\mathrm{constantd})=1-1+2-2=0\n\\]\nso\n\\[\n4 \\mathrm{rootval}=\\mathrm{constanta}+\\mathrm{constantb}+\\mathrm{constantc}+\\mathrm{constantd}\n\\]\n."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineapple",
        "r": "trapezoid",
        "a": "suitcase",
        "b": "marigold",
        "c": "lighthouse",
        "d": "harmonica"
      },
      "question": "11. \\( suitcase, marigold, lighthouse, harmonica \\) are distinct integers such that\n\\[\n(pineapple-suitcase)(pineapple-marigold)(pineapple-lighthouse)(pineapple-harmonica)-4=0\n\\]\nhas an integral root \\( trapezoid \\). Show that \\( 4 trapezoid=suitcase+marigold+lighthouse+harmonica \\).",
      "solution": "Solution. Since \\( trapezoid \\) is a root,\n\\[\n(trapezoid-suitcase)(trapezoid-marigold)(trapezoid-lighthouse)(trapezoid-harmonica)=4\n\\]\nand since \\( suitcase, marigold, lighthouse, harmonica \\) are distinct integers, \\( trapezoid-suitcase, trapezoid-marigold, trapezoid-lighthouse, trapezoid-harmonica \\) are distinct integers whose product is 4 . But the only set of four distinct integers whose product is 4 is the set \\{1,-1,2,-2\\}. Hence\n\\[\n(trapezoid-suitcase)+(trapezoid-marigold)+(trapezoid-lighthouse)+(trapezoid-harmonica)=1-1+2-2=0\n\\]\nso\n\\[\n4 trapezoid=suitcase+marigold+lighthouse+harmonica\n\\]\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "r": "leafpoint",
        "a": "variableone",
        "b": "variabletwo",
        "c": "variablethree",
        "d": "variablefour"
      },
      "question": "Problem:\n<<<\n11. \\( variableone, variabletwo, variablethree, variablefour \\) are distinct integers such that\n\\[\n(constantval-variableone)(constantval-variabletwo)(constantval-variablethree)(constantval-variablefour)-4=0\n\\]\nhas an integral root \\( leafpoint \\). Show that \\( 4 leafpoint=variableone+variabletwo+variablethree+variablefour \\).\n>>>",
      "solution": "Solution:\n<<<\nSolution. Since \\( leafpoint \\) is a root,\n\\[\n(leafpoint-variableone)(leafpoint-variabletwo)(leafpoint-variablethree)(leafpoint-variablefour)=4\n\\]\nand since \\( variableone, variabletwo, variablethree, variablefour \\) are distinct integers, \\( leafpoint-variableone, leafpoint-variabletwo, leafpoint-variablethree, leafpoint-variablefour \\) are distinct integers whose product is 4 . But the only set of four distinct integers whose product is 4 is the set \\{1,-1,2,-2\\}. Hence\n\\[\n(leafpoint-variableone)+(leafpoint-variabletwo)+(leafpoint-variablethree)+(leafpoint-variablefour)=1-1+2-2=0\n\\]\nso\n\\[\n4 leafpoint=variableone+variabletwo+variablethree+variablefour\n\\]\n>>>\n"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "r": "hjgrksla",
        "a": "povxtrnea",
        "b": "uyarqinfp",
        "c": "ldwsojemn",
        "d": "tgkmbzcia"
      },
      "question": "11. \\( povxtrnea, uyarqinfp, ldwsojemn, tgkmbzcia \\) are distinct integers such that\n\\[\n(qzxwvtnp-povxtrnea)(qzxwvtnp-uyarqinfp)(qzxwvtnp-ldwsojemn)(qzxwvtnp-tgkmbzcia)-4=0\n\\]\nhas an integral root \\( hjgrksla \\). Show that \\( 4 hjgrksla=povxtrnea+uyarqinfp+ldwsojemn+tgkmbzcia \\).",
