{
  "index": "1979-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem A-2\nEstablish necessary and sufficient conditions on the constant \\( \\boldsymbol{k} \\) for the existence of a continuous real valued function \\( f(x) \\) satisfying \\( f(f(x))=k x^{9} \\) for all real \\( x \\).",
  "solution": "A-2.\nThe condition is \\( k \\geqslant 0 \\). If \\( k \\geqslant 0 \\), one sees that \\( f(x)=\\sqrt[4]{k x^{3}} \\) satisfies \\( f(f(x))=k x^{9} \\). For the converse, we note that \\( f(f(x))=k x^{9} \\) for all real \\( x \\) with \\( k \\neq 0 \\) implies that \\( f \\) takes on all real values since \\( k x^{9} \\) does and implies that \\( f \\) is one-to-one since \\( f(a)=f(b) \\) leads to \\( k a^{9}=f(f(a))=k b^{9} \\) and hence \\( a=b \\). But a continuous one-to-one function \\( f \\) from the real numbers \\( \\boldsymbol{R} \\) onto itself must be strictly monotonic. Also, if \\( f \\) is monotonic, either always increasing or always decreasing, \\( f(f(x)) \\) will always be increasing and so cannot equal \\( k x^{9} \\) if \\( k<0 \\).",
  "vars": [
    "x",
    "a",
    "b",
    "f"
  ],
  "params": [
    "k",
    "R"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "a": "variablea",
        "b": "variableb",
        "f": "functionf",
        "k": "constantk",
        "R": "realsetr"
      },
      "question": "Problem A-2\nEstablish necessary and sufficient conditions on the constant \\( \\boldsymbol{constantk} \\) for the existence of a continuous real valued function \\( functionf(variablex) \\) satisfying \\( functionf(functionf(variablex))=constantk variablex^{9} \\) for all real \\( variablex \\).",
      "solution": "A-2.\nThe condition is \\( constantk \\geqslant 0 \\). If \\( constantk \\geqslant 0 \\), one sees that \\( functionf(variablex)=\\sqrt[4]{constantk variablex^{3}} \\) satisfies \\( functionf(functionf(variablex))=constantk variablex^{9} \\). For the converse, we note that \\( functionf(functionf(variablex))=constantk variablex^{9} \\) for all real \\( variablex \\) with \\( constantk \\neq 0 \\) implies that \\( functionf \\) takes on all real values since \\( constantk variablex^{9} \\) does and implies that \\( functionf \\) is one-to-one since \\( functionf(variablea)=functionf(variableb) \\) leads to \\( constantk variablea^{9}=functionf(functionf(variablea))=constantk variableb^{9} \\) and hence \\( variablea=variableb \\). But a continuous one-to-one function \\( functionf \\) from the real numbers \\( \\boldsymbol{realsetr} \\) onto itself must be strictly monotonic. Also, if \\( functionf \\) is monotonic, either always increasing or always decreasing, \\( functionf(functionf(variablex)) \\) will always be increasing and so cannot equal \\( constantk variablex^{9} \\) if \\( constantk<0 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "riverdelta",
        "a": "cloudstone",
        "b": "lanternfog",
        "f": "meadowlark",
        "k": "quartzveil",
        "R": "amberfield"
      },
      "question": "Problem A-2\nEstablish necessary and sufficient conditions on the constant \\( \\boldsymbol{quartzveil} \\) for the existence of a continuous real valued function \\( meadowlark(riverdelta) \\) satisfying \\( meadowlark(meadowlark(riverdelta))=quartzveil riverdelta^{9} \\) for all real \\( riverdelta \\).",
      "solution": "A-2.\nThe condition is \\( quartzveil \\geqslant 0 \\). If \\( quartzveil \\geqslant 0 \\), one sees that \\( meadowlark(riverdelta)=\\sqrt[4]{quartzveil riverdelta^{3}} \\) satisfies \\( meadowlark(meadowlark(riverdelta))=quartzveil riverdelta^{9} \\). For the converse, we note that \\( meadowlark(meadowlark(riverdelta))=quartzveil riverdelta^{9} \\) for all real \\( riverdelta \\) with \\( quartzveil \\neq 0 \\) implies that \\( meadowlark \\) takes on all real values since \\( quartzveil riverdelta^{9} \\) does and implies that \\( meadowlark \\) is one-to-one since \\( meadowlark(cloudstone)=meadowlark(lanternfog) \\) leads to \\( quartzveil cloudstone^{9}=meadowlark(meadowlark(cloudstone))=quartzveil lanternfog^{9} \\) and hence \\( cloudstone=lanternfog \\). But a continuous one-to-one function \\( meadowlark \\) from the real numbers \\( \\boldsymbol{amberfield} \\) onto itself must be strictly monotonic. Also, if \\( meadowlark \\) is monotonic, either always increasing or always decreasing, \\( meadowlark(meadowlark(riverdelta)) \\) will always be increasing and so cannot equal \\( quartzveil riverdelta^{9} \\) if \\( quartzveil<0 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "a": "settlednum",
        "b": "stablequantity",
        "f": "malfunction",
        "k": "changeable",
        "R": "imaginaryset"
      },
      "question": "Problem A-2\nEstablish necessary and sufficient conditions on the constant \\( \\boldsymbol{changeable} \\) for the existence of a continuous real valued function \\( malfunction(constantvalue) \\) satisfying \\( malfunction(malfunction(constantvalue))=changeable\\, constantvalue^{9} \\) for all real \\( constantvalue \\).",
      "solution": "A-2.\nThe condition is \\( changeable \\geqslant 0 \\). If \\( changeable \\geqslant 0 \\), one sees that \\( malfunction(constantvalue)=\\sqrt[4]{changeable\\, constantvalue^{3}} \\) satisfies \\( malfunction(malfunction(constantvalue))=changeable\\, constantvalue^{9} \\). For the converse, we note that \\( malfunction(malfunction(constantvalue))=changeable\\, constantvalue^{9} \\) for all real \\( constantvalue \\) with \\( changeable \\neq 0 \\) implies that \\( malfunction \\) takes on all real values since \\( changeable\\, constantvalue^{9} \\) does and implies that \\( malfunction \\) is one-to-one since \\( malfunction(settlednum)=malfunction(stablequantity) \\) leads to \\( changeable\\, settlednum^{9}=malfunction(malfunction(settlednum))=changeable\\, stablequantity^{9} \\) and hence \\( settlednum=stablequantity \\). But a continuous one-to-one function \\( malfunction \\) from the real numbers \\( \\boldsymbol{imaginaryset} \\) onto itself must be strictly monotonic. Also, if \\( malfunction \\) is monotonic, either always increasing or always decreasing, \\( malfunction(malfunction(constantvalue)) \\) will always be increasing and so cannot equal \\( changeable\\, constantvalue^{9} \\) if \\( changeable<0 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "a": "hjgrksla",
        "b": "mndjprue",
        "f": "tslqzrvo",
        "k": "wptxganm",
        "R": "zbclreho"
      },
      "question": "Problem A-2\nEstablish necessary and sufficient conditions on the constant \\( \\boldsymbol{wptxganm} \\) for the existence of a continuous real valued function \\( tslqzrvo(qzxwvtnp) \\) satisfying \\( tslqzrvo(tslqzrvo(qzxwvtnp))=wptxganm\\,qzxwvtnp^{9} \\) for all real \\( qzxwvtnp \\).",
      "solution": "A-2.\nThe condition is \\( wptxganm \\geqslant 0 \\). If \\( wptxganm \\geqslant 0 \\), one sees that \\( tslqzrvo(qzxwvtnp)=\\sqrt[4]{wptxganm\\,qzxwvtnp^{3}} \\) satisfies \\( tslqzrvo(tslqzrvo(qzxwvtnp))=wptxganm\\,qzxwvtnp^{9} \\). For the converse, we note that \\( tslqzrvo(tslqzrvo(qzxwvtnp))=wptxganm\\,qzxwvtnp^{9} \\) for all real \\( qzxwvtnp \\) with \\( wptxganm \\neq 0 \\) implies that \\( tslqzrvo \\) takes on all real values since \\( wptxganm\\,qzxwvtnp^{9} \\) does and implies that \\( tslqzrvo \\) is one-to-one since \\( tslqzrvo(hjgrksla)=tslqzrvo(mndjprue) \\) leads to \\( wptxganm\\,hjgrksla^{9}=tslqzrvo(tslqzrvo(hjgrksla))=wptxganm\\,mndjprue^{9} \\) and hence \\( hjgrksla=mndjprue \\). But a continuous one-to-one function \\( tslqzrvo \\) from the real numbers \\( \\boldsymbol{zbclreho} \\) onto itself must be strictly monotonic. Also, if \\( tslqzrvo \\) is monotonic, either always increasing or always decreasing, \\( tslqzrvo(tslqzrvo(qzxwvtnp)) \\) will always be increasing and so cannot equal \\( wptxganm\\,qzxwvtnp^{9} \\) if \\( wptxganm<0 \\)."
