{
  "index": "1988-A-5",
  "type": "ALG",
  "tag": [
    "ALG",
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "Prove that there exists a \\emph{unique} function $f$ from the set\n$\\mathrm{R}^+$ of positive real numbers to $\\mathrm{R}^+$ such that\n\\[\nf(f(x)) = 6x-f(x)\n\\]\nand\n\\[\nf(x)>0\n\\]\nfor all $x>0$.",
  "solution": "Solution. We will show that \\( f(x)=2 x \\) for \\( x>0 \\). Fix \\( x>0 \\), and consider the sequence \\( \\left(a_{n}\\right) \\) of positive numbers defined by \\( a_{0}=x \\) and \\( a_{n+1}=f\\left(a_{n}\\right) \\). The given functional equation implies that \\( a_{n+2}+a_{n+1}-6 a_{n}=0 \\). The zeros of the characteristic polynomial \\( t^{2}+t-6 \\) of this linear recursion are -3 and 2 , so there exist real constants \\( c_{1}, c_{2} \\) such that \\( a_{n}=c_{1} 2^{n}+c_{2}(-3)^{n} \\) for all \\( n \\geq 0 \\). If \\( c_{2} \\neq 0 \\), then for large \\( n, a_{n} \\) has the same sign as \\( c_{2}(-3)^{n} \\), which alternates with \\( n \\); this contradicts \\( a_{n}>0 \\). Therefore \\( c_{2}=0 \\), and \\( a_{n}=c_{1} 2^{n} \\). In particular \\( f(x)=a_{1}=2 a_{0}=2 x \\). This holds for any \\( x \\). Finally, the function \\( f(x)=2 x \\) does satisfy the conditions of the problem.\n\nRelated question. The following, which was Problem 5 of the 1989 Asian-Pacific Mathematical Olympiad [APMO], can be solved using a similar method:\n\nDetermine all functions \\( f \\) from the reals to the reals for which\n(1) \\( f(x) \\) is strictly increasing,\n(2) \\( f(x)+g(x)=2 x \\) for all real \\( x \\),\nwhere \\( g(x) \\) is the composition inverse function to \\( f(x) \\). (Note: \\( f \\) and \\( g \\) are said to be composition inverses if \\( f(g(x))=x \\) and \\( g(f(x))=x \\) for all real \\( x \\).)\n\nRemark (linear recursive sequences with constant coefficients). We can describe the general sequence \\( x_{0}, x_{1}, x_{2}, \\ldots \\) satisfying the linear recursion\n\\[\nx_{n+k}+b_{k-1} x_{n+k-1}+\\cdots+b_{1} x_{n+1}+b_{0} x_{n}=0\n\\]\nfor all \\( n \\geq 0 \\), where \\( b_{0}, b_{1}, \\ldots, b_{k-1} \\) are constants, by considering the characteristic polynomial\n\\[\nf(t)=t^{k}+b_{k-1} t^{k-1}+\\cdots+b_{1} t+b_{0}\n\\]\nover a field large enough to contain all its zeros. If the zeros \\( r_{1}, \\ldots, r_{k} \\) of \\( f(t) \\) are distinct, then the general solution is \\( x_{n}=\\sum_{i=1}^{k} c_{i} r_{i}^{n} \\) where the \\( c_{i} \\) are arbitrary constants not depending on \\( n \\). More generally, if \\( f(t)=\\prod_{i=1}^{s}\\left(t-r_{i}\\right)^{m_{i}} \\), then, provided that we are working over a field of characteristic zero, the general solution is \\( x_{n}=\\sum_{i=1}^{s} c_{i}(n) r_{i}^{n} \\) where \\( c_{i}(n) \\) is a polynomial of degree less than \\( m_{i} \\). For any characteristic, the latter statement remains true if we replace the polynomial \\( c_{i}(n) \\) by a general linear combination of the binomial coefficients \\( \\binom{n}{0},\\binom{n}{1}, \\ldots,\\binom{n}{m_{i}-1} \\). (These combinations are the same as the polynomials in \\( n \\) of degree less than \\( m_{i} \\) if the characteristic is zero, or if the characteristic is at least as large as \\( m_{i} \\).) All of these statements can be proved by showing that the generating function \\( \\sum_{i=0}^{\\infty} x_{i} t^{i} \\) is a rational function of \\( t \\), and then decomposing it as a sum of partial fractions and expanding each partial fraction in a power series to get a formula for \\( x_{i} \\). For more discussion, see [NZM, Appendix A.4].\n\nFor example, these results can be used to show that the Fibonacci sequence, defined by \\( F_{0}=0, F_{1}=1 \\), and \\( F_{n+2}=F_{n+1}+F_{n} \\) for \\( n \\geq 0 \\), satisfies\n\\[\nF_{n}=\\frac{\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{n}-\\left(\\frac{1-\\sqrt{5}}{2}\\right)^{n}}{\\sqrt{5}}\n\\]\n\nThe formula for the general solution to linear recursions with constant coefficients can be thought of as the discrete analogue of the general solution to homogeneous linear ordinary differential equations with constant coefficients.",
  "vars": [
    "f",
    "g",
    "x",
    "n",
    "k",
    "s",
    "t",
    "i",
    "a_0",
    "a_n",
    "a_n+1",
    "a_n+2",
    "x_0",
    "x_1",
    "x_2",
    "x_n",
    "x_n+k",
    "x_n+k-1",
    "x_n+1",
    "x_i",
