{
  "index": "1955-A-1",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "1. Prove that there is no set of integers \\( m, n, p \\) except \\( 0,0,0 \\) for which \\( m \\) \\( +n \\sqrt{2}+p \\sqrt{3}=0 \\).",
  "solution": "Solution. We recall that (1) if \\( a \\) is a positive integer and \\( \\sqrt{a} \\) is not an integer, then \\( \\sqrt{a} \\) is irrational.\n\nSuppose that \\( m, n, p \\) are integers such that\n\\[\nm+n \\sqrt{2}+p \\sqrt{3}=0 .\n\\]\n\nIf both \\( n \\) and \\( p \\) are zero, so is \\( m \\). If just one of \\( n \\) and \\( p \\) is zero, we have either \\( \\sqrt{2}=-m / n \\) or \\( \\sqrt{3}=-m / p \\), both contrary to (1). If neither \\( n \\) nor \\( p \\) is zero, then\n\\[\nm^{2}=(n \\sqrt{2}+p \\sqrt{3})^{2}=2 n^{2}+3 p^{2}+2 n p \\sqrt{6}\n\\]\nand\n\\[\n\\sqrt{6}=\\left(m^{2}-2 n^{2}-3 p^{2}\\right) / 2 n p\n\\]\nagain contrary to (1). So the only integer triplet for which (2) is true is \\( m=0, n=0, p=0 \\).\n\nFor completeness we include a proof of (1). Suppose \\( a \\) is a positive integer and \\( \\sqrt{a}=b / c \\) where \\( b \\) and \\( c \\) are positive integers. We may assume \\( b \\) and \\( c \\) are relatively prime. Then \\( a c^{2}=b^{2} \\). Consider a prime \\( q \\) that divides \\( c \\). Then \\( q \\) divides \\( b^{2} \\) and hence \\( b \\); so if there is such a prime \\( q \\), then \\( b \\) and \\( c \\) have a common factor, which is a contradiction Hence \\( c \\) has no prime factors, so \\( c=1 \\), and \\( \\sqrt{a}=b \\), an integer.",
  "vars": [
    "m",
    "n",
    "p"
  ],
  "params": [
    "a",
    "b",
    "c",
    "q"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "m": "integerm",
        "n": "integern",
        "p": "integerp",
        "a": "parama",
        "b": "paramb",
        "c": "paramc",
        "q": "primeq"
      },
      "question": "1. Prove that there is no set of integers \\( integerm, integern, integerp \\) except \\( 0,0,0 \\) for which \\( integerm +integern \\sqrt{2}+integerp \\sqrt{3}=0 \\).",
      "solution": "Solution. We recall that (1) if \\( parama \\) is a positive integer and \\( \\sqrt{parama} \\) is not an integer, then \\( \\sqrt{parama} \\) is irrational.\n\nSuppose that \\( integerm, integern, integerp \\) are integers such that\n\\[\nintegerm+integern \\sqrt{2}+integerp \\sqrt{3}=0 .\n\\]\n\nIf both \\( integern \\) and \\( integerp \\) are zero, so is \\( integerm \\). If just one of \\( integern \\) and \\( integerp \\) is zero, we have either \\( \\sqrt{2}=-integerm / integern \\) or \\( \\sqrt{3}=-integerm / integerp \\), both contrary to (1). If neither \\( integern \\) nor \\( integerp \\) is zero, then\n\\[\nintegerm^{2}=(integern \\sqrt{2}+integerp \\sqrt{3})^{2}=2 integern^{2}+3 integerp^{2}+2 integern integerp \\sqrt{6}\n\\]\nand\n\\[\n\\sqrt{6}=\\left(integerm^{2}-2 integern^{2}-3 integerp^{2}\\right) / 2 integern integerp\n\\]\nagain contrary to (1). So the only integer triplet for which (2) is true is \\( integerm=0, integern=0, integerp=0 \\).\n\nFor completeness we include a proof of (1). Suppose \\( parama \\) is a positive integer and \\( \\sqrt{parama}=paramb / paramc \\) where \\( paramb \\) and \\( paramc \\) are positive integers. We may assume \\( paramb \\) and \\( paramc \\) are relatively prime. Then \\( parama paramc^{2}=paramb^{2} \\). Consider a prime \\( primeq \\) that divides \\( paramc \\). Then \\( primeq \\) divides \\( paramb^{2} \\) and hence \\( paramb \\); so if there is such a prime \\( primeq \\), then \\( paramb \\) and \\( paramc \\) have a common factor, which is a contradiction Hence \\( paramc \\) has no prime factors, so \\( paramc=1 \\), and \\( \\sqrt{parama}=paramb \\), an integer."
