{
  "index": "1976-A-2",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "A-2. Let \\( P(x, y)=x^{2} y+x y^{2} \\) and \\( Q(x, y)=x^{2}+x y+y^{2} \\). For \\( n=1,2,3, \\ldots \\), let \\( F_{n}(x, y)=(x+y)^{n}-x^{n}-y^{n} \\) and \\( G_{n}(x, y)=(x+y)^{n}+x^{n}+y^{n} \\). One observes that \\( G_{2}=2 Q, F_{3}=3 P, G_{4}=2 Q^{2}, F_{5}=5 P Q, G_{6}=2 Q^{3}+3 P^{2} \\). Prove that, in fact, for each \\( n \\) either \\( F_{n} \\) or \\( G_{n} \\) is expressible as a polynomial in \\( P \\) and \\( Q \\) with integer coefficients.",
  "solution": "A-2.\nOne easily verifies that\n\\[\n\\begin{array}{c}\n(x+y)^{n}=(x+y)^{n-2} Q+(x+y)^{n-3} P, \\\\\nx^{n}+y^{n}=\\left(x^{n-2}+y^{n-2}\\right) Q-\\left(x^{n-3}+y^{n-3}\\right) P .\n\\end{array}\n\\]\n\nSubtracting or adding corresponding sides gives\n\\[\nF_{n}=Q F_{n-2}+P G_{n-3}, G_{n}=Q G_{n-2}+P F_{n-3} .\n\\]\n\nThe desired results now follow by strong mathematical induction using the given results for \\( G_{2}, F_{3} \\), \\( G_{4}, F_{5} \\), and \\( G_{6} \\) and (R).",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "n",
    "P",
    "Q",
    "F_n",
    "F_n-2",
    "F_n-3",
    "G_n",
    "G_n-2",
    "G_n-3"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "y": "ordinate",
        "n": "indexer",
        "P": "polyfirst",
        "Q": "polysecond",
        "F_n": "fsequence",
        "F_n-2": "fsequencemtwo",
        "F_n-3": "fsequencemthree",
        "G_n": "gsequence",
        "G_n-2": "gsequencemtwo",
        "G_n-3": "gsequencemthree"
      },
      "question": "A-2. Let \\( polyfirst(abscissa, ordinate)=abscissa^{2} \\, ordinate+abscissa \\, ordinate^{2} \\) and \\( polysecond(abscissa, ordinate)=abscissa^{2}+abscissa \\, ordinate+ordinate^{2} \\). For \\( indexer=1,2,3, \\ldots \\), let \\( fsequence(abscissa, ordinate)=(abscissa+ordinate)^{indexer}-abscissa^{indexer}-ordinate^{indexer} \\) and \\( gsequence(abscissa, ordinate)=(abscissa+ordinate)^{indexer}+abscissa^{indexer}+ordinate^{indexer} \\). One observes that \\( G_{2}=2 \\, polysecond, fsequence_{3}=3 \\, polyfirst, G_{4}=2 \\, polysecond^{2}, fsequence_{5}=5 \\, polyfirst \\, polysecond, G_{6}=2 \\, polysecond^{3}+3 \\, polyfirst^{2} \\). Prove that, in fact, for each \\( indexer \\) either \\( fsequence \\) or \\( gsequence \\) is expressible as a polynomial in \\( polyfirst \\) and \\( polysecond \\) with integer coefficients.",
      "solution": "A-2.\n\nOne easily verifies that\n\\[\n\\begin{array}{c}\n(abscissa+ordinate)^{indexer}=(abscissa+ordinate)^{indexer-2} \\, polysecond+(abscissa+ordinate)^{indexer-3} \\, polyfirst, \\\\\nabscissa^{indexer}+ordinate^{indexer}=\\left(abscissa^{indexer-2}+ordinate^{indexer-2}\\right) \\, polysecond-\\left(abscissa^{indexer-3}+ordinate^{indexer-3}\\right) \\, polyfirst .\n\\end{array}\n\\]\n\nSubtracting or adding corresponding sides gives\n\\[\nfsequence=polysecond \\, fsequencemtwo+polyfirst \\, gsequencemthree, \\quad gsequence=polysecond \\, gsequencemtwo+polyfirst \\, fsequencemthree .\n\\]\n\nThe desired results now follow by strong mathematical induction using the given results for \\( G_{2}, fsequence_{3}, G_{4}, fsequence_{5}, \\) and \\( G_{6} \\) and (R)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marshmallow",