    "F_0",
    "F_1",
    "F_n"
  ],
  "params": [
    "c_1",
    "c_2",
    "c_i",
    "r_1",
    "r_k",
    "r_i",
    "b_0",
    "b_1",
    "b_k-1",
    "m_i"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "transform",
        "g": "inverse",
        "x": "realpos",
        "n": "indexer",
        "k": "ordernum",
        "s": "rootcount",
        "t": "variable",
        "i": "summation",
        "a_0": "seqstart",
        "a_n": "seqgeneral",
        "a_n+1": "seqnext",
        "a_n+2": "seqnexttwo",
        "x_0": "genstart",
        "x_1": "genfirst",
        "x_2": "gensecond",
        "x_n": "gengeneral",
        "x_n+k": "gengeneralshift",
        "x_n+k-1": "gengeneralprev",
        "x_n+1": "gengeneralnext",
        "x_i": "genspecific",
        "F_0": "fibstart",
        "F_1": "fibfirst",
        "F_n": "fibgeneral",
        "c_1": "coeffone",
        "c_2": "coefftwo",
        "c_i": "coeffgeneric",
        "r_1": "rootone",
        "r_k": "rootlast",
        "r_i": "rootgeneric",
        "b_0": "constzero",
        "b_1": "constone",
        "b_k-1": "constlast",
        "m_i": "multindex"
      },
      "question": "Prove that there exists a \\emph{unique} function $\\mathrm{transform}$ from the set $\\mathrm{R}^+$ of positive real numbers to $\\mathrm{R}^+$ such that\n\\[\n\\mathrm{transform}\\bigl(\\mathrm{transform}(\\mathrm{realpos})\\bigr)=6\\mathrm{realpos}-\\mathrm{transform}(\\mathrm{realpos})\n\\]\nand\n\\[\n\\mathrm{transform}(\\mathrm{realpos})>0\n\\]\nfor all $\\mathrm{realpos}>0$.",
      "solution": "Solution. We will show that $\\mathrm{transform}(\\mathrm{realpos})=2\\,\\mathrm{realpos}$ for $\\mathrm{realpos}>0$. Fix $\\mathrm{realpos}>0$, and consider the sequence $(\\mathrm{seqgeneral})$ of positive numbers defined by $\\mathrm{seqstart}=\\mathrm{realpos}$ and $\\mathrm{seqnext}=\\mathrm{transform}(\\mathrm{seqgeneral})$. The given functional equation implies that $\\mathrm{seqnexttwo}+\\mathrm{seqnext}-6\\,\\mathrm{seqgeneral}=0$. The zeros of the characteristic polynomial $\\mathrm{variable}^{2}+\\mathrm{variable}-6$ of this linear recursion are $-3$ and $2$, so there exist real constants $\\mathrm{coeffone},\\mathrm{coefftwo}$ such that\n\\[\n\\mathrm{seqgeneral}=\\mathrm{coeffone}\\,2^{\\mathrm{indexer}}+\\mathrm{coefftwo}(-3)^{\\mathrm{indexer}}\\qquad(\\mathrm{indexer}\\ge0).\n\\]\nIf $\\mathrm{coefftwo}\\neq0$, then for large $\\mathrm{indexer}$ the term $\\mathrm{seqgeneral}$ has the same sign as $\\mathrm{coefftwo}(-3)^{\\mathrm{indexer}}$, which alternates with $\\mathrm{indexer}$; this contradicts $\\mathrm{seqgeneral}>0$. Therefore $\\mathrm{coefftwo}=0$ and $\\mathrm{seqgeneral}=\\mathrm{coeffone}\\,2^{\\mathrm{indexer}}$. In particular\n\\[\n\\mathrm{transform}(\\mathrm{realpos})=a_{1}=2\\,\\mathrm{seqstart}=2\\,\\mathrm{realpos}.\n\\]\nThis holds for any $\\mathrm{realpos}$. Finally, the function $\\mathrm{transform}(\\mathrm{realpos})=2\\,\\mathrm{realpos}$ does satisfy the conditions of the problem.\n\nRelated question. The following, which was Problem 5 of the 1989 Asian–Pacific Mathematical Olympiad (APMO), can be solved using a similar method:\n\nDetermine all functions $\\mathrm{transform}$ from the reals to the reals for which\n(1) $\\mathrm{transform}(\\mathrm{realpos})$ is strictly increasing,\n(2) $\\mathrm{transform}(\\mathrm{realpos})+\\mathrm{inverse}(\\mathrm{realpos})=2\\,\\mathrm{realpos}$ for all real $\\mathrm{realpos}$,\nwhere $\\mathrm{inverse}(\\mathrm{realpos})$ is the composition inverse of $\\mathrm{transform}(\\mathrm{realpos})$. (Note: $\\mathrm{transform}$ and $\\mathrm{inverse}$ are composition inverses if $\\mathrm{transform}(\\mathrm{inverse}(\\mathrm{realpos}))=\\mathrm{realpos}$ and $\\mathrm{inverse}(\\mathrm{transform}(\\mathrm{realpos}))=\\mathrm{realpos}$ for all real $\\mathrm{realpos}$.)\n\nRemark (linear recursive sequences with constant coefficients). We can describe the general sequence $\\mathrm{genstart},\\mathrm{genfirst},\\mathrm{gensecond},\\ldots$ satisfying the linear recursion\n\\[\n\\mathrm{gengeneralshift}+\\mathrm{constlast}\\,\\mathrm{gengeneralprev}+\\cdots+\\mathrm{constone}\\,\\mathrm{gengeneralnext}+\\mathrm{constzero}\\,\\mathrm{gengeneral}=0\n\\]\nfor all $\\mathrm{indexer}\\ge0$, where $\\mathrm{constzero},\\mathrm{constone},\\ldots,\\mathrm{constlast}$ are constants, by considering the characteristic polynomial\n\\[\n\\mathrm{transform}(\\mathrm{variable})=\\mathrm{variable}^{\\mathrm{ordernum}}+\\mathrm{constlast}\\,\\mathrm{variable}^{\\mathrm{ordernum}-1}+\\cdots+\\mathrm{constone}\\,\\mathrm{variable}+\\mathrm{constzero}.