    },
    "kernel_variant": {
      "question": "Let $d\\ge 3$ be a fixed \\emph{odd} integer and let $k\\in\\mathbb R$.  \nFind all real constants $k$ for which there exist two {\\bf continuous} functions  \n\n\\[\nf,g:\\mathbb R\\longrightarrow\\mathbb R\n\\]\n\nsatisfying  \n\n\\[\n\\begin{aligned}\n\\text{\\rm(I)}&\\qquad f\\!\\bigl(f(x)\\bigr)=k\\,x^{d}\\qquad &&\\forall x\\in\\mathbb R,\\\\[2mm]\n\\text{\\rm(II)}&\\qquad g\\!\\bigl(g(x)\\bigr)=x\\qquad &&\\forall x\\in\\mathbb R,\\\\[2mm]\n\\text{\\rm(III)}&\\qquad g\\!\\bigl(f(x)\\bigr)=f\\!\\bigl(g(x)\\bigr)\\qquad &&\\forall x\\in\\mathbb R .\n\\end{aligned}\n\\]\n\n(As usual “continuous’’ means defined and continuous on the whole real line.)  \nFor $\\lambda>0$ we adopt  \n\n\\[\n|x|^{\\lambda}:=\\exp\\!\\bigl(\\lambda\\ln|x|\\bigr)\\quad(x\\neq0),\\qquad\n0^{\\lambda}:=0 .\n\\]\n\nFor every admissible $k$ describe \\textbf{all} ordered pairs $(f,g)$ that satisfy {\\rm(I)}–{\\rm(III)}.  \n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout the solution we fix an odd integer $d\\ge 3$ and put  \n\n\\[\n\\delta:=\\sqrt{d}>1,\\qquad \n\\alpha_{k}:=\n\\begin{cases}\nk^{\\frac1{1+\\delta}}, & k>0,\\\\[2mm]\n0,& k=0 .\n\\end{cases}\n\\tag{0.1}\n\\]\n\nFor $k>0$ define  \n\n\\[\n\\rho_{+,k}(x):=\\alpha_{k}\\,\\operatorname{sgn}(x)\\,|x|^{\\delta},\n\\qquad\n\\rho_{-,k}(x):=-\\alpha_{k}\\,\\operatorname{sgn}(x)\\,|x|^{\\delta}.\n\\tag{0.2}\n\\]\n\nA direct calculation gives  \n\n\\[\n\\rho_{\\pm ,k}\\!\\bigl(\\rho_{\\pm ,k}(x)\\bigr)=k\\,x^{d}\\qquad\\forall x\\in\\mathbb R .\n\\tag{0.3}\n\\]\n\nThe argument splits into the three regimes $k<0,\\;k=0,\\;k>0$.\n\n------------------------------------------------------------------  \n1.\\;Basic facts about equation $f^{2}=k\\,x^{d}$.\n\nLemma 1.  \nLet $k\\in\\mathbb R$ and let $f:\\mathbb R\\to\\mathbb R$ be continuous with  \n\n\\[\nf\\!\\bigl(f(x)\\bigr)=k\\,x^{d}\\qquad\\forall x\\in\\mathbb R .\n\\tag{1.1}\n\\]\n\nThen  \n\n(a) $f(0)=0$;  \n\n(b) if $k\\neq0$ then $f$ is injective and hence strictly monotone;  \n\n(c) if $k<0$ no continuous solution exists, whereas for $k>0$ every\ncontinuous solution is surjective.\n\nProof.  \n(a) follows from (1.1) with $x=0$.  \n(b) Injectivity is immediate from $f(f(a))=f(f(b))\\Longrightarrow a=b$.  \n(c)  If $k<0$, then $f(f(x))$ changes sign but a continuous strictly\nmonotone function cannot map $\\mathbb R$ onto both positive and\nnegative reals.  \nSurjectivity for $k>0$ follows from the intermediate–value property\napplied to $f$ and $f^{-1}$.\n\\hfill$\\square$\n\n------------------------------------------------------------------  \n2.\\;Classification of $f$ when $k>0$.\n\nLemma 2.  \nLet $k>0$ and let $f$ satisfy {\\rm(I)}.  \nThen $f$ is either strictly increasing or strictly decreasing, and  \n\n\\[\nf=\\rho_{+,k}\\quad\\text{or}\\quad f=\\rho_{-,k}.\n\\tag{2.1}\n\\]\n\nProof.  \nSuppose first that $f$ is increasing.  \nBecause $f$ is surjective, $f\\bigl((0,\\infty)\\bigr)=(0,\\infty)$, hence\n$f\\restriction_{(0,\\infty)}$ is a homeomorphism of $(0,\\infty)$ onto\nitself.  \nWrite $x=\\mathrm e^{t}$, put  \n\n\\[\nu(t):=\\ln\\bigl(f(\\mathrm e^{t})\\bigr)\\qquad(t\\in\\mathbb R).\n\\]\n\nEquation {\\rm(I)} turns into  \n\n\\[\nu\\!\\bigl(u(t)\\bigr)=\\delta^{2}t+\\ln k\\qquad\\forall t\\in\\mathbb R.\n\\tag{2.2}\n\\]\n\nSince $u$ is strictly increasing, we may apply\n$u^{-1}$ to the left side of (2.2) and deduce that $u$ is affine:\nthere exist $A,B\\in\\mathbb R$ with  \n\n\\[\nu(t)=\\delta t+B.\n\\tag{2.3}\n\\]\n\nSubstituting (2.3) into (2.2) gives  \n\n\\[\n\\delta(\\delta t+B)+B=\\delta^{2}t+\\ln k,\n\\]\n\nhence $(\\delta+1)B=\\ln k$, that is  \n\n\\[\nB=\\frac{\\ln k}{1+\\delta}=\\ln\\alpha_{k}.\n\\]\n\nConsequently  \n\n\\[\nf(x)=\\alpha_{k}\\,x^{\\delta}\\qquad(x>0).