      "solution": "Solution. Since \\( hjgrksla \\) is a root,\n\\[\n(hjgrksla-povxtrnea)(hjgrksla-uyarqinfp)(hjgrksla-ldwsojemn)(hjgrksla-tgkmbzcia)=4\n\\]\nand since \\( povxtrnea, uyarqinfp, ldwsojemn, tgkmbzcia \\) are distinct integers, \\( hjgrksla-povxtrnea, hjgrksla-uyarqinfp, hjgrksla-ldwsojemn, hjgrksla-tgkmbzcia \\) are distinct integers whose product is 4. But the only set of four distinct integers whose product is 4 is the set \\( \\{1,-1,2,-2\\} \\). Hence\n\\[\n(hjgrksla-povxtrnea)+(hjgrksla-uyarqinfp)+(hjgrksla-ldwsojemn)+(hjgrksla-tgkmbzcia)=1-1+2-2=0\n\\]\nso\n\\[\n4 hjgrksla=povxtrnea+uyarqinfp+ldwsojemn+tgkmbzcia\n\\]"
    },
    "kernel_variant": {
      "question": "Let n ≥ 4 be an even positive integer and let  \na₁ , a₂ , … , aₙ be n pair-wise distinct integers.  \nDefine  \n\n  Pₙ(x) = (x – a₁)(x – a₂)⋯(x – aₙ) – 2^{\\,n–2} .\n\nAssume that Pₙ(x) possesses an integral root r; i.e.  \n\n  (r – a₁)(r – a₂)⋯(r – aₙ) = 2^{\\,n–2}.        (★)\n\n(a) Prove that necessarily n = 4.\n\n(b) Show that, after a suitable re-ordering,\n\n  { a₁ , a₂ , a₃ , a₄ } = { r – 2 , r – 1 , r + 1 , r + 2 } ,\n\nand therefore a₁ + a₂ + a₃ + a₄ = 4r.\n\n(c) Conclude that for every even n ≠ 4 the equation Pₙ(x)=0 has no integral solutions.",
      "solution": "Step 1.  Structure of the factors (r – aᵢ).  \nBecause the right–hand side of (★) is a power of 2, every factor r – aᵢ is an integer whose prime divisors are all equal to 2; hence each factor can be written uniquely in the form  \n\n  r – aᵢ = εᵢ · 2^{eᵢ},            εᵢ ∈ {–1, +1}, eᵢ ∈ ℕ₀.        (1)\n\nSince the aᵢ are distinct, the signed numbers εᵢ·2^{eᵢ} are distinct as well.\n\nLet S = {e₁, e₂, … , eₙ}.  Relation (★) gives  \n\n  ∑_{i=1}^{n} eᵢ = n – 2.                                     (2)\n\nStep 2.  A lower bound for  ∑ eᵢ  when n ≥ 6.  \nWrite n = 2m (m ≥ 2).  Because the numbers in (1) are distinct, at most two factors can share the same absolute value (they would have opposite signs).  To minimise the sum of the exponents while keeping 2m distinct signed powers of two, we take\n\n  ±1 ( exponent 0 ),  \n  ±2 ( exponent 1 ),  \n  ±4 ( exponent 2 ), … ,  \n  ±2^{m–1} ( exponent m–1 ).\n\nHence the minimal possible value of the left–hand side of (2) is\n\n  Σ₍min₎(n) = 2·(0 + 1 + 2 + … + (m – 1))  \n     = 2 · [ m(m – 1)/2 ] = m(m – 1).                     (3)\n\nStep 3.  Impossibility when n ≥ 6.  \nFor m ≥ 3 (i.e. n ≥ 6) we have\n\n  Σ₍min₎(n) = m(m – 1) ≥ 3·2 = 6 > 2m – 2 = n – 2,\n\ncontradicting (2).  Consequently (★) is impossible for n ≥ 6.\n\nThus n can only be 4.  From now on put n = 4 (hence m = 2).\n\nStep 4.  Determining the exponents when n = 4.  \nEquation (2) now reads e₁ + e₂ + e₃ + e₄ = 2.  \nBecause the eᵢ are non-negative integers and the signed 2^{eᵢ} are distinct, the only possibility is\n\n  {e₁, e₂, e₃, e₄} = {0, 0, 1, 1}.                           (4)\n\nStep 5.  Determining the signs.  \nTo obtain a positive product 2^{2} = 4, we need an even number of negative factors.  \nWith the multiset (4) there are exactly two ways:\n\n  { +1, –1, +2, –2 } or { +1, –1, –2, +2 }.\n\nIn either case the unordered set of factors is\n\n  { –2, –1, 1, 2 }.                                           (5)\n\nStep 6.  Expressing the aᵢ.  \nFrom (1) and (5) we have\n\n  { r – a₁ , r – a₂ , r – a₃ , r – a₄ } = { –2, –1, 1, 2 }.\n\nAdding r to each element yields\n\n  { a₁ , a₂ , a₃ , a₄ } = { r – 2, r – 1, r + 1, r + 2 }.    (6)\n\nStep 7.  Conclusion for n = 4.  \nSumming the four equalities in (6) gives\n\n  a₁ + a₂ + a₃ + a₄ = (r – 2) + (r – 1) + (r + 1) + (r + 2) = 4r.    (7)\n\nTherefore every admissible quadruple consists of the four consecutive integers r – 2, r – 1, r + 1, r + 2 (in any order), and conversely any such quadruple indeed satisfies (★).  