\n\\]\nIf the zeros $\\mathrm{rootone},\\ldots,\\mathrm{rootlast}$ of $\\mathrm{transform}(\\mathrm{variable})$ are distinct, then the general solution is $\\mathrm{gengeneral}=\\sum_{\\mathrm{summation}=1}^{\\mathrm{ordernum}} \\mathrm{coeffgeneric}\\,\\mathrm{rootgeneric}^{\\mathrm{indexer}}$ where the $\\mathrm{coeffgeneric}$ are arbitrary constants independent of $\\mathrm{indexer}$. More generally, if $\\mathrm{transform}(\\mathrm{variable})=\\prod_{\\mathrm{summation}=1}^{\\mathrm{rootcount}}(\\mathrm{variable}-\\mathrm{rootgeneric})^{\\mathrm{multindex}}$, then, provided we are working over a field of characteristic $0$, the general solution is $\\mathrm{gengeneral}=\\sum_{\\mathrm{summation}=1}^{\\mathrm{rootcount}} \\mathrm{coeffgeneric}(\\mathrm{indexer})\\,\\mathrm{rootgeneric}^{\\mathrm{indexer}}$ where $\\mathrm{coeffgeneric}(\\mathrm{indexer})$ is a polynomial of degree less than $\\mathrm{multindex}$. For any characteristic, the latter statement remains true if we replace the polynomial $\\mathrm{coeffgeneric}(\\mathrm{indexer})$ by an arbitrary linear combination of the binomial coefficients $\\binom{\\mathrm{indexer}}{0},\\binom{\\mathrm{indexer}}{1},\\ldots,\\binom{\\mathrm{indexer}}{\\mathrm{multindex}-1}$. All of these statements can be proved by showing that the generating function $\\sum_{\\mathrm{summation}=0}^{\\infty} \\mathrm{genspecific}\\,\\mathrm{variable}^{\\mathrm{summation}}$ is a rational function of $\\mathrm{variable}$, then decomposing it into partial fractions and expanding each partial fraction into a power series.\n\nFor example, these results imply that the Fibonacci sequence, defined by $\\mathrm{fibstart}=0,\\;\\mathrm{fibfirst}=1$ and $F_{n+2}=F_{n+1}+F_{n}$ for $\\mathrm{indexer}\\ge0$, satisfies\n\\[\n\\mathrm{fibgeneral}=\\frac{\\bigl(\\tfrac{1+\\sqrt{5}}{2}\\bigr)^{\\mathrm{indexer}}-\\bigl(\\tfrac{1-\\sqrt{5}}{2}\\bigr)^{\\mathrm{indexer}}}{\\sqrt{5}}.\n\\]\n\nThe formula for the general solution to linear recursions with constant coefficients can be viewed as a discrete analogue of the general solution to homogeneous linear ordinary differential equations with constant coefficients."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "skyscraper",
        "g": "locomotive",
        "x": "blueberry",
        "n": "strawberry",
        "k": "butterscotch",
        "s": "marshmallow",
        "t": "cheesecake",
        "i": "pineapple",
        "a_0": "watermelon",
        "a_n": "tangerine",
        "a_n+1": "cloudberry",
        "a_n+2": "pomegranate",
        "x_0": "blacksmith",
        "x_1": "bricklayer",
        "x_2": "craftsman",
        "x_n": "lighthouse",
        "x_n+k": "sailboat",
        "x_n+k-1": "starboard",
        "x_n+1": "wheelhouse",
        "x_i": "gearshift",
        "F_0": "hydrangea",
        "F_1": "magnolia",
        "F_n": "honeysuckle",
        "c_1": "chandelier",
        "c_2": "periscope",
        "c_i": "windjammer",
        "r_1": "moonscape",
        "r_k": "starflower",
        "r_i": "seaplane",
        "b_0": "driftwood",
        "b_1": "hearthstone",
        "b_k-1": "ironclad",
        "m_i": "stingray"
      },
      "question": "Prove that there exists a \\emph{unique} function $skyscraper$ from the set\n$\\mathrm{R}^+$ of positive real numbers to $\\mathrm{R}^+$ such that\n\\[\nskyscraper(skyscraper(blueberry)) = 6 blueberry - skyscraper(blueberry)\n\\]\nand\n\\[\nskyscraper(blueberry)>0\n\\]\nfor all $blueberry>0$.",