        "y": "blacksmith",
        "n": "highwayman",
        "P": "starling",
        "Q": "moonlight",
        "F_n": "dreamcatch",
        "F_n-2": "dreamcatchprevtwo",
        "F_n-3": "dreamcatchprevtri",
        "G_n": "windchaser",
        "G_n-2": "windchaserprevtwo",
        "G_n-3": "windchaserprevtri"
      },
      "question": "A-2. Let \\( starling(marshmallow, blacksmith)=marshmallow^{2} blacksmith+marshmallow blacksmith^{2} \\) and \\( moonlight(marshmallow, blacksmith)=marshmallow^{2}+marshmallow blacksmith+blacksmith^{2} \\). For \\( highwayman=1,2,3, \\ldots \\), let \\( dreamcatch_{highwayman}(marshmallow, blacksmith)=(marshmallow+blacksmith)^{highwayman}-marshmallow^{highwayman}-blacksmith^{highwayman} \\) and \\( windchaser_{highwayman}(marshmallow, blacksmith)=(marshmallow+blacksmith)^{highwayman}+marshmallow^{highwayman}+blacksmith^{highwayman} \\). One observes that \\( windchaser_{2}=2 moonlight, dreamcatch_{3}=3 starling, windchaser_{4}=2 moonlight^{2}, dreamcatch_{5}=5 starling moonlight, windchaser_{6}=2 moonlight^{3}+3 starling^{2} \\). Prove that, in fact, for each \\( highwayman \\) either \\( dreamcatch_{highwayman} \\) or \\( windchaser_{highwayman} \\) is expressible as a polynomial in \\( starling \\) and \\( moonlight \\) with integer coefficients.",
      "solution": "A-2.\nOne easily verifies that\n\\[\n\\begin{array}{c}\n(marshmallow+blacksmith)^{highwayman}=(marshmallow+blacksmith)^{highwayman-2} moonlight+(marshmallow+blacksmith)^{highwayman-3} starling, \\\\\nmarshmallow^{highwayman}+blacksmith^{highwayman}=\\left(marshmallow^{highwayman-2}+blacksmith^{highwayman-2}\\right) moonlight-\\left(marshmallow^{highwayman-3}+blacksmith^{highwayman-3}\\right) starling .\n\\end{array}\n\\]\n\nSubtracting or adding corresponding sides gives\n\\[\ndreamcatch_{highwayman}=moonlight\\, dreamcatchprevtwo + starling\\, windchaserprevtri, \\qquad windchaser_{highwayman}=moonlight\\, windchaserprevtwo + starling\\, dreamcatchprevtri .\n\\]\n\nThe desired results now follow by strong mathematical induction using the given results for \\( windchaser_{2}, dreamcatch_{3} \\), \\( windchaser_{4}, dreamcatch_{5} \\), and \\( windchaser_{6} \\) and (R)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "stationary",
        "y": "immutable",
        "n": "fractional",
        "P": "transcend",
        "Q": "irrational",
        "F_n": "stillness",
        "F_n-2": "stillnesslag",
        "F_n-3": "stillnesslagg",
        "G_n": "quietude",
        "G_n-2": "quietudelag",
        "G_n-3": "quietudelagg"
      },
