{
  "index": "1977-A-3",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem A-3\nLet \\( u, f \\), and \\( g \\) be functions, defined for all real numbers \\( x \\), such that\n\\[\n\\frac{u(x+1)+u(x-1)}{2}=f(x) \\text { and } \\frac{u(x+4)+u(x-4)}{2}=g(x) .\n\\]\n\nDetermine \\( u(x) \\) in terms of \\( f \\) and \\( g \\).",
  "solution": "A-3.\nWe show that there are an infinite number of expressions for \\( u(x) \\) in terms of \\( f \\) and \\( g \\); some of the simpler ones are:\n\\[\n\\begin{aligned}\nu(x) & =g(x)-f(x+3)+f(x+1)+f(x-1)-f(x-3) \\\\\n& =-g(x+2)+f(x+5)-f(x+3)+f(x+1)+f(x-1) \\\\\n& =g(x+4)-f(x+7)+f(x+5)-f(x+3)+f(x+1)\n\\end{aligned}\n\\]\n\nLet \\( E \\) be the shift operator on functions \\( A \\) defined by \\( E A(x)=A(x+1) \\). Then \\( \\left(E+E^{-1}\\right) u(x)= \\) \\( 2 f(x) \\) and \\( \\left(E^{4}+E^{-4}\\right) u(x)=2 g(x) \\) are given. Thus \\( \\left(E^{2}+1\\right) u(x)=2 E f(x) \\) and \\( \\left(E^{8}+1\\right) u(x)= \\) \\( 2 E^{4} g(x) \\). Motivated by the fact that \\( E^{2}+1 \\) and \\( E^{8}+1 \\) are relatively prime polynomials in \\( E \\), one finds that\n\\[\n\\begin{aligned}\n1 & =\\frac{1}{2}\\left(E^{8}+1\\right)-\\frac{1}{2}\\left(E^{6}-E^{4}+E^{2}-1\\right)\\left(E^{2}+1\\right), \\\\\nu(x) & =\\frac{1}{2}\\left(E^{8}+1\\right) u(x)-\\frac{1}{2}\\left(E^{6}-E^{4}+E^{2}-1\\right)\\left(E^{2}+1\\right) u(x), \\\\\nu(x) & =E^{4} g(x)-\\left(E^{6}-E^{4}+E^{2}-1\\right) E f(x), \\\\\nu(x) & =E^{4} g(x)+\\left(-E^{7}+E^{5}-E^{3}+E\\right) f(x), \\\\\nu(x) & =g(x+4)-f(x+7)+f(x+5)-f(x+3)+f(x+1)\n\\end{aligned}\n\\]\n\nOther expressions are obtained using\n\\[\n\\begin{aligned}\ng(y) & =-g(y-2)+f(y+3)+f(y-5) \\\\\n& =-g(y+2)+f(y+5)+f(y-3) .\n\\end{aligned}\n\\]",
  "vars": [
    "A",
    "E",
    "f",
    "g",
    "u",
    "x",
    "y"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "auxfun",
        "E": "shiftop",
        "f": "basefun",
        "g": "secfun",
        "u": "solnfun",
        "x": "realvar",
        "y": "othervar"
      },
      "question": "Problem A-3\nLet \\( solnfun, basefun \\), and \\( secfun \\) be functions, defined for all real numbers \\( realvar \\), such that\n\\[\n\\frac{solnfun(realvar+1)+solnfun(realvar-1)}{2}=basefun(realvar) \\text { and } \\frac{solnfun(realvar+4)+solnfun(realvar-4)}{2}=secfun(realvar) .\n\\]\n\nDetermine \\( solnfun(realvar) \\) in terms of \\( basefun \\) and \\( secfun \\).",
      "solution": "A-3.\nWe show that there are an infinite number of expressions for \\( solnfun(realvar) \\) in terms of \\( basefun \\) and \\( secfun \\); some of the simpler ones are:\n\\[\n\\begin{aligned}\nsolnfun(realvar) & =secfun(realvar)-basefun(realvar+3)+basefun(realvar+1)+basefun(realvar-1)-basefun(realvar-3) \\\\\n& =-secfun(realvar+2)+basefun(realvar+5)-basefun(realvar+3)+basefun(realvar+1)+basefun(realvar-1) \\\\\n& =secfun(realvar+4)-basefun(realvar+7)+basefun(realvar+5)-basefun(realvar+3)+basefun(realvar+1)\n\\end{aligned}\n\\]\n\nLet \\( shiftop \\) be the shift operator on functions \\( auxfun \\) defined by \\( shiftop auxfun(realvar)=auxfun(realvar+1) \\). Then \\( \\left(shiftop+shiftop^{-1}\\right) solnfun(realvar)= \\) \\( 2 basefun(realvar) \\) and \\( \\left(shiftop^{4}+shiftop^{-4}\\right) solnfun(realvar)=2 secfun(realvar) \\) are given. Thus \\( \\left(shiftop^{2}+1\\right) solnfun(realvar)=2 shiftop basefun(realvar) \\) and \\( \\left(shiftop^{8}+1\\right) solnfun(realvar)= \\) \\( 2 shiftop^{4} secfun(realvar) \\). Motivated by the fact that \\( shiftop^{2}+1 \\) and \\( shiftop^{8}+1 \\) are relatively prime polynomials in \\( shiftop \\), one finds that\n\\[\n\\begin{aligned}\n1 & =\\frac{1}{2}\\left(shiftop^{8}+1\\right)-\\frac{1}{2}\\left(shiftop^{6}-shiftop^{4}+shiftop^{2}-1\\right)\\left(shiftop^{2}+1\\right), \\\\\nsolnfun(realvar) & =\\frac{1}{2}\\left(shiftop^{8}+1\\right) solnfun(realvar)-\\frac{1}{2}\\left(shiftop^{6}-shiftop^{4}+shiftop^{2}-1\\right)\\left(shiftop^{2}+1\\right) solnfun(realvar), \\\\\nsolnfun(realvar) & =shiftop^{4} secfun(realvar)-\\left(shiftop^{6}-shiftop^{4}+shiftop^{2}-1\\right) shiftop basefun(realvar), \\\\\nsolnfun(realvar) & =shiftop^{4} secfun(realvar)+\\left(-shiftop^{7}+shiftop^{5}-shiftop^{3}+shiftop\\right) basefun(realvar), \\\\\nsolnfun(realvar) & =secfun(realvar+4)-basefun(realvar+7)+basefun(realvar+5)-basefun(realvar+3)+basefun(realvar+1)\n\\end{aligned}\n\\]\n\nOther expressions are obtained using\n\\[\n\\begin{aligned}\nsecfun(othervar) & =-secfun(othervar-2)+basefun(othervar+3)+basefun(othervar-5) \\\\\n& =-secfun(othervar+2)+basefun(othervar+5)+basefun(othervar-3) .\n\\end{aligned}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "snowflake",
        "E": "blueberry",
        "f": "lighthouse",
        "g": "pineapple",
        "u": "butterfly",
        "x": "driftwood",
        "y": "moonlight"
      },
      "question": "Problem A-3\nLet \\( butterfly, lighthouse \\), and \\( pineapple \\) be functions, defined for all real numbers \\( driftwood \\), such that\n\\[\n\\frac{butterfly(driftwood+1)+butterfly(driftwood-1)}{2}=lighthouse(driftwood) \\text { and } \\frac{butterfly(driftwood+4)+butterfly(driftwood-4)}{2}=pineapple(driftwood) .\n\\]\n\nDetermine \\( butterfly(driftwood) \\) in terms of \\( lighthouse \\) and \\( pineapple \\).",
      "solution": "A-3.\nWe show that there are an infinite number of expressions for \\( butterfly(driftwood) \\) in terms of \\( lighthouse \\) and \\( pineapple \\); some of the simpler ones are:\n\\[\n\\begin{aligned}\nbutterfly(driftwood) & =pineapple(driftwood)-lighthouse(driftwood+3)+lighthouse(driftwood+1)+lighthouse(driftwood-1)-lighthouse(driftwood-3) \\\\\n& =-pineapple(driftwood+2)+lighthouse(driftwood+5)-lighthouse(driftwood+3)+lighthouse(driftwood+1)+lighthouse(driftwood-1) \\\\\n& =pineapple(driftwood+4)-lighthouse(driftwood+7)+lighthouse(driftwood+5)-lighthouse(driftwood+3)+lighthouse(driftwood+1)\n\\end{aligned}\n\\]\n\nLet \\( blueberry \\) be the shift operator on functions \\( snowflake \\) defined by \\( blueberry\\, snowflake(driftwood)=snowflake(driftwood+1) \\). Then \\( \\left(blueberry+blueberry^{-1}\\right) butterfly(driftwood)= \\) \\( 2 lighthouse(driftwood) \\) and \\( \\left(blueberry^{4}+blueberry^{-4}\\right) butterfly(driftwood)=2 pineapple(driftwood) \\) are given. Thus \\( \\left(blueberry^{2}+1\\right) butterfly(driftwood)=2 blueberry\\, lighthouse(driftwood) \\) and \\( \\left(blueberry^{8}+1\\right) butterfly(driftwood)= \\) \\( 2 blueberry^{4} pineapple(driftwood) \\). Motivated by the fact that \\( blueberry^{2}+1 \\) and \\( blueberry^{8}+1 \\) are relatively prime polynomials in \\( blueberry \\), one finds that\n\\[\n\\begin{aligned}\n1 & =\\frac{1}{2}\\left(blueberry^{8}+1\\right)-\\frac{1}{2}\\left(blueberry^{6}-blueberry^{4}+blueberry^{2}-1\\right)\\left(blueberry^{2}+1\\right), \\\\\nbutterfly(driftwood) & =\\frac{1}{2}\\left(blueberry^{8}+1\\right) butterfly(driftwood)-\\frac{1}{2}\\left(blueberry^{6}-blueberry^{4}+blueberry^{2}-1\\right)\\left(blueberry^{2}+1\\right) butterfly(driftwood), \\\\\nbutterfly(driftwood) & =blueberry^{4} pineapple(driftwood)-\\left(blueberry^{6}-blueberry^{4}+blueberry^{2}-1\\right) blueberry\\, lighthouse(driftwood), \\\\\nbutterfly(driftwood) & =blueberry^{4} pineapple(driftwood)+\\left(-blueberry^{7}+blueberry^{5}-blueberry^{3}+blueberry\\right) lighthouse(driftwood), \\\\\nbutterfly(driftwood) & =pineapple(driftwood+4)-lighthouse(driftwood+7)+lighthouse(driftwood+5)-lighthouse(driftwood+3)+lighthouse(driftwood+1)\n\\end{aligned}\n\\]\n\nOther expressions are obtained using\n\\[\n\\begin{aligned}\npineapple(moonlight) & =-pineapple(moonlight-2)+lighthouse(moonlight+3)+lighthouse(moonlight-5) \\\\\n& =-pineapple(moonlight+2)+lighthouse(moonlight+5)+lighthouse(moonlight-3) .\n\\end{aligned}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "voidfunction",
        "E": "stationary",
        "f": "fixedvalue",
        "g": "immobilemap",
        "u": "knownvalue",
        "x": "solidconst",
        "y": "steadyconst"
      },
      "question": "Problem A-3\nLet \\( knownvalue, fixedvalue \\), and \\( immobilemap \\) be functions, defined for all real numbers \\( solidconst \\), such that\n\\[\n\\frac{knownvalue(solidconst+1)+knownvalue(solidconst-1)}{2}=fixedvalue(solidconst) \\text { and } \\frac{knownvalue(solidconst+4)+knownvalue(solidconst-4)}{2}=immobilemap(solidconst) .\n\\]\n\nDetermine \\( knownvalue(solidconst) \\) in terms of \\( fixedvalue \\) and \\( immobilemap \\).",