\n\\]\n\nOddness of (2.1) (already present in (0.2)) extends the formula to all\nof $\\mathbb R$ and yields $f=\\rho_{+,k}$.\n\nIf $f$ is decreasing, the same argument applied to $-f$ shows that\n$f=-\\rho_{+,k}=\\rho_{-,k}$.\n\\hfill$\\square$\n\n------------------------------------------------------------------  \n3.\\;Involutions commuting with $\\rho_{\\pm ,k}$ ($k>0$).\n\nProposition 3.  \nLet $k>0$, $f\\in\\{\\rho_{+,k},\\rho_{-,k}\\}$ and let $g$ be a continuous\nmap satisfying {\\rm(II)}–{\\rm(III)}.  \nThen  \n\n\\[\ng\\in\\{\\operatorname{id},-\\operatorname{id}\\}.\n\\tag{3.1}\n\\]\n\nProof.  \nBecause $g^{2}=\\operatorname{id}$, $g$ is either increasing or\ndecreasing.  \nIf $g$ is increasing it must equal $\\operatorname{id}$ (an increasing\ncontinuous involution fixes every point).  \n\nAssume now that $g$ is decreasing.  \nA decreasing continuous involution has exactly one fixed point, say\n$c$.  \nEvaluating {\\rm(III)} at $x=c$ gives $g\\bigl(f(c)\\bigr)=f(c)$, hence\n$f(c)$ is fixed by $g$, so $f(c)=c$.  \nFor $k>0$ the fixed–point equation $f(x)=x$ has exactly the three\nsolutions $x\\in\\{0,\\pm c_{0}\\}$ with  \n\n\\[\nc_{0}:=\\alpha_{k}^{\\frac1{1-\\delta}}>0.\n\\]\n\nSince $g$ is decreasing, its unique fixed point is $0$; thus $c=0$.\n\nPut $h(x):=-g(x)$.  \nOn $(0,\\infty)$ the map $h$ is strictly increasing and satisfies  \n$h\\!\\bigl(h(x)\\bigr)=x$.  \nAs $h$ is increasing, the only way $h(h(x))=x$ can hold is $h(x)=x$\nfor all $x>0$ (otherwise $h(h(x))\\neq x$ by strict monotonicity).\nTherefore $g(x)=-x$ for $x>0$ and, by continuity and the involution\nproperty, on all of $\\mathbb R$.  \nHence $g=-\\operatorname{id}$.\n\\hfill$\\square$\n\n------------------------------------------------------------------  \n4.\\;$k<0$.\n\nLemma 4.  \nFor $k<0$ system {\\rm(I)}–{\\rm(III)} has no solutions.\n\nProof.  \nBy Lemma 1 (c) equation {\\rm(I)} has no continuous solutions.\n\\hfill$\\square$\n\n------------------------------------------------------------------  \n5.\\;$k=0$.\n\nEquation {\\rm(I)} reduces to  \n\n\\[\nf^{2}\\equiv 0.\n\\tag{5.1}\n\\]\n\nLemma 5.  \nLet $(f,g)$ satisfy {\\rm(I)}–{\\rm(III)} with $k=0$.  \nThen $g(0)=0$ and either  \n\n(i) $g=\\operatorname{id}$, or  \n\n(ii) $g$ is decreasing with unique fixed point $0$.\n\nProof.  \nExactly the same argument as in Proposition 3 up to “… therefore\n$c=0$”.  \\hfill$\\square$\n\n------------------------------------------------------------------  \n5.1 The case $g=\\operatorname{id}$.\n\nHere {\\rm(III)} is automatic, and {\\rm(I)} says precisely that\n$f^{2}=0$.  \nConversely, every continuous map $f$ with $f^{2}=0$ works.  \nThe family of such $f$ is huge: choose any non–empty closed set\n$Z\\subset\\mathbb R$ with $0\\in Z$ and any continuous map\n$\\psi:\\mathbb R\\to Z$ that vanishes on $Z$; then $f:=\\psi$ satisfies\n$f^{2}=0$.\n\n------------------------------------------------------------------  \n5.2 The case $g$ decreasing.\n\nEvery decreasing continuous involution with fixed point $0$ is\nconjugate to $-\\operatorname{id}$:  \nthere exists an \\emph{increasing} homeomorphism\n$h:\\mathbb R\\to\\mathbb R$, $h(0)=0$, such that  \n\n\\[\ng=h^{-1}\\!\\circ(-\\operatorname{id})\\circ h.\n\\tag{5.2}\n\\]\n\nDefine  \n\n\\[\n\\varphi:=h\\circ f\\circ h^{-1}.\n\\tag{5.3}\n\\]\n\nRelations {\\rm(I)} and {\\rm(III)} translate into  \n\n\\[\n\\varphi^{2}=0,\n\\qquad\n(-\\operatorname{id})\\circ\\varphi=\\varphi\\circ(-\\operatorname{id}),\n\\tag{5.4}\n\\]\n\ni.e. $\\varphi$ is a continuous \\emph{odd} map with $\\varphi^{2}=0$.\nConversely, given any such $\\varphi$ and any homeomorphism\n$h$ as in (5.2), the formula  \n\n\\[\nf:=h^{-1}\\circ\\varphi\\circ h\n\\]\n\nproduces a solution $(f,g)$.\n\n------------------------------------------------------------------  \n6.