      "solution": "Solution. We will show that \\( skyscraper(blueberry)=2 blueberry \\) for \\( blueberry>0 \\). Fix \\( blueberry>0 \\), and consider the sequence \\( \\left(tangerine_{strawberry}\\right) \\) of positive numbers defined by \\( watermelon=blueberry \\) and \\( cloudberry=skyscraper\\left(tangerine_{strawberry}\\right) \\). The given functional equation implies that \\( pomegranate+cloudberry-6 tangerine_{strawberry}=0 \\). The zeros of the characteristic polynomial \\( cheesecake^{2}+cheesecake-6 \\) of this linear recursion are -3 and 2, so there exist real constants \\( chandelier, periscope \\) such that \\( tangerine_{strawberry}=chandelier 2^{strawberry}+periscope(-3)^{strawberry} \\) for all \\( strawberry \\geq 0 \\). If \\( periscope \\neq 0 \\), then for large \\( strawberry, tangerine_{strawberry} \\) has the same sign as \\( periscope(-3)^{strawberry} \\), which alternates with \\( strawberry \\); this contradicts \\( tangerine_{strawberry}>0 \\). Therefore \\( periscope=0 \\), and \\( tangerine_{strawberry}=chandelier 2^{strawberry} \\). In particular \\( skyscraper(blueberry)=a_{1}=2 watermelon=2 blueberry \\). This holds for any \\( blueberry \\). Finally, the function \\( skyscraper(blueberry)=2 blueberry \\) does satisfy the conditions of the problem.\n\nRelated question. The following, which was Problem 5 of the 1989 Asian-Pacific Mathematical Olympiad [APMO], can be solved using a similar method:\n\nDetermine all functions \\( skyscraper \\) from the reals to the reals for which\n(1) \\( skyscraper(blueberry) \\) is strictly increasing,\n(2) \\( skyscraper(blueberry)+locomotive(blueberry)=2 blueberry \\) for all real \\( blueberry \\),\nwhere \\( locomotive(blueberry) \\) is the composition inverse function to \\( skyscraper(blueberry) \\). (Note: \\( skyscraper \\) and \\( locomotive \\) are said to be composition inverses if \\( skyscraper(locomotive(blueberry))=blueberry \\) and \\( locomotive(skyscraper(blueberry))=blueberry \\) for all real \\( blueberry \\).)\n\nRemark (linear recursive sequences with constant coefficients). We can describe the general sequence \\( blacksmith, bricklayer, craftsman, \\ldots \\) satisfying the linear recursion\n\\[\nsailboat+ironclad_{butterscotch-1}\\, starboard+\\cdots+hearthstone\\, wheelhouse+driftwood\\, lighthouse=0\n\\]\nfor all \\( strawberry \\geq 0 \\), where \\( driftwood, hearthstone, \\ldots, ironclad_{butterscotch-1} \\) are constants, by considering the characteristic polynomial\n\\[\nskyscraper(cheesecake)=cheesecake^{butterscotch}+ironclad_{butterscotch-1} cheesecake^{butterscotch-1}+\\cdots+hearthstone cheesecake+driftwood\n\\]\nover a field large enough to contain all its zeros. If the zeros \\( moonscape, \\ldots, starflower \\) of \\( skyscraper(cheesecake) \\) are distinct, then the general solution is \\( lighthouse_{strawberry}=\\sum_{pineapple=1}^{butterscotch} windjammer\\, seaplane^{strawberry} \\) where the \\( windjammer \\) are arbitrary constants not depending on \\( strawberry \\). More generally, if \\( skyscraper(cheesecake)=\\prod_{pineapple=1}^{marshmallow}\\left(cheesecake-seaplane\\right)^{stingray} \\), then, provided that we are working over a field of characteristic zero, the general solution is \\( lighthouse_{strawberry}=\\sum_{pineapple=1}^{marshmallow} windjammer(strawberry)\\, seaplane^{strawberry} \\) where \\( windjammer(strawberry) \\) is a polynomial of degree less than \\( stingray \\). For any characteristic, the latter statement remains true if we replace the polynomial \\( windjammer(strawberry) \\) by a general linear combination of the binomial coefficients \\( \\binom{strawberry}{0},\\binom{strawberry}{1}, \\ldots,\\binom{strawberry}{stingray-1} \\). (These combinations are the same as the polynomials in \\( strawberry \\) of degree less than \\( stingray \\) if the characteristic is zero, or if the characteristic is at least as large as \\( stingray \\).) All of these statements can be proved by showing that the generating function \\( \\sum_{pineapple=0}^{\\infty} gearshift\\, cheesecake^{pineapple} \\) is a rational function of \\( cheesecake \\), and then decomposing it as a sum of partial fractions and expanding each partial fraction in a power series to get a formula for \\( gearshift \\). For more discussion, see [NZM, Appendix A.4].\n\nFor example, these results can be used to show that the Fibonacci sequence, defined by \\( hydrangea=0, magnolia=1 \\), and \\( F_{strawberry+2}=F_{strawberry+1}+honeysuckle \\) for \\( strawberry \\geq 0 \\), satisfies\n\\[\nhoneysuckle=\\frac{\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{strawberry}-\\left(\\frac{1-\\sqrt{5}}{2}\\right)^{strawberry}}{\\sqrt{5}}\n\\]\n\nThe formula for the general solution to linear recursions with constant coefficients can be thought of as the discrete analogue of the general solution to homogeneous linear ordinary differential equations with constant coefficients."
