{
  "index": "1991-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Suppose $f$ and $g$ are non-constant, differentiable, real-valued\nfunctions defined on $(-\\infty, \\infty)$. Furthermore, suppose that for\neach pair of real numbers $x$ and $y$,\n\\begin{align*}\nf(x+y) &= f(x)f(y) - g(x)g(y), \\\\\ng(x+y) &= f(x)g(y) + g(x)f(y).\n\\end{align*}\nIf $f'(0) = 0$, prove that $(f(x))^2 + (g(x))^2 = 1$ for all $x$.",
  "solution": "Solution 1. Differentiate both sides with respect to \\( y \\) to obtain\n\\[\n\\begin{aligned}\nf^{\\prime}(x+y) & =f(x) f^{\\prime}(y)-g(x) g^{\\prime}(y) \\\\\ng^{\\prime}(x+y) & =f(x) g^{\\prime}(y)+g(x) f^{\\prime}(y)\n\\end{aligned}\n\\]\n\nSetting \\( y=0 \\) yields\n\\[\nf^{\\prime}(x)=-g^{\\prime}(0) g(x) \\quad \\text { and } \\quad g^{\\prime}(x)=g^{\\prime}(0) f(x)\n\\]\n\nThus\n\\[\n2 f(x) f^{\\prime}(x)+2 g(x) g^{\\prime}(x)=0,\n\\]\nand therefore\n\\[\nf(x)^{2}+g(x)^{2}=C\n\\]\nfor some constant \\( C \\). Since \\( f \\) and \\( g \\) are nonconstant, \\( C \\neq 0 \\). The identity\n\\[\nf(x+y)^{2}+g(x+y)^{2}=\\left(f(x)^{2}+g(x)^{2}\\right)\\left(f(y)^{2}+g(y)^{2}\\right),\n\\]\nimplies \\( C=C^{2} \\). But \\( C \\neq 0 \\), so \\( C=1 \\).\nSolution 2. Define \\( h: \\mathbb{R} \\rightarrow \\mathbb{C} \\) by \\( h(x)=f(x)+i g(x) \\). Then \\( h \\) is differentiable, and \\( h^{\\prime}(0)=b i \\) for some \\( b \\in \\mathbb{R} \\). The two given functional equations imply \\( h(x+y)= \\) \\( h(x) h(y) \\). Differentiating with respect to \\( y \\) and substituting \\( y=0 \\) yields \\( h^{\\prime}(x)= \\) \\( h(x) h^{\\prime}(0)=b i \\cdot h(x) \\), so \\( h(x)=C e^{b i x} \\) for some \\( C \\in \\mathbb{C} \\). From \\( h(0+0)=h(0) h(0) \\) we obtain \\( C=C^{2} \\). If \\( C=0 \\), then \\( h \\) would be identically zero, and \\( f \\) and \\( g \\) would be constant, contradiction. Thus \\( C=1 \\). Finally, for any \\( x \\in \\mathbb{R} \\),\n\\[\nf(x)^{2}+g(x)^{2}=|h(x)|^{2}=\\left|e^{b i x}\\right|^{2}=1 .\n\\]",
  "vars": [
    "x",
    "y",
    "f",
    "g",
    "h"
  ],
  "params": [
    "C",
    "b"
  ],
  "sci_consts": [
    "e",
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "realvarx",
        "y": "realvary",
        "f": "mapfunction",
        "g": "auxfunction",
        "h": "complexfunc",
        "C": "constvalue",
        "b": "slopeconst"
      },
      "question": "Suppose $mapfunction$ and $auxfunction$ are non-constant, differentiable, real-valued\nfunctions defined on $(-\\infty, \\infty)$. Furthermore, suppose that for\neach pair of real numbers $realvarx$ and $realvary$,\n\\begin{align*}\nmapfunction(realvarx+realvary) &= mapfunction(realvarx)mapfunction(realvary) - auxfunction(realvarx)auxfunction(realvary), \\\\\nauxfunction(realvarx+realvary) &= mapfunction(realvarx)auxfunction(realvary) + auxfunction(realvarx)mapfunction(realvary).\n\\end{align*}\nIf $mapfunction'(0) = 0$, prove that $(mapfunction(realvarx))^2 + (auxfunction(realvarx))^2 = 1$ for all $realvarx$.",