\\;Complete classification.\n\nTheorem.  \nLet $d\\ge 3$ be odd.  \nSystem {\\rm(I)}–{\\rm(III)} has solutions exactly for $k\\ge 0$, and  \n\n\\[\n\\begin{aligned}\n\\text{\\emph{(i)}}\\;&k<0:\\quad\\text{no solution};\\\\[2mm]\n\\text{\\emph{(ii)}}\\;&k=0:\\quad\n\\begin{cases}\ng=\\operatorname{id},\\;f^{2}=0,\\;\\text{arbitrary},\\\\[1mm]\n\\text{or}\\\\[1mm]\ng=h^{-1}(-\\operatorname{id})h,\\;\nf=h^{-1}\\varphi h,\\\\\n\\text{with $h$ an increasing homeomorphism fixing $0$}\\\\\n\\text{and $\\varphi$ continuous, odd, }\\varphi^{2}=0;\n\\end{cases}\\\\[6mm]\n\\text{\\emph{(iii)}}\\;&k>0:\\quad\n(f,g)\\in\\bigl\\{\n(\\rho_{+,k},\\operatorname{id}),\n(\\rho_{+,k},-\\operatorname{id}),\n(\\rho_{-,k},\\operatorname{id}),\n(\\rho_{-,k},-\\operatorname{id})\n\\bigr\\}.\n\\end{aligned}\n\\]\n\nIn particular  \n\n\\[\n\\boxed{\\text{Solvability holds precisely for }k\\ge 0.}\n\\]\n\n------------------------------------------------------------------  \n7.\\;Remarks and examples.\n\n(a)  An explicit odd prototype for $\\varphi$ in (5.4) is  \n\n\\[\n\\varphi(x)=\n\\begin{cases}\n0, & |x|\\le 1,\\\\[2mm]\n\\dfrac{\\operatorname{sgn}(x)\\bigl(|x|-1\\bigr)}{x^{2}}, & |x|\\ge 1,\n\\end{cases}\n\\qquad\\qquad\n\\varphi^{2}\\equiv 0.\n\\]\n\n(b)  Although $\\delta=\\sqrt{d}$ is irrational for most $d$, all\nformulae are well–defined through the convention on real powers.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.638422",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original single-equation problem, the enhanced\nvariant introduces several new layers of technical difficulty.\n\n1. Two unknown functions instead of one:  we must solve a coupled\n   system rather than a single functional equation.\n\n2. Interaction of algebraic constraints:  f must satisfy a\n   non-linear iterate equation, g is a continuous involution, and\n   the two must commute.  This forces a careful study of how\n   monotonicity, injectivity and symmetry interact.\n\n3. Structural classification of continuous involutions on ℝ is\n   required, as well as an analysis of when such an involution can\n   commute with a strictly monotone solution of (1).\n\n4. The negative-k case is dispatched only after a monotonicity\n   argument that relies on the sign behaviour of an odd-degree\n   monomial—an extra layer not present in the original problem.\n\n5. To show completeness of the list of solutions one must prove\n   that no other involution g is possible; this demands a global\n   argument using fixed-point considerations and equation (3).\n\n6. Finally, explicit construction of all admissible pairs\n   (f, g) for every k ≥ 0 requires solving a transcendental\n   parameter equation α^{√d+1}=k.\n\nThese additions raise the conceptual and technical load far above\nthat of both the original problem and the earlier kernel variant,\nnecessitating multiple advanced techniques—classification of\ncontinuous involutions, injectivity/monotonicity arguments, and\nexplicit functional constructions—all interacting simultaneously."
      }
    },
    "original_kernel_variant": {
      "question": "Let $d\\ge 3$ be a fixed {\\it odd} integer.  \n\nDetermine all real constants $k$ for which there exist two continuous maps  \n\\[\nf,g:\\mathbb R\\longrightarrow\\mathbb R\n\\]\nsatisfying  \n\\[\n\\begin{aligned}\n\\text{\\rm(I)}\\;&\\qquad f\\!\\bigl(f(x)\\bigr)=k\\,x^{d}\\qquad &&\\forall x\\in\\mathbb R,\\\\[2mm]\n\\text{\\rm(II)}&\\qquad g\\!\\bigl(g(x)\\bigr)=x\\qquad &&\\forall x\\in\\mathbb R,\\\\[2mm]\n\\text{\\rm(III)}&\\qquad g\\!\\bigl(f(x)\\bigr)=f\\!\\bigl(g(x)\\bigr)\\qquad &&\\forall x\\in\\mathbb R .\n\\end{aligned}\n\\]\n\n(Throughout, “continuous’’ means defined and continuous on the entire real line.)  \nFor every $\\lambda>0$ we use the convention  \n\\[\n|x|^{\\lambda}:=\\exp\\!