    },
    "descriptive_long_confusing": {
      "map": {
        "m": "pineapple",
        "n": "suitcase",
        "p": "classroom",
        "a": "elephant",
        "b": "hamburger",
        "c": "butterfly",
        "q": "strawberry"
      },
      "question": "1. Prove that there is no set of integers \\( pineapple, suitcase, classroom \\) except \\( 0,0,0 \\) for which \\( pineapple +suitcase \\sqrt{2}+classroom \\sqrt{3}=0 \\).",
      "solution": "Solution. We recall that (1) if \\( elephant \\) is a positive integer and \\( \\sqrt{elephant} \\) is not an integer, then \\( \\sqrt{elephant} \\) is irrational.\n\nSuppose that \\( pineapple, suitcase, classroom \\) are integers such that\n\\[\npineapple+suitcase \\sqrt{2}+classroom \\sqrt{3}=0 .\n\\]\n\nIf both \\( suitcase \\) and \\( classroom \\) are zero, so is \\( pineapple \\). If just one of \\( suitcase \\) and \\( classroom \\) is zero, we have either \\( \\sqrt{2}=-pineapple / suitcase \\) or \\( \\sqrt{3}=-pineapple / classroom \\), both contrary to (1). If neither \\( suitcase \\) nor \\( classroom \\) is zero, then\n\\[\npineapple^{2}=(suitcase \\sqrt{2}+classroom \\sqrt{3})^{2}=2 suitcase^{2}+3 classroom^{2}+2 suitcase classroom \\sqrt{6}\n\\]\nand\n\\[\n\\sqrt{6}=\\left(pineapple^{2}-2 suitcase^{2}-3 classroom^{2}\\right) / 2 suitcase classroom\n\\]\nagain contrary to (1). So the only integer triplet for which (2) is true is \\( pineapple=0, suitcase=0, classroom=0 \\).\n\nFor completeness we include a proof of (1). Suppose \\( elephant \\) is a positive integer and \\( \\sqrt{elephant}=hamburger / butterfly \\) where \\( hamburger \\) and \\( butterfly \\) are positive integers. We may assume \\( hamburger \\) and \\( butterfly \\) are relatively prime. Then \\( elephant \\, butterfly^{2}=hamburger^{2} \\). Consider a prime \\( strawberry \\) that divides \\( butterfly \\). Then \\( strawberry \\) divides \\( hamburger^{2} \\) and hence \\( hamburger \\); so if there is such a prime \\( strawberry \\), then \\( hamburger \\) and \\( butterfly \\) have a common factor, which is a contradiction Hence \\( butterfly \\) has no prime factors, so \\( butterfly=1 \\), and \\( \\sqrt{elephant}=hamburger \\), an integer."