      "solution": "Solution 1. Differentiate both sides with respect to \\( realvary \\) to obtain\n\\[\n\\begin{aligned}\nmapfunction^{\\prime}(realvarx+realvary) & =mapfunction(realvarx) mapfunction^{\\prime}(realvary)-auxfunction(realvarx) auxfunction^{\\prime}(realvary) \\\\\nauxfunction^{\\prime}(realvarx+realvary) & =mapfunction(realvarx) auxfunction^{\\prime}(realvary)+auxfunction(realvarx) mapfunction^{\\prime}(realvary)\n\\end{aligned}\n\\]\n\nSetting \\( realvary=0 \\) yields\n\\[\nmapfunction^{\\prime}(realvarx)=-auxfunction^{\\prime}(0) auxfunction(realvarx) \\quad \\text { and } \\quad auxfunction^{\\prime}(realvarx)=auxfunction^{\\prime}(0) mapfunction(realvarx)\n\\]\n\nThus\n\\[\n2 mapfunction(realvarx) mapfunction^{\\prime}(realvarx)+2 auxfunction(realvarx) auxfunction^{\\prime}(realvarx)=0,\n\\]\nand therefore\n\\[\nmapfunction(realvarx)^{2}+auxfunction(realvarx)^{2}=constvalue\n\\]\nfor some constant \\( constvalue \\). Since \\( mapfunction \\) and \\( auxfunction \\) are nonconstant, \\( constvalue \\neq 0 \\). The identity\n\\[\nmapfunction(realvarx+realvary)^{2}+auxfunction(realvarx+realvary)^{2}=\\left(mapfunction(realvarx)^{2}+auxfunction(realvarx)^{2}\\right)\\left(mapfunction(realvary)^{2}+auxfunction(realvary)^{2}\\right),\n\\]\nimplies \\( constvalue=constvalue^{2} \\). But \\( constvalue \\neq 0 \\), so \\( constvalue=1 \\).\n\nSolution 2. Define \\( complexfunc: \\mathbb{R} \\rightarrow \\mathbb{C} \\) by \\( complexfunc(realvarx)=mapfunction(realvarx)+i auxfunction(realvarx) \\). Then \\( complexfunc \\) is differentiable, and \\( complexfunc^{\\prime}(0)=slopeconst i \\) for some \\( slopeconst \\in \\mathbb{R} \\). The two given functional equations imply \\( complexfunc(realvarx+realvary)= \\) \\( complexfunc(realvarx) complexfunc(realvary) \\). Differentiating with respect to \\( realvary \\) and substituting \\( realvary=0 \\) yields \\( complexfunc^{\\prime}(realvarx)= \\) \\( complexfunc(realvarx) complexfunc^{\\prime}(0)=slopeconst i \\cdot complexfunc(realvarx) \\), so \\( complexfunc(realvarx)=constvalue e^{slopeconst i realvarx} \\) for some \\( constvalue \\in \\mathbb{C} \\). From \\( complexfunc(0+0)=complexfunc(0) complexfunc(0) \\) we obtain \\( constvalue=constvalue^{2} \\). If \\( constvalue=0 \\), then \\( complexfunc \\) would be identically zero, and \\( mapfunction \\) and \\( auxfunction \\) would be constant, contradiction. Thus \\( constvalue=1 \\). Finally, for any \\( realvarx \\in \\mathbb{R} \\),\n\\[\nmapfunction(realvarx)^{2}+auxfunction(realvarx)^{2}=|complexfunc(realvarx)|^{2}=\\left|e^{slopeconst i realvarx}\\right|^{2}=1 .\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sunflower",
        "y": "moonlight",