      "question": "A-2. Let \\( transcend(stationary, immutable)=stationary^{2} immutable+stationary immutable^{2} \\) and \\( irrational(stationary, immutable)=stationary^{2}+stationary immutable+immutable^{2} \\). For \\( fractional=1,2,3, \\ldots \\), let \\( stillness(stationary, immutable)=(stationary+immutable)^{fractional}-stationary^{fractional}-immutable^{fractional} \\) and \\( quietude(stationary, immutable)=(stationary+immutable)^{fractional}+stationary^{fractional}+immutable^{fractional} \\). One observes that \\( G_{2}=2 irrational, F_{3}=3 transcend, G_{4}=2 irrational^{2}, F_{5}=5 transcend irrational, G_{6}=2 irrational^{3}+3 transcend^{2} \\). Prove that, in fact, for each \\( fractional \\) either \\( stillness \\) or \\( quietude \\) is expressible as a polynomial in \\( transcend \\) and \\( irrational \\) with integer coefficients.",
      "solution": "A-2.\nOne easily verifies that\n\\[\n\\begin{array}{c}\n(stationary+immutable)^{fractional}=(stationary+immutable)^{fractional-2} irrational+(stationary+immutable)^{fractional-3} transcend, \\\\\nstationary^{fractional}+immutable^{fractional}=\\left(stationary^{fractional-2}+immutable^{fractional-2}\\right) irrational-\\left(stationary^{fractional-3}+immutable^{fractional-3}\\right) transcend .\n\\end{array}\n\\]\n\nSubtracting or adding corresponding sides gives\n\\[\nstillness=irrational stillnesslag+transcend quietudelagg, \\quad quietude=irrational quietudelag+transcend stillnesslagg .\n\\]\n\nThe desired results now follow by strong mathematical induction using the given results for \\( G_{2}, F_{3} \\), \\( G_{4}, F_{5} \\), and \\( G_{6} \\) and (R)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "n": "lskdjfgh",
        "P": "mznqptgh",
        "Q": "rplkshjd",
        "F_n": "sjkahdlf",
        "F_n-2": "vbncxzlk",
        "F_n-3": "gfdsaqwe",
        "G_n": "oiuytrew",
        "G_n-2": "nmjhgfds",
        "G_n-3": "plokijuh"
      },
      "question": "A-2. Let \\( mznqptgh(qzxwvtnp, hjgrksla)=qzxwvtnp^{2} hjgrksla+qzxwvtnp hjgrksla^{2} \\) and \\( rplkshjd(qzxwvtnp, hjgrksla)=qzxwvtnp^{2}+qzxwvtnp hjgrksla+hjgrksla^{2} \\). For \\( lskdjfgh=1,2,3, \\ldots \\), let \\( sjkahdlf(qzxwvtnp, hjgrksla)=(qzxwvtnp+hjgrksla)^{lskdjfgh}-qzxwvtnp^{lskdjfgh}-hjgrksla^{lskdjfgh} \\) and \\( oiuytrew(qzxwvtnp, hjgrksla)=(qzxwvtnp+hjgrksla)^{lskdjfgh}+qzxwvtnp^{lskdjfgh}+hjgrksla^{lskdjfgh} \\). One observes that \\( G_{2}=2 rplkshjd, F_{3}=3 mznqptgh, G_{4}=2 rplkshjd^{2}, F_{5}=5 mznqptgh rplkshjd, G_{6}=2 rplkshjd^{3}+3 mznqptgh^{2} \\). Prove that, in fact, for each \\( lskdjfgh \\) either \\( sjkahdlf \\) or \\( oiuytrew \\) is expressible as a polynomial in \\( mznqptgh \\) and \\( rplkshjd \\) with integer coefficients.",
      "solution": "A-2.\nOne easily verifies that\n\\[\n\\begin{array}{c}\n(qzxwvtnp+hjgrksla)^{lskdjfgh}=(qzxwvtnp+hjgrksla)^{lskdjfgh-2} rplkshjd+(qzxwvtnp+hjgrksla)^{lskdjfgh-3} mznqptgh, \\\\\nqzxwvtnp^{lskdjfgh}+hjgrksla^{lskdjfgh}=\\left(qzxwvtnp^{lskdjfgh-2}+hjgrksla^{lskdjfgh-2}\\right) rplkshjd-\\left(qzxwvtnp^{lskdjfgh-3}+hjgrksla^{lskdjfgh-3}\\right) mznqptgh .\n\\end{array}\n\\]\n\nSubtracting or adding corresponding sides gives\n\\[\nsjkahdlf=rplkshjd vbncxzlk+mznqptgh plokijuh, \\quad oiuytrew=rplkshjd nmjhgfds+mznqptgh gfdsaqwe .\n\\]\n\nThe desired results now follow by strong mathematical induction using the given results for \\( G_{2}, F_{3} \\), \\( G_{4}, F_{5} \\), and \\( G_{6} \\) and (R)."