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "unordered",
        "g": "noninvert",
        "x": "immutable",
        "n": "boundednum",
        "k": "singleton",
        "s": "wholeone",
        "t": "standstill",
        "i": "collective",
        "a_0": "terminalval",
        "a_n": "specificval",
        "a_n+1": "precedingval",
        "a_n+2": "antecedentval",
        "x_0": "endpointval",
        "x_1": "penultimate",
        "x_2": "prefinalval",
        "x_n": "definiteval",
        "x_n+k": "ancillaryval",
        "x_n+k-1": "proximateval",
        "x_n+1": "priorval",
        "x_i": "memberval",
        "F_0": "nonfibstart",
        "F_1": "nonfibnext",
        "F_n": "nonfibgen",
        "c_1": "variableone",
        "c_2": "variabletwo",
        "c_i": "variablegen",
        "r_1": "leafoneval",
        "r_k": "leafkval",
        "r_i": "leafival",
        "b_0": "freezero",
        "b_1": "freefirst",
        "b_k-1": "freekprev",
        "m_i": "uniquenessi"
      },
      "question": "Prove that there exists a \\emph{unique} function $unordered$ from the set\n$\\mathrm{R}^+$ of positive real numbers to $\\mathrm{R}^+$ such that\n\\[\nunordered(unordered(immutable)) = 6immutable-unordered(immutable)\n\\]\nand\n\\[\nunordered(immutable)>0\n\\]\nfor all $immutable>0$.",
      "solution": "Solution. We will show that \\( unordered(immutable)=2 immutable \\) for \\( immutable>0 \\). Fix \\( immutable>0 \\), and consider the sequence \\( \\left(specificval\\right) \\) of positive numbers defined by \\( terminalval=immutable \\) and \\( precedingval=unordered\\left(specificval\\right) \\). The given functional equation implies that \\( antecedentval+precedingval-6 specificval=0 \\). The zeros of the characteristic polynomial \\( standstill^{2}+standstill-6 \\) of this linear recursion are -3 and 2 , so there exist real constants \\( variableone, variabletwo \\) such that \\( specificval=variableone 2^{boundednum}+variabletwo(-3)^{boundednum} \\) for all \\( boundednum \\geq 0 \\). If \\( variabletwo \\neq 0 \\), then for large \\( boundednum, specificval \\) has the same sign as \\( variabletwo(-3)^{boundednum} \\), which alternates with \\( boundednum \\); this contradicts \\( specificval>0 \\). Therefore \\( variabletwo=0 \\), and \\( specificval=variableone 2^{boundednum} \\). In particular \\( unordered(immutable)=a_{1}=2 terminalval=2 immutable \\). This holds for any \\( immutable \\). Finally, the function \\( unordered(immutable)=2 immutable \\) does satisfy the conditions of the problem.\n\nRelated question. The following, which was Problem 5 of the 1989 Asian-Pacific Mathematical Olympiad [APMO], can be solved using a similar method:\n\nDetermine all functions \\( unordered \\) from the reals to the reals for which\n(1) \\( unordered(immutable) \\) is strictly increasing,\n(2) \\( unordered(immutable)+noninvert(immutable)=2 immutable \\) for all real \\( immutable \\),\nwhere \\( noninvert(immutable) \\) is the composition inverse function to \\( unordered(immutable) \\). (Note: \\( unordered \\) and \\( noninvert \\) are said to be composition inverses if \\( unordered(noninvert(immutable))=immutable \\) and \\( noninvert(unordered(immutable))=immutable \\) for all real \\( immutable \\).)\n\nRemark (linear recursive sequences with constant coefficients). We can describe the general sequence \\( endpointval, penultimate, prefinalval, \\ldots \\) satisfying the linear recursion\n\\[\nancillaryval+freekprev proximateval+\\cdots+freefirst priorval+freezero definiteval=0\n\\]\nfor all \\( boundednum \\geq 0 \\), where \\( freezero, freefirst, \\ldots, freekprev \\) are constants, by considering the characteristic polynomial\n\\[\nunordered(standstill)=standstill^{singleton}+freekprev standstill^{singleton-1}+\\cdots+freefirst standstill+freezero\n\\]\nover a field large enough to contain all its zeros. If the zeros \\( leafoneval, \\ldots, leafkval \\) of \\( unordered(standstill) \\) are distinct, then