        "f": "pinegrove",
        "g": "blueberry",
        "h": "sandcastle",
        "C": "waterspout",
        "b": "firebrand"
      },
      "question": "Suppose $pinegrove$ and $blueberry$ are non-constant, differentiable, real-valued\nfunctions defined on $(-\\infty, \\infty)$. Furthermore, suppose that for\neach pair of real numbers $sunflower$ and $moonlight$,\n\\begin{align*}\npinegrove(sunflower+moonlight) &= pinegrove(sunflower)pinegrove(moonlight) - blueberry(sunflower)blueberry(moonlight), \\\\\nblueberry(sunflower+moonlight) &= pinegrove(sunflower)blueberry(moonlight) + blueberry(sunflower)pinegrove(moonlight).\n\\end{align*}\nIf $pinegrove'(0) = 0$, prove that $(pinegrove(sunflower))^2 + (blueberry(sunflower))^2 = 1$ for all $sunflower$.",
      "solution": "Solution 1. Differentiate both sides with respect to \\( moonlight \\) to obtain\n\\[\n\\begin{aligned}\npinegrove^{\\prime}(sunflower+moonlight) & = pinegrove(sunflower) \\, pinegrove^{\\prime}(moonlight) - blueberry(sunflower) \\, blueberry^{\\prime}(moonlight) \\\\\nblueberry^{\\prime}(sunflower+moonlight) & = pinegrove(sunflower) \\, blueberry^{\\prime}(moonlight) + blueberry(sunflower) \\, pinegrove^{\\prime}(moonlight)\n\\end{aligned}\n\\]\n\nSetting \\( moonlight = 0 \\) yields\n\\[\npinegrove^{\\prime}(sunflower) = -\\, blueberry^{\\prime}(0)\\, blueberry(sunflower) \\quad \\text{and} \\quad blueberry^{\\prime}(sunflower) = \\, blueberry^{\\prime}(0)\\, pinegrove(sunflower)\n\\]\n\nThus\n\\[\n2\\, pinegrove(sunflower)\\, pinegrove^{\\prime}(sunflower) + 2\\, blueberry(sunflower)\\, blueberry^{\\prime}(sunflower) = 0,\n\\]\nand therefore\n\\[\npinegrove(sunflower)^{2} + blueberry(sunflower)^{2} = waterspout\n\\]\nfor some constant \\( waterspout \\). Since \\( pinegrove \\) and \\( blueberry \\) are nonconstant, \\( waterspout \\neq 0 \\). The identity\n\\[\npinegrove(sunflower+moonlight)^{2} + blueberry(sunflower+moonlight)^{2} =\n\\left(pinegrove(sunflower)^{2} + blueberry(sunflower)^{2}\\right)\n\\left(pinegrove(moonlight)^{2} + blueberry(moonlight)^{2}\\right),\n\\]\nimplies \\( waterspout = waterspout^{2} \\). But \\( waterspout \\neq 0 \\), so \\( waterspout = 1 \\).\n\nSolution 2. Define \\( sandcastle: \\mathbb{R} \\rightarrow \\mathbb{C} \\) by \\( sandcastle(sunflower) = pinegrove(sunflower) + i\\, blueberry(sunflower) \\). Then \\( sandcastle \\) is differentiable, and \\( sandcastle^{\\prime}(0) = firebrand i \\) for some \\( firebrand \\in \\mathbb{R} \\). The two given functional equations imply \\( sandcastle(sunflower+moonlight) = sandcastle(sunflower)\\, sandcastle(moonlight) \\). Differentiating with respect to \\( moonlight \\) and substituting \\( moonlight = 0 \\) yields\n\\( sandcastle^{\\prime}(sunflower) = sandcastle(sunflower)\\, sandcastle^{\\prime}(0) = firebrand i \\cdot sandcastle(sunflower) \\),\nso \\( sandcastle(sunflower) = waterspout e^{\\, firebrand i \\, sunflower} \\) for some \\( waterspout \\in \\mathbb{C} \\). From \\( sandcastle(0+0) = sandcastle(0)\\, sandcastle(0) \\) we obtain \\( waterspout = waterspout^{2} \\). If \\( waterspout = 0 \\), then \\( sandcastle \\) would be identically zero, and \\( pinegrove \\) and \\( blueberry \\) would be constant, contradiction. Thus \\( waterspout = 1 \\). Finally, for any \\( sunflower \\in \\mathbb{R} \\),\n\\[\npinegrove(sunflower)^{2} + blueberry(sunflower)^{2} = |sandcastle(sunflower)|^{2} = \\left|e^{\\, firebrand i \\, sunflower}\\right|^{2} = 1 .\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constant",
        "y": "fixedvalue",
        "f": "nonfunction",
        "g": "malfunction",
        "h": "nonactive",
        "C": "variable",
        "b": "unstable"
      },
      "question": "Suppose $nonfunction$ and $malfunction$ are non-constant, differentiable, real-valued\nfunctions defined on $(-\\infty, \\infty)$. Furthermore, suppose that for\neach pair of real numbers $constant$ and $fixedvalue$,\n\\begin{align*}\nnonfunction(constant+fixedvalue) &= nonfunction(constant)nonfunction(fixedvalue) - malfunction(constant)malfunction(fixedvalue), \\\\\nmalfunction(constant+fixedvalue) &= nonfunction(constant)malfunction(fixedvalue) + malfunction(constant)nonfunction(fixedvalue).\n\\end{align*}\nIf $nonfunction'(0) = 0$, prove that $(nonfunction(constant))^2 + (malfunction(constant))^2 = 1$ for all $constant$.",
      "solution": "Solution 1. Differentiate both sides with respect to \\( fixedvalue \\) to obtain\n\\[\n\\begin{aligned}\nnonfunction^{\\prime}(constant+fixedvalue) & =nonfunction(constant) nonfunction^{\\prime}(fixedvalue)-malfunction(constant) malfunction^{\\prime}(fixedvalue) \\\\\nmalfunction^{\\prime}(constant+fixedvalue) & =nonfunction(constant) malfunction^{\\prime}(fixedvalue)+malfunction(constant) nonfunction^{\\prime}(fixedvalue)\n\\end{aligned}\n\\]\n\nSetting \\( fixedvalue=0 \\) yields\n\\[\nnonfunction^{\\prime}(constant)=-malfunction^{\\prime}(0) malfunction(constant) \\quad \\text { and } \\quad malfunction^{\\prime}(constant)=malfunction^{\\prime}(0) nonfunction(constant)\n\\]\n\nThus\n\\[\n2 nonfunction(constant) nonfunction^{\\prime}(constant)+2 malfunction(constant) malfunction^{\\prime}(constant)=0,\n\\],\nand therefore\n\\[\nnonfunction(constant)^{2}+malfunction(constant)^{2}=variable\n\\]\nfor some constant \\( variable \\). Since \\( nonfunction \\) and \\( malfunction \\) are nonconstant, \\( variable \\neq 0 \\). The identity\n\\[\nnonfunction(constant+fixedvalue)^{2}+malfunction(constant+fixedvalue)^{2}=\\left(nonfunction(constant)^{2}+malfunction(constant)^{2}\\right)\\left(nonfunction(fixedvalue)^{2}+malfunction(fixedvalue)^{2}\\right),\n\\],\nimplies \\( variable=variable^{2} \\). But \\( variable \\neq 0 \\), so \\( variable=1 \\).