      "solution": "A-3.\nWe show that there are an infinite number of expressions for \\( knownvalue(solidconst) \\) in terms of \\( fixedvalue \\) and \\( immobilemap \\); some of the simpler ones are:\n\\[\n\\begin{aligned}\nknownvalue(solidconst) & =immobilemap(solidconst)-fixedvalue(solidconst+3)+fixedvalue(solidconst+1)+fixedvalue(solidconst-1)-fixedvalue(solidconst-3) \\\\\n& =-immobilemap(solidconst+2)+fixedvalue(solidconst+5)-fixedvalue(solidconst+3)+fixedvalue(solidconst+1)+fixedvalue(solidconst-1) \\\\\n& =immobilemap(solidconst+4)-fixedvalue(solidconst+7)+fixedvalue(solidconst+5)-fixedvalue(solidconst+3)+fixedvalue(solidconst+1)\n\\end{aligned}\n\\]\n\nLet \\( stationary \\) be the shift operator on functions \\( voidfunction \\) defined by \\( stationary\\ voidfunction(solidconst)=voidfunction(solidconst+1) \\). Then \\( \\left(stationary+stationary^{-1}\\right) knownvalue(solidconst)= \\) \\( 2 fixedvalue(solidconst) \\) and \\( \\left(stationary^{4}+stationary^{-4}\\right) knownvalue(solidconst)=2 immobilemap(solidconst) \\) are given. Thus \\( \\left(stationary^{2}+1\\right) knownvalue(solidconst)=2 stationary\\ fixedvalue(solidconst) \\) and \\( \\left(stationary^{8}+1\\right) knownvalue(solidconst)= \\) \\( 2 stationary^{4} immobilemap(solidconst) \\). Motivated by the fact that \\( stationary^{2}+1 \\) and \\( stationary^{8}+1 \\) are relatively prime polynomials in \\( stationary \\), one finds that\n\\[\n\\begin{aligned}\n1 & =\\frac{1}{2}\\left(stationary^{8}+1\\right)-\\frac{1}{2}\\left(stationary^{6}-stationary^{4}+stationary^{2}-1\\right)\\left(stationary^{2}+1\\right), \\\\\nknownvalue(solidconst) & =\\frac{1}{2}\\left(stationary^{8}+1\\right) knownvalue(solidconst)-\\frac{1}{2}\\left(stationary^{6}-stationary^{4}+stationary^{2}-1\\right)\\left(stationary^{2}+1\\right) knownvalue(solidconst), \\\\\nknownvalue(solidconst) & =stationary^{4} immobilemap(solidconst)-\\left(stationary^{6}-stationary^{4}+stationary^{2}-1\\right) stationary\\ fixedvalue(solidconst), \\\\\nknownvalue(solidconst) & =stationary^{4} immobilemap(solidconst)+\\left(-stationary^{7}+stationary^{5}-stationary^{3}+stationary\\right) fixedvalue(solidconst), \\\\\nknownvalue(solidconst) & =immobilemap(solidconst+4)-fixedvalue(solidconst+7)+fixedvalue(solidconst+5)-fixedvalue(solidconst+3)+fixedvalue(solidconst+1)\n\\end{aligned}\n\\]\n\nOther expressions are obtained using\n\\[\n\\begin{aligned}\nimmobilemap(steadyconst) & =-immobilemap(steadyconst-2)+fixedvalue(steadyconst+3)+fixedvalue(steadyconst-5) \\\\\n& =-immobilemap(steadyconst+2)+fixedvalue(steadyconst+5)+fixedvalue(steadyconst-3) .\n\\end{aligned}\n\\]"
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "E": "klmsdpto",
        "f": "hjgrksla",
        "g": "bvlqcdnr",
        "u": "spxtrnqo",
        "x": "wzjplhea",
        "y": "mdcvprte"
      },
      "question": "Problem A-3\nLet \\( spxtrnqo, hjgrksla \\), and \\( bvlqcdnr \\) be functions, defined for all real numbers \\( wzjplhea \\), such that\n\\[\n\\frac{spxtrnqo(wzjplhea+1)+spxtrnqo(wzjplhea-1)}{2}=hjgrksla(wzjplhea) \\text { and } \\frac{spxtrnqo(wzjplhea+4)+spxtrnqo(wzjplhea-4)}{2}=bvlqcdnr(wzjplhea) .\n\\]\n\nDetermine \\( spxtrnqo(wzjplhea) \\) in terms of \\( hjgrksla \\) and \\( bvlqcdnr \\).",