the general solution is \\( definiteval=\\sum_{collective=1}^{singleton} variablegen leafival^{boundednum} \\) where the \\( variablegen \\) are arbitrary constants not depending on \\( boundednum \\). More generally, if \\( unordered(standstill)=\\prod_{collective=1}^{wholeone}\\left(standstill-leafival\\right)^{uniquenessi} \\), then, provided that we are working over a field of characteristic zero, the general solution is \\( definiteval=\\sum_{collective=1}^{wholeone} variablegen(boundednum) leafival^{boundednum} \\) where \\( variablegen(boundednum) \\) is a polynomial of degree less than \\( uniquenessi \\). For any characteristic, the latter statement remains true if we replace the polynomial \\( variablegen(boundednum) \\) by a general linear combination of the binomial coefficients \\( \\binom{boundednum}{0},\\binom{boundednum}{1}, \\ldots,\\binom{boundednum}{uniquenessi-1} \\). (These combinations are the same as the polynomials in \\( boundednum \\) of degree less than \\( uniquenessi \\) if the characteristic is zero, or if the characteristic is at least as large as \\( uniquenessi \\).) All of these statements can be proved by showing that the generating function \\( \\sum_{collective=0}^{\\infty} memberval standstill^{collective} \\) is a rational function of \\( standstill \\), and then decomposing it as a sum of partial fractions and expanding each partial fraction in a power series to get a formula for \\( memberval \\). For more discussion, see [NZM, Appendix A.4].\n\nFor example, these results can be used to show that the Fibonacci sequence, defined by \\( nonfibstart=0, nonfibnext=1 \\), and \\( F_{boundednum+2}=F_{boundednum+1}+F_{boundednum} \\) for \\( boundednum \\geq 0 \\), satisfies\n\\[\nnonfibgen=\\frac{\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{boundednum}-\\left(\\frac{1-\\sqrt{5}}{2}\\right)^{boundednum}}{\\sqrt{5}}\n\\]\n\nThe formula for the general solution to linear recursions with constant coefficients can be thought of as the discrete analogue of the general solution to homogeneous linear ordinary differential equations with constant coefficients."
    },
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        "x_n+1": "rqehztnc",
        "x_i": "ofrndplk",
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        "F_n": "veqltugs",
        "c_1": "tbqlsnev",
        "c_2": "pkwdhram",
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        "m_i": "ulkcdyso"
      },
      "question": "Prove that there exists a \\emph{unique} function $qzxwvtnp$ from the set $\\mathrm{R}^+$ of positive real numbers to $\\mathrm{R}^+$ such that\n\\[\nqzxwvtnp(qzxwvtnp(vrtosnke)) = 6vrtosnke-qzxwvtnp(vrtosnke)\n\\]\nand\n\\[\nqzxwvtnp(vrtosnke)>0\n\\]\nfor all $vrtosnke>0$.",
      "solution": "Solution. We will show that $qzxwvtnp(vrtosnke)=2\\,vrtosnke$ for $vrtosnke>0$. Fix $vrtosnke>0$, and consider the sequence $(dfzmhlsa)$ of positive numbers defined by $ryxqmntu=vrtosnke$ and $xqsgrbwe=qzxwvtnp(dfzmhlsa)$. The given functional equation implies that $mlrkhpov+xqsgrbwe-6\\,dfzmhlsa=0$. The zeros of the characteristic polynomial $bqjdehfa^{2}+bqjdehfa-6$ of this linear recursion are $-3$ and $2$, so there exist real constants $tbqlsnev,pkwdhram$ such that $dfzmhlsa=tbqlsnev\\,2^{alpqwzen}+pkwdhram(-3)^{alpqwzen}$ for all $alpqwzen\\ge 0$. If $pkwdhram\\neq0$, then for large $alpqwzen$, $dfzmhlsa$ has the same sign as $pkwdhram(-3)^{alpqwzen}$, which alternates with $alpqwzen$; this contradicts $dfzmhlsa>0$. Therefore $pkwdhram=0$, and $dfzmhlsa=tbqlsnev\\,2^{alpqwzen}$. In particular $qzxwvtnp(vrtosnke)=a_{1}=2\\,ryxqmntu=2\\,vrtosnke$. This holds for any $vrtosnke$. Finally, the function $qzxwvtnp(vrtosnke)=2\\,vrtosnke$ does satisfy the conditions of the problem.