\nSolution 2. Define \\( nonactive: \\mathbb{R} \\rightarrow \\mathbb{C} \\) by \\( nonactive(constant)=nonfunction(constant)+i malfunction(constant) \\). Then \\( nonactive \\) is differentiable, and \\( nonactive^{\\prime}(0)=unstable i \\) for some \\( unstable \\in \\mathbb{R} \\). The two given functional equations imply \\( nonactive(constant+fixedvalue)= \\) \\( nonactive(constant) nonactive(fixedvalue) \\). Differentiating with respect to \\( fixedvalue \\) and substituting \\( fixedvalue=0 \\) yields \\( nonactive^{\\prime}(constant)= \\) \\( nonactive(constant) nonactive^{\\prime}(0)=unstable i \\cdot nonactive(constant) \\), so \\( nonactive(constant)=variable e^{unstable i constant} \\) for some \\( variable \\in \\mathbb{C} \\). From \\( nonactive(0+0)=nonactive(0) nonactive(0) \\) we obtain \\( variable=variable^{2} \\). If \\( variable=0 \\), then \\( nonactive \\) would be identically zero, and \\( nonfunction \\) and \\( malfunction \\) would be constant, contradiction. Thus \\( variable=1 \\). Finally, for any \\( constant \\in \\mathbb{R} \\),\n\\[\nnonfunction(constant)^{2}+malfunction(constant)^{2}=|nonactive(constant)|^{2}=\\left|e^{unstable i constant}\\right|^{2}=1 .\n\\]\n"
    },
    "garbled_string": {
      "map": {
        "x": "ntrbqplk",
        "y": "zvxsgmdh",
        "f": "qzxwvtnp",
        "g": "hjgrksla",
        "h": "ptfkgqbn",
        "C": "mclqrdzs",
        "b": "wvprxjkn"
      },
      "question": "Suppose $qzxwvtnp$ and $hjgrksla$ are non-constant, differentiable, real-valued\nfunctions defined on $(-\\infty, \\infty)$. Furthermore, suppose that for\neach pair of real numbers $ntrbqplk$ and $zvxsgmdh$,\n\\begin{align*}\nqzxwvtnp(ntrbqplk+zvxsgmdh) &= qzxwvtnp(ntrbqplk)qzxwvtnp(zvxsgmdh) - hjgrksla(ntrbqplk)hjgrksla(zvxsgmdh), \\\\\nhjgrksla(ntrbqplk+zvxsgmdh) &= qzxwvtnp(ntrbqplk)hjgrksla(zvxsgmdh) + hjgrksla(ntrbqplk)qzxwvtnp(zvxsgmdh).\n\\end{align*}\nIf $qzxwvtnp'(0) = 0$, prove that $(qzxwvtnp(ntrbqplk))^2 + (hjgrksla(ntrbqplk))^2 = 1$ for all $ntrbqplk$.",
      "solution": "Solution 1. Differentiate both sides with respect to \\( zvxsgmdh \\) to obtain\n\\[\n\\begin{aligned}\nqzxwvtnp^{\\prime}(ntrbqplk+zvxsgmdh) & =qzxwvtnp(ntrbqplk) qzxwvtnp^{\\prime}(zvxsgmdh)-hjgrksla(ntrbqplk) hjgrksla^{\\prime}(zvxsgmdh) \\\\\nhjgrksla^{\\prime}(ntrbqplk+zvxsgmdh) & =qzxwvtnp(ntrbqplk) hjgrksla^{\\prime}(zvxsgmdh)+hjgrksla(ntrbqplk) qzxwvtnp^{\\prime}(zvxsgmdh)\n\\end{aligned}\n\\]\n\nSetting \\( zvxsgmdh=0 \\) yields\n\\[\nqzxwvtnp^{\\prime}(ntrbqplk)=-hjgrksla^{\\prime}(0) hjgrksla(ntrbqplk) \\quad \\text { and } \\quad hjgrksla^{\\prime}(ntrbqplk)=hjgrksla^{\\prime}(0) qzxwvtnp(ntrbqplk)\n\\]\n\nThus\n\\[\n2 qzxwvtnp(ntrbqplk) qzxwvtnp^{\\prime}(ntrbqplk)+2 hjgrksla(ntrbqplk) hjgrksla^{\\prime}(ntrbqplk)=0,\n\\]\nand