      "solution": "A-3.\nWe show that there are an infinite number of expressions for \\( spxtrnqo(wzjplhea) \\) in terms of \\( hjgrksla \\) and \\( bvlqcdnr \\); some of the simpler ones are:\n\\[\n\\begin{aligned}\nspxtrnqo(wzjplhea) & =bvlqcdnr(wzjplhea)-hjgrksla(wzjplhea+3)+hjgrksla(wzjplhea+1)+hjgrksla(wzjplhea-1)-hjgrksla(wzjplhea-3) \\\\\n& =-bvlqcdnr(wzjplhea+2)+hjgrksla(wzjplhea+5)-hjgrksla(wzjplhea+3)+hjgrksla(wzjplhea+1)+hjgrksla(wzjplhea-1) \\\\\n& =bvlqcdnr(wzjplhea+4)-hjgrksla(wzjplhea+7)+hjgrksla(wzjplhea+5)-hjgrksla(wzjplhea+3)+hjgrksla(wzjplhea+1)\n\\end{aligned}\n\\]\n\nLet \\( klmsdpto \\) be the shift operator on functions \\( qzxwvtnp \\) defined by \\( klmsdpto qzxwvtnp(wzjplhea)=qzxwvtnp(wzjplhea+1) \\). Then \\( \\left(klmsdpto+klmsdpto^{-1}\\right) spxtrnqo(wzjplhea)=2 hjgrksla(wzjplhea) \\) and \\( \\left(klmsdpto^{4}+klmsdpto^{-4}\\right) spxtrnqo(wzjplhea)=2 bvlqcdnr(wzjplhea) \\) are given. Thus \\( \\left(klmsdpto^{2}+1\\right) spxtrnqo(wzjplhea)=2 klmsdpto hjgrksla(wzjplhea) \\) and \\( \\left(klmsdpto^{8}+1\\right) spxtrnqo(wzjplhea)=2 klmsdpto^{4} bvlqcdnr(wzjplhea) \\). Motivated by the fact that \\( klmsdpto^{2}+1 \\) and \\( klmsdpto^{8}+1 \\) are relatively prime polynomials in \\( klmsdpto \\), one finds that\n\\[\n\\begin{aligned}\n1 & =\\frac{1}{2}\\left(klmsdpto^{8}+1\\right)-\\frac{1}{2}\\left(klmsdpto^{6}-klmsdpto^{4}+klmsdpto^{2}-1\\right)\\left(klmsdpto^{2}+1\\right), \\\\\nspxtrnqo(wzjplhea) & =\\frac{1}{2}\\left(klmsdpto^{8}+1\\right) spxtrnqo(wzjplhea)-\\frac{1}{2}\\left(klmsdpto^{6}-klmsdpto^{4}+klmsdpto^{2}-1\\right)\\left(klmsdpto^{2}+1\\right) spxtrnqo(wzjplhea), \\\\\nspxtrnqo(wzjplhea) & =klmsdpto^{4} bvlqcdnr(wzjplhea)-\\left(klmsdpto^{6}-klmsdpto^{4}+klmsdpto^{2}-1\\right) klmsdpto hjgrksla(wzjplhea), \\\\\nspxtrnqo(wzjplhea) & =klmsdpto^{4} bvlqcdnr(wzjplhea)+\\left(-klmsdpto^{7}+klmsdpto^{5}-klmsdpto^{3}+klmsdpto\\right) hjgrksla(wzjplhea), \\\\\nspxtrnqo(wzjplhea) & =bvlqcdnr(wzjplhea+4)-hjgrksla(wzjplhea+7)+hjgrksla(wzjplhea+5)-hjgrksla(wzjplhea+3)+hjgrksla(wzjplhea+1)\n\\end{aligned}\n\\]\n\nOther expressions are obtained using\n\\[\n\\begin{aligned}\nbvlqcdnr(mdcvprte) & =-bvlqcdnr(mdcvprte-2)+hjgrksla(mdcvprte+3)+hjgrksla(mdcvprte-5) \\\\\n& =-bvlqcdnr(mdcvprte+2)+hjgrksla(mdcvprte+5)+hjgrksla(mdcvprte-3) .\n\\end{aligned}\n\\]"
    },
    "kernel_variant": {
      "question": "Let u, f, g be real–valued functions defined for every real number x.  \nThey satisfy the two coupled symmetric–difference relations  \n (1) (u(x+1)+u(x−1)+u(x+2)+u(x−2)) ⁄ 4  =  f(x),  \n (2) (u(x+3)+u(x−3)+u(x+5)+u(x−5)) ⁄ 4  =  g(x).  \n\nGive an explicit formula for u(x) that involves only finitely many shifts of f and g and contains **no occurrence of u** on the right–hand side.",