This completes parts (a) and (b).\n\nStep 8.  Non-existence for the remaining even n.  \nPart (a) already shows that no solution is possible when n ≠ 4 (with n even).  Hence Pₙ(x)=0 has no integral roots for all even n ≥ 6, proving part (c).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.413905",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension / more variables  \n   • The problem is extended from 4 unknown integers to an arbitrary even number n of unknowns.  \n   • One must treat all n simultaneously and then single out the only viable case n = 4.\n\n2. Additional constraints  \n   • The constant term is now 2^{n–2}, forcing every factor to be a signed power of 2 and introducing exponent calculations.\n\n3. More sophisticated structures  \n   • The argument involves combinatorial optimisation (minimal sum of exponents) and parity considerations on signed powers of two.\n\n4. Deeper theoretical requirements  \n   • One needs to combine number-theoretic factor-structure analysis with inequalities in order to rule out large classes of n in a single stroke.\n\n5. Multiple interacting concepts  \n   • Distinctness of the aᵢ leads to uniqueness of signed powers of two;  \n   • Prime-power factorisation controls the magnitude of exponents;  \n   • A counting argument (Step 2) links combinatorics with Diophantine constraints;  \n   • Only after this global obstruction is settled does one descend to the classical quartic situation and recover the original identity.\n\nBecause it demands an argument valid for arbitrarily large n, careful optimisation, and a two-tiered proof (first ruling out n ≥ 6, then characterising n = 4), this enhanced variant is far more intricate than the original problem, which needed only a short case-check on four small factors."
      }
    },
    "original_kernel_variant": {
      "question": "Let n ≥ 4 be an even positive integer and let  \na₁ , a₂ , … , aₙ be n pair-wise distinct integers.  \nDefine  \n\n  Pₙ(x) = (x – a₁)(x – a₂)⋯(x – aₙ) – 2^{\\,n–2} .\n\nAssume that Pₙ(x) possesses an integral root r; i.e.  \n\n  (r – a₁)(r – a₂)⋯(r – aₙ) = 2^{\\,n–2}.        (★)\n\n(a) Prove that necessarily n = 4.\n\n(b) Show that, after a suitable re-ordering,\n\n  { a₁ , a₂ , a₃ , a₄ } = { r – 2 , r – 1 , r + 1 , r + 2 } ,\n\nand therefore a₁ + a₂ + a₃ + a₄ = 4r.\n\n(c) Conclude that for every even n ≠ 4 the equation Pₙ(x)=0 has no integral solutions.",
      "solution": "Step 1.  Structure of the factors (r – aᵢ).  \nBecause the right–hand side of (★) is a power of 2, every factor r – aᵢ is an integer whose prime divisors are all equal to 2; hence each factor can be written uniquely in the form  \n\n  r – aᵢ = εᵢ · 2^{eᵢ},            εᵢ ∈ {–1, +1}, eᵢ ∈ ℕ₀.        (1)\n\nSince the aᵢ are distinct, the signed numbers εᵢ·2^{eᵢ} are distinct as well.\n\nLet S = {e₁, e₂, … , eₙ}.  Relation (★) gives  \n\n  ∑_{i=1}^{n} eᵢ = n – 2.                                     (2)\n\nStep 2.  A lower bound for  ∑ eᵢ  when n ≥ 6.  \nWrite n = 2m (m ≥ 2).  Because the numbers in (1) are distinct, at most two factors can share the same absolute value (they would have opposite signs).  To minimise the sum of the exponents while keeping 2m distinct signed powers of two, we take\n\n  ±1 ( exponent 0 ),  \n  ±2 ( exponent 1 ),  \n  ±4 ( exponent 2 ), … ,  \n  ±2^{m–1} ( exponent m–1 ).