therefore\n\\[\nqzxwvtnp(ntrbqplk)^{2}+hjgrksla(ntrbqplk)^{2}=mclqrdzs\n\\]\nfor some constant \\( mclqrdzs \\). Since \\( qzxwvtnp \\) and \\( hjgrksla \\) are nonconstant, \\( mclqrdzs \\neq 0 \\). The identity\n\\[\nqzxwvtnp(ntrbqplk+zvxsgmdh)^{2}+hjgrksla(ntrbqplk+zvxsgmdh)^{2}=\\left(qzxwvtnp(ntrbqplk)^{2}+hjgrksla(ntrbqplk)^{2}\\right)\\left(qzxwvtnp(zvxsgmdh)^{2}+hjgrksla(zvxsgmdh)^{2}\\right),\n\\]\nimplies \\( mclqrdzs=mclqrdzs^{2} \\). But \\( mclqrdzs \\neq 0 \\), so \\( mclqrdzs=1 \\).\n\nSolution 2. Define \\( ptfkgqbn: \\mathbb{R} \\rightarrow \\mathbb{C} \\) by \\( ptfkgqbn(ntrbqplk)=qzxwvtnp(ntrbqplk)+i hjgrksla(ntrbqplk) \\). Then \\( ptfkgqbn \\) is differentiable, and \\( ptfkgqbn^{\\prime}(0)=wvprxjkn i \\) for some \\( wvprxjkn \\in \\mathbb{R} \\). The two given functional equations imply \\( ptfkgqbn(ntrbqplk+zvxsgmdh)= \\) \\( ptfkgqbn(ntrbqplk) ptfkgqbn(zvxsgmdh) \\). Differentiating with respect to \\( zvxsgmdh \\) and substituting \\( zvxsgmdh=0 \\) yields \\( ptfkgqbn^{\\prime}(ntrbqplk)= \\) \\( ptfkgqbn(ntrbqplk) ptfkgqbn^{\\prime}(0)=wvprxjkn i \\cdot ptfkgqbn(ntrbqplk) \\), so \\( ptfkgqbn(ntrbqplk)=mclqrdzs e^{wvprxjkn i ntrbqplk} \\) for some \\( mclqrdzs \\in \\mathbb{C} \\). From \\( ptfkgqbn(0+0)=ptfkgqbn(0) ptfkgqbn(0) \\) we obtain \\( mclqrdzs=mclqrdzs^{2} \\). If \\( mclqrdzs=0 \\), then \\( ptfkgqbn \\) would be identically zero, and \\( qzxwvtnp \\) and \\( hjgrksla \\) would be constant, contradiction. Thus \\( mclqrdzs=1 \\). Finally, for any \\( ntrbqplk \\in \\mathbb{R} \\),\n\\[\nqzxwvtnp(ntrbqplk)^{2}+hjgrksla(ntrbqplk)^{2}=|ptfkgqbn(ntrbqplk)|^{2}=\\left|e^{wvprxjkn i ntrbqplk}\\right|^{2}=1 .\n\\]"
    },
    "kernel_variant": {
      "question": "Let I be an open interval containing 0.  \nLet a,b,c,d : I → ℝ be differentiable, non-constant functions that satisfy, for every x,y with x, y, x+y ∈ I, the quaternion–addition system  \n\n  a(x+y)=a(x)a(y)−b(x)b(y)−c(x)c(y)−d(x)d(y)  \n  b(x+y)=a(x)b(y)+b(x)a(y)+c(x)d(y)−d(x)c(y)  \n  c(x+y)=a(x)c(y)−b(x)d(y)+c(x)a(y)+d(x)b(y)  \n  d(x+y)=a(x)d(y)+b(x)c(y)−c(x)b(y)+d(x)a(y).     (★)\n\nAssume that  \n  a′(0)=0 and (b′(0))²+(c′(0))²+(d′(0))²≠0.                             (†)\n\nProve that  \n\n  a(x)²+b(x)²+c(x)²+d(x)²=1 for every x∈I.",
      "solution": "Step 1.  Quaternion reformulation.  \nIntroduce the quaternion–valued function  \n  Q : I → ℍ, Q(x)=a(x)+b(x)i+c(x)j+d(x)k.  \nSystem (★) is exactly the statement  \n\n  Q(x+y)=Q(x)Q(y) for all x,y,x+y∈I.                                    (1)\n\n(1) is the Cauchy–type functional equation for a map into the non-commutative group (ℍ\\{0},·).\n\nStep 2.  The value of Q(0).  \nPutting x=y=0 in (1) gives Q(0)=Q(0)².  \nHence Q(0) is an idempotent quaternion; in ℍ that means Q(0)=0 or Q(0)=1.  \nIf Q(0)=0 then with y=0 in (1) we would get Q(x)=0 for every x, contradicting non-constancy. Thus  \n\n  Q(0)=1.                                                                (2)\n\nConsequently a(0)=1 and b(0)=c(0)=d(0)=0.\n\nStep 3.  First derivative of Q.  \nDifferentiate (1) with respect to y and then put y=0:\n\n  ∂/∂y|_{y=0}Q(x+y)=Q(x)Q′(0).  \nBut the left side is Q′(x), so\n\n  Q′(x)=Q(x)Q′(0) for every x∈I.                                        (3)\n\nWrite Q′(0)=α+βi+γj+δk.  From hypothesis (†) we have α=a′(0)=0 and (β,γ,δ)≠(0,0,0).  \nThus  \n\n  Q′(0)=V:=βi+γj+δk,         V²=−‖v‖²<0, ‖v‖:=√(β²+γ²+δ²).            (4)\n\nV is a fixed purely imaginary quaternion.\n\nStep 4.  A linear first–order ODE in ℍ.  \nEquation (3) is the constant-coefficient linear ordinary differential equation\n\n  Q′(x)=Q(x)V, Q(0)=1.                                                 (5)\n\nBecause V is constant, (5) is solved by the quaternionic exponential:\n\n  Q(x)=exp(xV)=∑_{n=0}^{∞} (xV)ⁿ/n!.                                   (6)\n\nUsing V²=−‖v‖², the usual identity for exponentials of purely imaginary quaternions yields\n\n  exp(xV)=cos(‖v‖x)+ (V/‖v‖) sin(‖v‖x).                               (7)\n\nTherefore  \n\n  a(x)=cos(‖v‖x),  (b(x),c(x),d(x))=(β,γ,δ)·(sin(‖v‖x)/‖v‖).        (8)\n\nStep 5.  Norm preservation.  \nFor any quaternion q, |q|²=q \\overline{q}.  Using (6) and the fact that exp(xV) is a unit quaternion (because V is purely imaginary),\n\n  |Q(x)|=1 for every x.                                                (9)\n\nBut |Q(x)|²=a(x)²+b(x)²+c(x)²+d(x)², so\n\n  a(x)²+b(x)²+c(x)²+d(x)²=1 for every x∈I.                            (10)\n\nThis is exactly the desired conclusion.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.719195",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension – four inter-dependent real functions instead of two, governed by the full quaternion addition law, enlarges the algebraic system from 2 to 4 equations and from the circle S¹ to the 3-sphere S³.  \n• Non-commutativity – the target group (ℍ\\{0},·) is non-abelian, so standard complex-analysis shortcuts no longer apply; one must manage ordering carefully.  \n• Group-theoretic insight – recognising (★) as a homomorphism into the quaternions and exploiting Lie-group/ Lie-algebra ideas (equation (5)) are essential.  \n• Continuous Cauchy theory in a non-commutative setting – differentiating (1) and solving the resulting quaternionic ODE requires familiarity with quaternion exponentials and their trigonometric form.  \n• Additional constraints – the derivative condition (†) forces the real part of Q′(0) to vanish, preventing trivial solutions and ensuring the generator V is purely imaginary, a subtle point absent from the original problem.  \n\nAll these features introduce layers of algebraic and analytic complexity far beyond the original sine-cosine style system, making the enhanced variant significantly harder."