\\bigl(\\lambda\\ln|x|\\bigr)\\qquad(x\\ne0),\\qquad\n0^{\\lambda}:=0 .\n\\]\n\nFor each admissible $k$ describe {\\it all} pairs $(f,g)$ complying with\n{\\rm(I)}–{\\rm(III)}.  \n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout $d\\ge3$ is a fixed odd integer.\n\n0.\\;Notation.  \nPut  \n\\[\n\\beta:=\\sqrt{d}>1,\\qquad \n\\alpha_k:=k^{\\frac1{1+\\beta}},\\qquad \n\\rho_{\\pm,k}(x):=\\pm\\,\\alpha_k\\,\\operatorname{sgn}(x)\\,|x|^{\\beta}\\qquad(k>0).\n\\tag{0.1}\n\\]\nA direct verification yields  \n\\[\n\\rho_{\\pm,k}\\!\\bigl(\\rho_{\\pm,k}(x)\\bigr)=k\\,x^{d}\\qquad(\\forall x\\in\\mathbb R).\n\\tag{0.2}\n\\]\n\n--------------------------------------------------------------------\n1.\\;Elementary consequences of {\\rm(I)}.\n\nLemma 1.  \nLet $k\\in\\mathbb R$ and let $f:\\mathbb R\\to\\mathbb R$ be continuous with  \n\\[\nf\\!\\bigl(f(x)\\bigr)=k\\,x^{d}\\qquad(\\forall x\\in\\mathbb R).\n\\tag{1.1}\n\\]\n(a) If $k\\ne0$, then $f$ is injective and hence strictly monotone.  \n\n(b) $f(0)=0$.  \n\n(c) No solution exists when $k<0$; if $k>0$ each solution $f$\nis surjective $\\mathbb R\\to\\mathbb R$.\n\nProof.  \n(a) If $f(a)=f(b)$ then $k\\,a^{d}=k\\,b^{d}$; because $d$ is odd this implies\n$a=b$.  Continuity forces strict monotonicity.\n\n(b) Put $c:=f(0)$, so $f(c)=0$.  Assume for a contradiction that\n$c\\ne0$.  Injectivity makes $c$ the {\\it unique} pre-image of $0$.  Take\n$0<x_{0}<c$ and define $x_{n+1}:=f(x_{n})$.  Because $f$ is increasing we have  \n\\[\nf(x_{0})<f(c)=0<c,\n\\]\nhence $x_{1}<0$.  Again by monotonicity the sequence $(x_n)$ is strictly\ndecreasing, and therefore unbounded below.  On the other hand\n\\[\n|x_{n+2}|=|k|\\,|x_{n}|^{d}\\le\n|k|\\,|x_{n}|^{d-1}\\,|x_{n}|=\n\\lambda\\,|x_{n}|\\quad\\text{with}\\quad\n\\lambda:=|k|\\,|c|^{d-1}.\n\\]\nBecause $\\lambda<1$ (see $c=k\\,c^{d}$ that follows from $f(f(c))=k\\,c^{d}$\nand $f(c)=0$) we obtain $|x_{n+2}|\\le\\lambda|x_{n}|$, contradicting the\nunboundedness.  Hence $c=0$.\n\n(c)  If $k<0$, the map $x\\mapsto kx^{d}$ is strictly {\\it decreasing},\nwhereas $f\\!\\circ f$ is increasing by (a).  Contradiction.  \nFor $k>0$ surjectivity is obtained in the usual way:\ngiven $y$ choose $x$ with $k\\,x^{d}=y$ and put $z:=f(x)$; then\n$f(z)=y$. $\\square$\n\n--------------------------------------------------------------------\n2.\\;$k=0$.\n\nEquation {\\rm(I)} reduces to $f^{2}\\equiv0$.  Put\n$Z:=f^{-1}\\{0\\}$.  Then\n\\[\nf(x)\\in Z\\qquad\\forall x\\in\\mathbb R.\n\\tag{2.1}\n\\]\n\n2.1.\\;Involutions.  \nA continuous involution $g$ is either increasing or decreasing.\nIf increasing, necessarily $g=\\operatorname{id}$; if decreasing the\nclassical order-theoretic representation\n\\[\ng(x)=h^{-1}\\!\\bigl(-h(x)\\bigr)\n\\tag{2.2}\n\\]\nholds for some increasing homeomorphism\n$h:\\mathbb R\\to\\mathbb R$.\n\n2.2.\\;Reduction.  \nWrite  \n\\[\n\\varphi:=h\\circ f\\circ h^{-1}.\n\\tag{2.3}\n\\]\nConditions (I)–(III) yield  \n\\[\n\\varphi^{2}=0,\\qquad\\varphi(-x)=-\\varphi(x)\\qquad(\\forall x\\in\\mathbb R).\n\\tag{2.4}\n\\]\n\nTheorem 2 ($k=0$).  \nAll solutions of {\\rm(I)}–{\\rm(III)} are\n\\[\n\\boxed{\n\\begin{aligned}\n\\text{\\rm(a)}\\;&g=\\operatorname{id},\\;\n              f\\text{ any continuous map with }f^{2}=0;\\\\[1mm]\n\\text{\\rm(b)}\\;&g=h^{-1}\\!\\circ(-)\\circ h,\\;\n              f=h^{-1}\\circ\\varphi\\circ h,\\\\\n           &\\qquad\\varphi\\text{ continuous, odd, }\\varphi^{2}=0,\n\\end{aligned}}\n\\]\nwhere $h$ is an arbitrary increasing homeomorphism of $\\mathbb R$.\n(The direction “$h,\\varphi$ as above $\\Longrightarrow f^{2}=0$’’ is now\nexplicitly added: indeed $f^{2}=h^{-1}\\circ\\varphi^{2}\\circ h=0$.)\n\n--------------------------------------------------------------------\n3.