      "solution": "Step 1 – Introduce the shift operator.  \nLet E be the shift operator (E h)(x)=h(x+1).  Then E⁻¹ is the shift by −1.  \nWith this notation the two given relations become  \n\n P₁(E) u = 4 f,  P₂(E) u = 4 g,  \n\nwhere  \n P₁(E)=E+E⁻¹+E²+E⁻²,  P₂(E)=E³+E⁻³+E⁵+E⁻⁵.\n\nStep 2 – Pass to the Chebyshev variable T=E+E⁻¹.  \nBecause Eⁿ+E⁻ⁿ is a polynomial in T, we rewrite\n\n E²+E⁻²=T²−2, E³+E⁻³=T³−3T, E⁵+E⁻⁵=T⁵−5T³+5T.\n\nHence  \n P₁(E)=T²+T−2         (= (T−1)(T+2)),  \n P₂(E)=T⁵−4T³+2T   (=T(T⁴−4T²+2)).\n\nCall these polynomials P₁(T) and P₂(T).\n\nStep 3 – A Bézout identity for P₁ and P₂.  \nUsing the extended Euclidean algorithm in ℚ[T]:\n\n T⁵−4T³+2T = (T³−T²−T−1)(T²+T−2) + (T−2),  \n T²+T−2     = T(T−2) + (3T−2),  \n T−2         = (1/3)(3T−2) − 4/3.\n\nBack–substitution yields the Bézout relation\n\n 1 = C₁(T)·P₁(T) + C₂(T)·P₂(T),\n\nwith\n C₁(T)=¼T⁴+½T³−T²−T−½,  C₂(T)=−(3+T)/4.\n\nStep 4 – Return to the shift operator.  \nMultiplying the Bézout identity by u and using P₁(E)u=4f, P₂(E)u=4g gives\n\n u = 4C₁(E+E⁻¹) f + 4C₂(E+E⁻¹) g.\n\nExpand C₁, C₂ back into shifts of order ≤4:\n\n 4C₁(E+E⁻¹) = E⁴+2E³+2E+2E⁻¹+2E⁻³+E⁻⁴ − 4,  \n 4C₂(E+E⁻¹) = −(3+E+E⁻¹).\n\nThus a concrete expression for u is\n\n u(x)=f(x+4)+2f(x+3)+2f(x+1)+2f(x−1)+2f(x−3)+f(x−4)  \n   −4f(x) − 3g(x) − g(x+1) − g(x−1).\n\nAny other expression obtained by adding suitable multiples of the two original relations is equally valid, but the one above is already completely explicit and contains no occurrence of u.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.630066",
        "was_fixed": false,
        "difficulty_analysis": "1. Longer-range interactions: P₁ involves shifts of ±1 and ±2, while P₂ involves ±3 and ±5, forcing the solver to juggle eight distinct translations instead of the two in the original problem.\n\n2. Higher-degree operator polynomials: After rewriting with T=E+E⁻¹ one must handle polynomials of degree 2 and 5 (versus degrees 1 and 4 originally).  Their relative primeness is less obvious and the extended Euclidean algorithm is noticeably more laborious.\n\n3. Non-uniform denominators and asymmetry: The averaging factors coincide (4) but the sets of shifts are not harmonically related (1,2 versus 3,5).  This destroys the simple “telescoping” tricks that work for the original problem and forces a systematic operator-theoretic approach.\n\n4. Deeper algebraic manipulation: Producing an explicit Bézout identity now requires several rounds of polynomial division and careful back-substitution, followed by translating the result back into shift notation.  Each of these steps is more intricate than in the original A-3 solution.\n\n5. More terms in the final formula: The explicit expression for u(x) uses nine shifted values of f and g, compared with at most five in the original, reflecting the greater combinatorial complexity of the operators involved."