      }
    },
    "original_kernel_variant": {
      "question": "Let I be an open interval containing 0.  \nLet a,b,c,d : I → ℝ be differentiable, non-constant functions that satisfy, for every x,y with x, y, x+y ∈ I, the quaternion–addition system  \n\n  a(x+y)=a(x)a(y)−b(x)b(y)−c(x)c(y)−d(x)d(y)  \n  b(x+y)=a(x)b(y)+b(x)a(y)+c(x)d(y)−d(x)c(y)  \n  c(x+y)=a(x)c(y)−b(x)d(y)+c(x)a(y)+d(x)b(y)  \n  d(x+y)=a(x)d(y)+b(x)c(y)−c(x)b(y)+d(x)a(y).     (★)\n\nAssume that  \n  a′(0)=0 and (b′(0))²+(c′(0))²+(d′(0))²≠0.                             (†)\n\nProve that  \n\n  a(x)²+b(x)²+c(x)²+d(x)²=1 for every x∈I.",
      "solution": "Step 1.  Quaternion reformulation.  \nIntroduce the quaternion–valued function  \n  Q : I → ℍ, Q(x)=a(x)+b(x)i+c(x)j+d(x)k.  \nSystem (★) is exactly the statement  \n\n  Q(x+y)=Q(x)Q(y) for all x,y,x+y∈I.                                    (1)\n\n(1) is the Cauchy–type functional equation for a map into the non-commutative group (ℍ\\{0},·).\n\nStep 2.  The value of Q(0).  \nPutting x=y=0 in (1) gives Q(0)=Q(0)².  \nHence Q(0) is an idempotent quaternion; in ℍ that means Q(0)=0 or Q(0)=1.  \nIf Q(0)=0 then with y=0 in (1) we would get Q(x)=0 for every x, contradicting non-constancy. Thus  \n\n  Q(0)=1.                                                                (2)\n\nConsequently a(0)=1 and b(0)=c(0)=d(0)=0.\n\nStep 3.  First derivative of Q.  \nDifferentiate (1) with respect to y and then put y=0:\n\n  ∂/∂y|_{y=0}Q(x+y)=Q(x)Q′(0).  \nBut the left side is Q′(x), so\n\n  Q′(x)=Q(x)Q′(0) for every x∈I.                                        (3)\n\nWrite Q′(0)=α+βi+γj+δk.  From hypothesis (†) we have α=a′(0)=0 and (β,γ,δ)≠(0,0,0).  \nThus  \n\n  Q′(0)=V:=βi+γj+δk,         V²=−‖v‖²<0, ‖v‖:=√(β²+γ²+δ²).            (4)\n\nV is a fixed purely imaginary quaternion.\n\nStep 4.  A linear first–order ODE in ℍ.  \nEquation (3) is the constant-coefficient linear ordinary differential equation\n\n  Q′(x)=Q(x)V, Q(0)=1.                                                 (5)\n\nBecause V is constant, (5) is solved by the quaternionic exponential:\n\n  Q(x)=exp(xV)=∑_{n=0}^{∞} (xV)ⁿ/n!.                                   (6)\n\nUsing V²=−‖v‖², the usual identity for exponentials of purely imaginary quaternions yields\n\n  exp(xV)=cos(‖v‖x)+ (V/‖v‖) sin(‖v‖x).                               (7)\n\nTherefore  \n\n  a(x)=cos(‖v‖x),  (b(x),c(x),d(x))=(β,γ,δ)·(sin(‖v‖x)/‖v‖).        (8)\n\nStep 5.  Norm preservation.  \nFor any quaternion q, |q|²=q \\overline{q}.  Using (6) and the fact that exp(xV) is a unit quaternion (because V is purely imaginary),\n\n  |Q(x)|=1 for every x.                                                (9)\n\nBut |Q(x)|²=a(x)²+b(x)²+c(x)²+d(x)², so\n\n  a(x)²+b(x)²+c(x)²+d(x)²=1 for every x∈I.                            (10)\n\nThis is exactly the desired conclusion.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.559634",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension – four inter-dependent real functions instead of two, governed by the full quaternion addition law, enlarges the algebraic system from 2 to 4 equations and from the circle S¹ to the 3-sphere S³.  \n• Non-commutativity – the target group (ℍ\\{0},·) is non-abelian, so standard complex-analysis shortcuts no longer apply; one must manage ordering carefully.  \n• Group-theoretic insight – recognising (★) as a homomorphism into the quaternions and exploiting Lie-group/ Lie-algebra ideas (equation (5)) are essential.  \n• Continuous Cauchy theory in a non-commutative setting – differentiating (1) and solving the resulting quaternionic ODE requires familiarity with quaternion exponentials and their trigonometric form.  \n• Additional constraints – the derivative condition (†) forces the real part of Q′(0) to vanish, preventing trivial solutions and ensuring the generator V is purely imaginary, a subtle point absent from the original problem.  \n\nAll these features introduce layers of algebraic and analytic complexity far beyond the original sine-cosine style system, making the enhanced variant significantly harder."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}