\n\nRelated question. The following, which was Problem 5 of the 1989 Asian-Pacific Mathematical Olympiad [APMO], can be solved using a similar method:\n\nDetermine all functions $qzxwvtnp$ from the reals to the reals for which\n(1) $qzxwvtnp(vrtosnke)$ is strictly increasing,\n(2) $qzxwvtnp(vrtosnke)+hjgrksla(vrtosnke)=2\\,vrtosnke$ for all real $vrtosnke$, where $hjgrksla(vrtosnke)$ is the composition inverse function to $qzxwvtnp(vrtosnke)$. (Note: $qzxwvtnp$ and $hjgrksla$ are said to be composition inverses if $qzxwvtnp(hjgrksla(vrtosnke))=vrtosnke$ and $hjgrksla(qzxwvtnp(vrtosnke))=vrtosnke$ for all real $vrtosnke$.)\n\nRemark (linear recursive sequences with constant coefficients). We can describe the general sequence $cenfgrst,whdqmpai,ayrvlsue,\\ldots$ satisfying the linear recursion\n\\[\nkygzradc+gqshvpnm\\,vbsloxmh+\\cdots+wzdfheyj\\,rqehztnc+npxbrlve\\,uhpkewza=0\n\\]\nfor all $alpqwzen\\ge 0$, where $npxbrlve,wzdfheyj,\\ldots,gqshvpnm$ are constants, by considering the characteristic polynomial\n\\[\nqzxwvtnp(bqjdehfa)=bqjdehfa^{cydhmrux}+gqshvpnm\\,bqjdehfa^{cydhmrux-1}+\\cdots+wzdfheyj\\,bqjdehfa+npxbrlve\n\\]\nover a field large enough to contain all its zeros. If the zeros $vbajhqps,\\ldots,dyxnlcua$ of $qzxwvtnp(bqjdehfa)$ are distinct, then the general solution is $uhpkewza=\\sum_{zpowklen=1}^{cydhmrux} scvomgye\\,mafhsdqt^{alpqwzen}$ where the $scvomgye$ are arbitrary constants not depending on $alpqwzen$. More generally, if $qzxwvtnp(bqjdehfa)=\\prod_{zpowklen=1}^{xavlgpzo}(bqjdehfa-mafhsdqt)^{ulkcdyso}$, then, provided that we are working over a field of characteristic zero, the general solution is $uhpkewza=\\sum_{zpowklen=1}^{xavlgpzo} scvomgye(alpqwzen)\\,mafhsdqt^{alpqwzen}$ where $scvomgye(alpqwzen)$ is a polynomial of degree less than $ulkcdyso$. For any characteristic, the latter statement remains true if we replace the polynomial $scvomgye(alpqwzen)$ by a general linear combination of the binomial coefficients $\\binom{alpqwzen}{0},\\binom{alpqwzen}{1},\\ldots,\\binom{alpqwzen}{ulkcdyso-1}$. (These combinations are the same as the polynomials in $alpqwzen$ of degree less than $ulkcdyso$ if the characteristic is zero, or if the characteristic is at least as large as $ulkcdyso$.) All of these statements can be proved by showing that the generating function $\\sum_{zpowklen=0}^{\\infty} ofrndplk\\,bqjdehfa^{zpowklen}$ is a rational function of $bqjdehfa$, and then decomposing it as a sum of partial fractions and expanding each partial fraction in a power series to get a formula for $ofrndplk$. For more discussion, see [NZM, Appendix A.4].\n\nFor example, these results can be used to show that the Fibonacci sequence, defined by $jzwqspan=0,\\,atynhkfd=1$, and $F_{n+2}=F_{n+1}+veqltugs$ for $alpqwzen\\ge 0$, satisfies\n\\[\nveqltugs=\\frac{\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{alpqwzen}-\\left(\\frac{1-\\sqrt{5}}{2}\\right)^{alpqwzen}}{\\sqrt{5}}\n\\]\n\nThe formula for the general solution to linear recursions with constant coefficients can be thought of as the discrete analogue of the general solution to homogeneous linear ordinary differential equations with constant coefficients."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nP(t)=t^{4}-10t^{3}+35t^{2}-50t+24=(t-1)(t-2)(t-3)(t-4)\n\\]\nand, for every natural number \\(n\\), let \\(f^{\\circ n}\\) denote the\n\\(n\\)-fold self–composition of a map \\(f\\).\nDetermine all strictly increasing {\\it affine} maps  \n\\[\nf:(0,\\infty)\\longrightarrow(0,\\infty),\\qquad \nf(x)=\\kappa x+c\\;(x>0),\n\\]\nthat satisfy  \n\n\\[\nf^{\\circ 4}(x)-10\\,f^{\\circ 3}(x)+35\\,f^{\\circ 2}(x)-50\\,f(x)+24\\,x=0\n\\qquad(x>0).\\tag{$\\clubsuit$}\n\\]\nHere\n\\[\n\\kappa>0,\\qquad c\\ge 0 .\n\\]",