      }
    },
    "original_kernel_variant": {
      "question": "Let u, f, g be real–valued functions defined for every real number x.  \nThey satisfy the two coupled symmetric–difference relations  \n (1) (u(x+1)+u(x−1)+u(x+2)+u(x−2)) ⁄ 4  =  f(x),  \n (2) (u(x+3)+u(x−3)+u(x+5)+u(x−5)) ⁄ 4  =  g(x).  \n\nGive an explicit formula for u(x) that involves only finitely many shifts of f and g and contains **no occurrence of u** on the right–hand side.",
      "solution": "Step 1 – Introduce the shift operator.  \nLet E be the shift operator (E h)(x)=h(x+1).  Then E⁻¹ is the shift by −1.  \nWith this notation the two given relations become  \n\n P₁(E) u = 4 f,  P₂(E) u = 4 g,  \n\nwhere  \n P₁(E)=E+E⁻¹+E²+E⁻²,  P₂(E)=E³+E⁻³+E⁵+E⁻⁵.\n\nStep 2 – Pass to the Chebyshev variable T=E+E⁻¹.  \nBecause Eⁿ+E⁻ⁿ is a polynomial in T, we rewrite\n\n E²+E⁻²=T²−2, E³+E⁻³=T³−3T, E⁵+E⁻⁵=T⁵−5T³+5T.\n\nHence  \n P₁(E)=T²+T−2         (= (T−1)(T+2)),  \n P₂(E)=T⁵−4T³+2T   (=T(T⁴−4T²+2)).\n\nCall these polynomials P₁(T) and P₂(T).\n\nStep 3 – A Bézout identity for P₁ and P₂.  \nUsing the extended Euclidean algorithm in ℚ[T]:\n\n T⁵−4T³+2T = (T³−T²−T−1)(T²+T−2) + (T−2),  \n T²+T−2     = T(T−2) + (3T−2),  \n T−2         = (1/3)(3T−2) − 4/3.\n\nBack–substitution yields the Bézout relation\n\n 1 = C₁(T)·P₁(T) + C₂(T)·P₂(T),\n\nwith\n C₁(T)=¼T⁴+½T³−T²−T−½,  C₂(T)=−(3+T)/4.\n\nStep 4 – Return to the shift operator.  \nMultiplying the Bézout identity by u and using P₁(E)u=4f, P₂(E)u=4g gives\n\n u = 4C₁(E+E⁻¹) f + 4C₂(E+E⁻¹) g.\n\nExpand C₁, C₂ back into shifts of order ≤4:\n\n 4C₁(E+E⁻¹) = E⁴+2E³+2E+2E⁻¹+2E⁻³+E⁻⁴ − 4,  \n 4C₂(E+E⁻¹) = −(3+E+E⁻¹).\n\nThus a concrete expression for u is\n\n u(x)=f(x+4)+2f(x+3)+2f(x+1)+2f(x−1)+2f(x−3)+f(x−4)  \n   −4f(x) − 3g(x) − g(x+1) − g(x−1).\n\nAny other expression obtained by adding suitable multiples of the two original relations is equally valid, but the one above is already completely explicit and contains no occurrence of u.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.502734",
        "was_fixed": false,
        "difficulty_analysis": "1. Longer-range interactions: P₁ involves shifts of ±1 and ±2, while P₂ involves ±3 and ±5, forcing the solver to juggle eight distinct translations instead of the two in the original problem.\n\n2. Higher-degree operator polynomials: After rewriting with T=E+E⁻¹ one must handle polynomials of degree 2 and 5 (versus degrees 1 and 4 originally).  Their relative primeness is less obvious and the extended Euclidean algorithm is noticeably more laborious.\n\n3. Non-uniform denominators and asymmetry: The averaging factors coincide (4) but the sets of shifts are not harmonically related (1,2 versus 3,5).  This destroys the simple “telescoping” tricks that work for the original problem and forces a systematic operator-theoretic approach.\n\n4. Deeper algebraic manipulation: Producing an explicit Bézout identity now requires several rounds of polynomial division and careful back-substitution, followed by translating the result back into shift notation.  Each of these steps is more intricate than in the original A-3 solution.\n\n5. More terms in the final formula: The explicit expression for u(x) uses nine shifted values of f and g, compared with at most five in the original, reflecting the greater combinatorial complexity of the operators involved."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}