    },
    "descriptive_long_misleading": {
      "map": {
        "m": "knownvalue",
        "n": "constant",
        "p": "immutable",
        "a": "lastvalue",
        "b": "endvalue",
        "c": "finishline",
        "q": "composite"
      },
      "question": "1. Prove that there is no set of integers \\( knownvalue, constant, immutable \\) except \\( 0,0,0 \\) for which \\( knownvalue + constant \\sqrt{2}+ immutable \\sqrt{3}=0 \\).",
      "solution": "Solution. We recall that (1) if \\( lastvalue \\) is a positive integer and \\( \\sqrt{lastvalue} \\) is not an integer, then \\( \\sqrt{lastvalue} \\) is irrational.\n\nSuppose that \\( knownvalue, constant, immutable \\) are integers such that\n\\[\nknownvalue+constant \\sqrt{2}+immutable \\sqrt{3}=0 .\n\\]\n\nIf both \\( constant \\) and \\( immutable \\) are zero, so is \\( knownvalue \\). If just one of \\( constant \\) and \\( immutable \\) is zero, we have either \\( \\sqrt{2}=-knownvalue / constant \\) or \\( \\sqrt{3}=-knownvalue / immutable \\), both contrary to (1). If neither \\( constant \\) nor \\( immutable \\) is zero, then\n\\[\nknownvalue^{2}=(constant \\sqrt{2}+immutable \\sqrt{3})^{2}=2 constant^{2}+3 immutable^{2}+2 constant immutable \\sqrt{6}\n\\]\nand\n\\[\n\\sqrt{6}=\\left(knownvalue^{2}-2 constant^{2}-3 immutable^{2}\\right) / 2 constant immutable\n\\]\nagain contrary to (1). So the only integer triplet for which (2) is true is \\( knownvalue=0, constant=0, immutable=0 \\).\n\nFor completeness we include a proof of (1). Suppose \\( lastvalue \\) is a positive integer and \\( \\sqrt{lastvalue}=endvalue / finishline \\) where \\( endvalue \\) and \\( finishline \\) are positive integers. We may assume \\( endvalue \\) and \\( finishline \\) are relatively prime. Then \\( lastvalue finishline^{2}=endvalue^{2} \\). Consider a prime \\( composite \\) that divides \\( finishline \\). Then \\( composite \\) divides \\( endvalue^{2} \\) and hence \\( endvalue \\); so if there is such a prime \\( composite \\), then \\( endvalue \\) and \\( finishline \\) have a common factor, which is a contradiction Hence \\( finishline \\) has no prime factors, so \\( finishline=1 \\), and \\( \\sqrt{lastvalue}=endvalue \\), an integer."
    },
    "garbled_string": {
      "map": {
        "m": "qzxwvtnp",
        "n": "hjgrksla",
        "p": "ctvysqre",
        "a": "lpmzoqnf",
        "b": "xkierwut",
        "c": "dafmqush",
        "q": "rovyhpxn"
      },
      "question": "1. Prove that there is no set of integers \\( qzxwvtnp, hjgrksla, ctvysqre \\) except \\( 0,0,0 \\) for which \\( qzxwvtnp +hjgrksla \\sqrt{2}+ctvysqre \\sqrt{3}=0 \\).",