    },
    "kernel_variant": {
      "question": "Let\n  U(a,b)=a^{2}b+ab^{2},          V(a,b)=a^{2}+ab+b^{2}.\nFor each positive integer n define the symmetric polynomials\n  A_{n}(a,b)=(a+b)^{n}-a^{n}-b^{n},   B_{n}(a,b)=(a+b)^{n}+a^{n}+b^{n}.\n(The first few values are  B_{2}=2V,  A_{3}=3U,  B_{4}=2V^{2},  A_{5}=5UV,  B_{6}=2V^{3}+3U^{2} .)\nProve that for every positive integer n at least one of the two polynomials A_{n}, B_{n} can be written as an integer-coefficient polynomial in the basic symmetric polynomials U and V.",
      "solution": "Step 1.  Two useful identities.\n\nBecause (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} we can factor the middle terms and obtain\n   (a+b)^{3}=(a+b)(a^{2}+ab+b^{2})+(a^{2}b+ab^{2})=(a+b)V+U .\nMultiplying by (a+b)^{n-3} (valid for every n≥3) gives\n   (a+b)^{n}=(a+b)^{n-2}V+(a+b)^{n-3}U.                       (1)\n\nUsing the binomial theorem in the same way, but this time for a^{n}+b^{n}, we get\n   a^{n}+b^{n}=(a^{n-2}+b^{n-2})V-(a^{n-3}+b^{n-3})U.          (2)\n\nStep 2.  Recurrences for A_{n} and B_{n}.\nSubtracting (2) from (1) and then adding them we find\n   A_{n}=V A_{n-2}+U B_{n-3},\n   B_{n}=V B_{n-2}+U A_{n-3},                   (R)\nfor every n≥3.\n\nStep 3.  Initial values.\nA_{1}=0,\nB_{2}=2V,\nA_{3}=3U,\nB_{4}=2V^{2},\nA_{5}=5UV,\nB_{6}=2V^{3}+3U^{2},\nso for n=1,…,6 the following statement is true:\n   (†)  if n is odd then A_{n} is a polynomial in U,V,\n        if n is even then B_{n} is a polynomial in U,V.\n\nStep 4.  Strong induction.\nWe prove (†) for every n≥1; this clearly implies the problem statement, since in each parity one of the two required polynomials is guaranteed to have the desired form.\n\nInduction hypothesis.  Assume that for every positive integer k<n,   A_{k} is a polynomial in U,V when k is odd and B_{k} is such a polynomial when k is even.\n\nInductive step.\nCase 1: n is odd (n≥5).  Apply the first relation (R):\n      A_{n}=V A_{n-2}+U B_{n-3}.\nHere n−2 is still odd and n−3 is even, hence by the induction hypothesis A_{n-2} and B_{n-3} are both polynomials in U and V.  Multiplying by V and U (which themselves are polynomials in U,V) preserves this property, so A_{n} is a polynomial in U,V.\n\nCase 2: n is even (n≥4).  Apply the second relation (R):\n      B_{n}=V B_{n-2}+U A_{n-3}.\nNow n−2 is even and n−3 is odd; by the hypothesis B_{n-2} and A_{n-3} are polynomials in U,V.  Again the right‐hand side is therefore a polynomial in U,V, so B_{n} has the required form.\n\nThus (†) holds for n, completing the induction.\n\nBecause in every positive integer exactly one of the two conditions ‘n odd’ and ‘n even’ holds, we conclude that for every n at least one of A_{n},B_{n} is an integer-coefficient polynomial in U and V, as desired. ∎",
      "_meta": {
        "core_steps": [
          "Write (x+y)^n and x^n+y^n in terms of (x+y)^{n-2}, (x+y)^{n-3}, P and Q",
          "Add/ subtract the two formulas to obtain the coupled recurrences  F_n = Q·F_{n-2} + P·G_{n-3}  and  G_n = Q·G_{n-2} + P·F_{n-3}",
          "Verify a finite set of small n for which F_k or G_k is already a polynomial in P,Q (base cases)",
          "Use strong induction on n with the recurrences to propagate the property to every n ≥ 1"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Choice of concrete base-case indices as long as they let the induction start (any set covering every residue class mod lcm{2,3}=6 will do).",
            "original": "{G₂, F₃, G₄, F₅, G₆}"
          },
          "slot2": {
            "description": "Purely notational labels (variable names and polynomial symbols) can be altered without affecting the logic.",
            "original": "variables x,y and symbols F_n,G_n"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}