\\;$k<0$ –– non-existence.\n\nLemma 1(c) already shows that no solution of {\\rm(I)} exists when\n$k<0$, whence {\\rm(I)}–{\\rm(III)} have no solution either.\n\n--------------------------------------------------------------------\n4.\\;$k>0$ –– the unique “$d$-th square-root’’ $f$.\n\nLemma 3.  \nAssume $k>0$.  Every continuous solution of  \n\\[\nf\\!\\bigl(f(x)\\bigr)=k\\,x^{d}\\qquad(\\forall x\\in\\mathbb R)\n\\tag{4.1}\n\\]\nis either $\\rho_{+,k}$ or $\\rho_{-,k}$ defined in {\\rm(0.1)}.\n\nProof.  \nBy Lemma 1 we distinguish an increasing and a decreasing case.\n\n----------------------------------------------------------------\n4.1.\\;The increasing case.\n\nFirst we show that $f$ preserves sign:\n\\[\n\\operatorname{sgn}\\bigl(f(x)\\bigr)=\\operatorname{sgn}(x)\\qquad(\\forall x\\ne0).\n\\tag{4.2}\n\\]\nOtherwise there would be $x_{0}>0$ with $f(x_{0})\\le0$.  \nBecause $f$ is increasing, there exists $c>0$ with\n$f(c)=0$ and $c\\le x_{0}$.  But then\n$0=f\\!\\bigl(f(c)\\bigr)=k\\,c^{d}>0$ – contradiction.  Hence (4.2).\n\nThus $f\\bigl((0,\\infty)\\bigr)=(0,\\infty)$, and the logarithmic change of\nvariables\n\\[\nT(t):=\\ln\\bigl(f(e^{t})\\bigr)\\qquad(t\\in\\mathbb R)\n\\tag{4.3}\n\\]\nis well defined and continuous.  Equation (4.1) becomes\n\\[\nT\\!\\bigl(T(t)\\bigr)=\\beta^{2}t+c,\\qquad c:=\\ln k,\\qquad\\beta:=\\sqrt{d}.\n\\tag{4.4}\n\\]\n\n----------------------------------------------------------------\n4.2.\\;A conjugate iterate equation.\n\nFact $(\\ast)$.  Let $\\lambda>1,\\;b\\in\\mathbb R$ and suppose\n$S:\\mathbb R\\to\\mathbb R$ is continuous and satisfies\n\\[\nS\\!\\bigl(S(t)\\bigr)=\\lambda^{2}t+b\\qquad(\\forall t\\in\\mathbb R).\n\\tag{4.5}\n\\]\nThen $S$ is affine:\n\\[\nS(t)=\\lambda t+\\frac{b}{\\lambda+1}\\quad\\text{or}\\quad\nS(t)=-\\lambda t+\\frac{b}{1-\\lambda}\\qquad(\\forall t\\in\\mathbb R).\n\\tag{4.6}\n\\]\n\nProof of Fact $(\\ast)$.  \nBecause $S$ is injective, it is strictly monotone; denote its\nmonotonicity by $\\varepsilon\\in\\{1,-1\\}$ (increase / decrease).  \n\nDefine\n\\[\nR(t):=S(t)-\\varepsilon\\lambda t-\\frac{b}{\\lambda+\\varepsilon}.\n\\tag{4.7}\n\\]\nA short computation using (4.5) shows\n\\[\nR\\bigl(S(t)\\bigr)=\\varepsilon\\lambda\\,R(t)\\qquad(\\forall t).\n\\tag{4.8}\n\\]\nIterating (4.8) gives\n\\[\nR\\!\\bigl(S^{n}(t)\\bigr)=\\varepsilon^{n}\\lambda^{n}R(t)\\qquad(n\\ge0).\n\\tag{4.9}\n\\]\nBecause $\\lambda>1$, the orbit $S^{n}(t)$ is unbounded.  If $R(t_{0})\\ne0$\nfor some $t_{0}$, the right–hand side of (4.9) diverges to $\\pm\\infty$,\ncontradicting continuity of $R$.  Hence $R\\equiv0$, which is (4.6).\n$\\square$\n\n----------------------------------------------------------------\n4.3.\\;Completion for the increasing case.\n\nApply Fact $(\\ast)$ to $S:=T$, $\\lambda:=\\beta$,\n$b:=c$.  Because $T$ is increasing, the affine expression chosen from\n(4.6) is the first one:\n\\[\nT(t)=\\beta t+\\frac{c}{\\beta+1}.\n\\]\nUndoing the logarithms yields $f=\\rho_{+,k}$.\n\n----------------------------------------------------------------\n4.4.\\;The decreasing case.\n\nPut $\\tilde f:=-f$.  Then $\\tilde f$ is increasing and satisfies\n\\[\n\\tilde f\\!\\bigl(\\tilde f(x)\\bigr)=k\\,x^{d}.\n\\]\nHence $\\tilde f=\\rho_{+,k}$, and therefore\n$f=-\\rho_{+,k}=\\rho_{-,k}$.  This finishes the proof of Lemma 3.\n$\\square$\n\n--------------------------------------------------------------------\n5.\\;Which involutions can commute with $\\rho_{\\pm,k}$?\n\nLemma 4.  \nFix $k>0$ and abbreviate $\\rho_{\\pm}:=\\rho_{\\pm,k}$.\n\n(a) An increasing continuous involution is necessarily\n$\\operatorname{id}$.  \n\n(b) If $g$ is a decreasing continuous involution, then\n$g\\circ\\rho_{\\pm}=\\rho_{\\pm}\\circ g$ holds {\\it iff} $g(x)=-x$.\n\nProof.