      "solution": "Solution. We recall that (1) if \\( lpmzoqnf \\) is a positive integer and \\( \\sqrt{lpmzoqnf} \\) is not an integer, then \\( \\sqrt{lpmzoqnf} \\) is irrational.\n\nSuppose that \\( qzxwvtnp, hjgrksla, ctvysqre \\) are integers such that\n\\[\nqzxwvtnp+hjgrksla \\sqrt{2}+ctvysqre \\sqrt{3}=0 .\n\\]\n\nIf both \\( hjgrksla \\) and \\( ctvysqre \\) are zero, so is \\( qzxwvtnp \\). If just one of \\( hjgrksla \\) and \\( ctvysqre \\) is zero, we have either \\( \\sqrt{2}=-qzxwvtnp / hjgrksla \\) or \\( \\sqrt{3}=-qzxwvtnp / ctvysqre \\), both contrary to (1). If neither \\( hjgrksla \\) nor \\( ctvysqre \\) is zero, then\n\\[\nqzxwvtnp^{2}=(hjgrksla \\sqrt{2}+ctvysqre \\sqrt{3})^{2}=2 hjgrksla^{2}+3 ctvysqre^{2}+2 hjgrksla \\, ctvysqre \\sqrt{6}\n\\]\nand\n\\[\n\\sqrt{6}=\\left(qzxwvtnp^{2}-2 hjgrksla^{2}-3 ctvysqre^{2}\\right) / 2 hjgrksla ctvysqre\n\\]\nagain contrary to (1). So the only integer triplet for which (2) is true is \\( qzxwvtnp=0, hjgrksla=0, ctvysqre=0 \\).\n\nFor completeness we include a proof of (1). Suppose \\( lpmzoqnf \\) is a positive integer and \\( \\sqrt{lpmzoqnf}=xkierwut / dafmqush \\) where \\( xkierwut \\) and \\( dafmqush \\) are positive integers. We may assume \\( xkierwut \\) and \\( dafmqush \\) are relatively prime. Then \\( lpmzoqnf \\, dafmqush^{2}=xkierwut^{2} \\). Consider a prime \\( rovyhpxn \\) that divides \\( dafmqush \\). Then \\( rovyhpxn \\) divides \\( xkierwut^{2} \\) and hence \\( xkierwut \\); so if there is such a prime \\( rovyhpxn \\), then \\( xkierwut \\) and \\( dafmqush \\) have a common factor, which is a contradiction. Hence \\( dafmqush \\) has no prime factors, so \\( dafmqush=1 \\), and \\( \\sqrt{lpmzoqnf}=xkierwut \\), an integer."
    },
    "kernel_variant": {
      "question": "Let 𝑚, 𝑛, 𝑝, 𝑞 be integers satisfying  \n  𝑚 + 𝑛 √2 + 𝑝 √3 + 𝑞 √6 = 0. (★)  \nProve that the only possible quadruple is (𝑚,𝑛,𝑝,𝑞) = (0,0,0,0).",
      "solution": "Note first that √2 and √3 are irrational, hence so is √6 = √2 √3.  \nIntroduce the field 𝐾 = ℚ(√2, √3).  A standard result from Galois theory gives  \n [𝐾 : ℚ] = 4,  and a convenient ℚ-basis is  \n 𝔅 = {1, √2, √3, √6}.                                              (1)\n\nAssume integers (𝑚,𝑛,𝑝,𝑞) satisfy (★).  Consider the four ℚ-automorphisms of 𝐾:\n\nσ₁ : √2↦ √2, √3↦ √3  \nσ₂ : √2↦ −√2, √3↦ √3  \nσ₃ : √2↦ √2, √3↦ −√3  \nσ₄ : √2↦ −√2, √3↦ −√3.                                    (2)\n\nApplying each σᵢ to (★) produces a homogeneous linear system\n\n⎡1  √2  √3  √6⎤           ⎡𝑚⎤   ⎡0⎤  \n⎢1 −√2  √3 −√6⎥  ·  ⎢𝑛⎥ = ⎢0⎥, (3)  \n⎢1  √2 −√3 −√6⎥     ⎢𝑝⎥   ⎢0⎥  \n⎣1 −√2 −√3  √6⎦     ⎢𝑞⎥   ⎣0⎦  \n\nwhose coefficient matrix has determinant  \n\ndet = 16 √2 √3 √6 ≠ 0.                                       (4)\n\nSince the determinant is non-zero, the only solution of (3) over ℚ—and a fortiori over ℤ—is  \n\n  𝑚 = 𝑛 = 𝑝 = 𝑞 = 0.                                        (5)\n\nHence (★) admits no non-trivial integer solution.  Observe that the argument relied on the linear independence of the basis (1); an elementary alternative (though longer) squares (★) twice and reaches the same contradiction by forcing √2, √3, and √6 to be rational.  Either way, uniqueness is proved.",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.121282",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}