\n\nHence the minimal possible value of the left–hand side of (2) is\n\n  Σ₍min₎(n) = 2·(0 + 1 + 2 + … + (m – 1))  \n     = 2 · [ m(m – 1)/2 ] = m(m – 1).                     (3)\n\nStep 3.  Impossibility when n ≥ 6.  \nFor m ≥ 3 (i.e. n ≥ 6) we have\n\n  Σ₍min₎(n) = m(m – 1) ≥ 3·2 = 6 > 2m – 2 = n – 2,\n\ncontradicting (2).  Consequently (★) is impossible for n ≥ 6.\n\nThus n can only be 4.  From now on put n = 4 (hence m = 2).\n\nStep 4.  Determining the exponents when n = 4.  \nEquation (2) now reads e₁ + e₂ + e₃ + e₄ = 2.  \nBecause the eᵢ are non-negative integers and the signed 2^{eᵢ} are distinct, the only possibility is\n\n  {e₁, e₂, e₃, e₄} = {0, 0, 1, 1}.                           (4)\n\nStep 5.  Determining the signs.  \nTo obtain a positive product 2^{2} = 4, we need an even number of negative factors.  \nWith the multiset (4) there are exactly two ways:\n\n  { +1, –1, +2, –2 } or { +1, –1, –2, +2 }.\n\nIn either case the unordered set of factors is\n\n  { –2, –1, 1, 2 }.                                           (5)\n\nStep 6.  Expressing the aᵢ.  \nFrom (1) and (5) we have\n\n  { r – a₁ , r – a₂ , r – a₃ , r – a₄ } = { –2, –1, 1, 2 }.\n\nAdding r to each element yields\n\n  { a₁ , a₂ , a₃ , a₄ } = { r – 2, r – 1, r + 1, r + 2 }.    (6)\n\nStep 7.  Conclusion for n = 4.  \nSumming the four equalities in (6) gives\n\n  a₁ + a₂ + a₃ + a₄ = (r – 2) + (r – 1) + (r + 1) + (r + 2) = 4r.    (7)\n\nTherefore every admissible quadruple consists of the four consecutive integers r – 2, r – 1, r + 1, r + 2 (in any order), and conversely any such quadruple indeed satisfies (★).  This completes parts (a) and (b).\n\nStep 8.  Non-existence for the remaining even n.  \nPart (a) already shows that no solution is possible when n ≠ 4 (with n even).  Hence Pₙ(x)=0 has no integral roots for all even n ≥ 6, proving part (c).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.356373",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension / more variables  \n   • The problem is extended from 4 unknown integers to an arbitrary even number n of unknowns.  \n   • One must treat all n simultaneously and then single out the only viable case n = 4.\n\n2. Additional constraints  \n   • The constant term is now 2^{n–2}, forcing every factor to be a signed power of 2 and introducing exponent calculations.\n\n3. More sophisticated structures  \n   • The argument involves combinatorial optimisation (minimal sum of exponents) and parity considerations on signed powers of two.\n\n4. Deeper theoretical requirements  \n   • One needs to combine number-theoretic factor-structure analysis with inequalities in order to rule out large classes of n in a single stroke.\n\n5. Multiple interacting concepts  \n   • Distinctness of the aᵢ leads to uniqueness of signed powers of two;  \n   • Prime-power factorisation controls the magnitude of exponents;  \n   • A counting argument (Step 2) links combinatorics with Diophantine constraints;  \n   • Only after this global obstruction is settled does one descend to the classical quartic situation and recover the original identity.\n\nBecause it demands an argument valid for arbitrarily large n, careful optimisation, and a two-tiered proof (first ruling out n ≥ 6, then characterising n = 4), this enhanced variant is far more intricate than the original problem, which needed only a short case-check on four small factors."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}