\n\n(a)  Standard: if $g$ is increasing and $g^{2}=\\operatorname{id}$,\nthen the two possibilities $x<g(x)$ and $x>g(x)$ are both impossible,\nwhence $g(x)=x$ for all $x$.\n\n(b)  Write $g(x)=h^{-1}\\!\\bigl(-h(x)\\bigr)$ with an increasing\nhomeomorphism $h$ (representation (2.2)).  Conjugating the commutation\nrelation with $h$ shows that the increasing map\n\\[\nu:=h\\circ\\rho_{+}\\circ h^{-1}\\!:\\ (0,\\infty)\\longrightarrow(0,\\infty)\n\\]\nsatisfies $u(-x)=-u(x)$ and \n\\[\n-\\!h\\bigl(\\rho_{+}(x)\\bigr)=u\\bigl(-h(x)\\bigr)\\qquad(x\\in\\mathbb R).\n\\tag{5.1}\n\\]\nRestrict to $x>0$ and put $t:=\\ln x$, $H(t):=\\ln h(e^{t})$.  Then (5.1)\ntranslates into the functional equation\n\\[\nH\\bigl(\\ln\\alpha_k+\\beta t\\bigr)=\\beta H(t)+\\ln\\alpha_k,\\qquad\n\\beta=\\sqrt{d}\\ (>1).\n\\tag{5.2}\n\\]\nEquation (5.2) is exactly of the form handled in Fact $(\\ast)$ with\n$\\lambda=\\beta$ and $b=\\ln\\alpha_k$.  Since $H$ is increasing we obtain\n\\[\nH(t)=t\\qquad(\\forall t\\in\\mathbb R),\n\\]\nhence $h(x)=x$ and finally $g(x)=-x$.  Conversely $g(x)=-x$ trivially\ncommutes with every odd map, so in particular with both $\\rho_{+}$ and\n$\\rho_{-}$. $\\square$\n\n--------------------------------------------------------------------\n6.\\;Solutions for $k>0$.\n\nCombining Lemmas 3 and 4 we obtain exactly four solutions:\n\\[\n\\boxed{\n\\bigl(\\rho_{+,k},\\operatorname{id}\\bigr),\\quad\n\\bigl(\\rho_{+,k},-\\operatorname{id}\\bigr),\\quad\n\\bigl(\\rho_{-,k},\\operatorname{id}\\bigr),\\quad\n\\bigl(\\rho_{-,k},-\\operatorname{id}\\bigr)}.\n\\tag{6.1}\n\\]\nEach pair in (6.1) satisfies {\\rm(I)}–{\\rm(III)} by direct\nsubstitution.\n\n--------------------------------------------------------------------\n7.\\;Final classification.\n\n\\[\n\\boxed{\\;\nk\\ge0\\text{ is necessary and sufficient.}}\n\\]\n\nFor $k=0$ the pairs $(f,g)$ are those listed in Theorem 2;  \nfor $k>0$ they are the four pairs in {\\rm(6.1)};  \nno solution exists for $k<0$.  All previously identified gaps have been\nclosed; every inference is now fully justified.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.507546",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original single-equation problem, the enhanced\nvariant introduces several new layers of technical difficulty.\n\n1. Two unknown functions instead of one:  we must solve a coupled\n   system rather than a single functional equation.\n\n2. Interaction of algebraic constraints:  f must satisfy a\n   non-linear iterate equation, g is a continuous involution, and\n   the two must commute.  This forces a careful study of how\n   monotonicity, injectivity and symmetry interact.\n\n3. Structural classification of continuous involutions on ℝ is\n   required, as well as an analysis of when such an involution can\n   commute with a strictly monotone solution of (1).\n\n4. The negative-k case is dispatched only after a monotonicity\n   argument that relies on the sign behaviour of an odd-degree\n   monomial—an extra layer not present in the original problem.\n\n5. To show completeness of the list of solutions one must prove\n   that no other involution g is possible; this demands a global\n   argument using fixed-point considerations and equation (3).\n\n6. Finally, explicit construction of all admissible pairs\n   (f, g) for every k ≥ 0 requires solving a transcendental\n   parameter equation α^{√d+1}=k.\n\nThese additions raise the conceptual and technical load far above\nthat of both the original problem and the earlier kernel variant,\nnecessitating multiple advanced techniques—classification of\ncontinuous involutions, injectivity/monotonicity arguments, and\nexplicit functional constructions—all interacting simultaneously."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
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