      "solution": "For an affine map \\(f(x)=\\kappa x+c\\;(x>0)\\) we plainly have  \n\n\\[\nf^{\\circ n}(x)=\\kappa^{\\,n}x+c\\,\n   \\frac{\\kappa^{\\,n}-1}{\\kappa-1}\\qquad(n\\ge 1).\n\\tag{1}\n\\]\n\nInsert these expressions into \\((\\clubsuit)\\).  Collecting the\ncoefficients of \\(x\\) and of the constant term \\(c\\) separately we get  \n\n\\[\n\\begin{aligned}\nP(\\kappa)\\,x&=0\n\\qquad(x>0),\\\\\nQ(\\kappa)\\,c&=0,\n\\end{aligned}\\tag{2}\n\\]\nwhere  \n\\[\nQ(t)=t^{3}-9t^{2}+26t-24.\n\\tag{3}\n\\]\n\nBecause \\((2)\\) has to hold for all \\(x>0\\) we must have  \n\n\\[\nP(\\kappa)=0,\\qquad \n   \\bigl(c>0 \\Longrightarrow Q(\\kappa)=0\\bigr).\n\\tag{4}\n\\]\n\nComputation gives  \n\n\\[\nP(t)=(t-1)(t-2)(t-3)(t-4),\\qquad\nQ(t)=(t-2)(t-3)(t-4).\n\\tag{5}\n\\]\n\nHence  \n\n\\[\nP(\\kappa)=0 \\Longrightarrow \n   \\kappa\\in\\{1,2,3,4\\}.\n\\tag{6}\n\\]\n\nIf \\(\\kappa=1\\) then \\(Q(1)=-6\\neq0\\); consequently the second\nequation in \\((4)\\) forces \\(c=0\\).  \nIf \\(\\kappa\\in\\{2,3,4\\}\\) then \\(Q(\\kappa)=0\\), and no further\nrestriction is imposed on \\(c\\).\n\nIn every case the slope \\(\\kappa\\) is positive and the map is strictly\nincreasing for every non–negative value of \\(c\\).  Therefore the\ncomplete solution family is  \n\n\\[\n\\boxed{\\,f_{1,0}(x)=x,\\; f_{\\kappa,c}(x)=\\kappa x+c\n  \\;(\\kappa\\in\\{2,3,4\\},\\;c\\ge 0)\\,}.\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.704429",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-order composition: The relation involves the 4-fold iterate of f, turning the\n  two‐term second-order recursion of the original into a five-term fourth-order one.\n• Spectral theory:  Solving the problem now needs linear-algebraic machinery (companion\n  matrices, eigenvalues, Perron–Frobenius theorem) instead of an elementary quadratic\n  characteristic polynomial.\n• Multiple positive eigenvalues:  The roots 1,2,3,4 are all positive, so the\n  sign-alternation trick that dispatches the −3 root in the original problem no longer\n  works; one must analyse long-term growth and use limit arguments.\n• Dynamical-systems viewpoint:  Step 5 converts the functional equation into a statement\n  about the ω-limit set of a discrete dynamical system and shows that all orbits share\n  the same Lyapunov quotient.\n• Uniqueness via closure of quotient set:  Establishing that g(x)=f(x)/x is constant\n  requires a delicate combination of monotonicity, orbit analysis and Perron–Frobenius\n  asymptotics.\n\nThese layers of linear-algebraic and dynamical techniques render the enhanced\nvariant markedly more intricate than both the original and the former kernel\nversion, which were solvable with a single elementary recursion plus a sign\nargument."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nP(t)=t^{4}-10t^{3}+35t^{2}-50t+24=(t-1)(t-2)(t-3)(t-4).\n\\]\n\nFor a natural number $n$ write $f^{\\circ n}$ for the $n$–fold self–composition of $f$.  \n\nDetermine all functions  \n\\[\nf:(0,\\infty)\\longrightarrow(0,\\infty)\n\\]  \nthat satisfy simultaneously  \n\n(1) $f$ is strictly increasing and continuous on $(0,\\infty)$;  \n\n(2) the quartic composition identity  \n\\[\nf^{\\circ 4}(x)-10\\,f^{\\circ 3}(x)+35\\,f^{\\circ 2}(x)-50\\,f(x)+24\\,x=0\n\\qquad(x>0).\\tag{$\\ast$}\n\\]\n\nShow that every such function is affine, namely  \n\\[\nf(x)=\\kappa x+c\\qquad(x>0),\n\\]\nwhere the slope $\\kappa$ is one of the four zeros $1,2,3,4$ of $P$ and where the\ntranslation part $c$ is non–negative, with the extra restriction $c=0$ when $\\kappa=1$.  \nEquivalently, the complete list of solutions is  \n\\[\nf_{1,0}(x)=x,\\qquad   \nf_{2,c}(x)=2x+c\\;(c\\ge 0),\\qquad  \nf_{3,c}(x)=3x+c\\;(c\\ge 0),\\qquad  \nf_{4,c}(x)=4x+c\\;(c\\ge 0).\n\\]",
      "solution": "Fix a strictly increasing continuous function  \n\\[\nf:(0,\\infty)\\longrightarrow(0,\\infty)\n\\]  \nfulfilling $(\\ast)$.  \nFor $x>0$ put  \n\\[\na_{0}(x):=x,\\qquad a_{n+1}(x):=f(a_{n}(x))\\quad(n\\ge 0).\n\\]\nSince $f$ is increasing, every orbit  \n\\[\n\\mathcal O(x):=\\bigl(a_{n}(x)\\bigr)_{n\\ge 0}\n\\]\nis strictly increasing in $x$ at each coordinate.\n\n------------------------------------------------------------------\n1.  A linear recurrence for every orbit\n------------------------------------------------------------------\nSubstituting $x\\mapsto a_{n}(x)$ in $(\\ast)$ yields  \n\\[\na_{n+4}-10a_{n+3}+35a_{n+2}-50a_{n+1}+24a_{n}=0\\qquad(n\\ge 0).\\tag{1}\n\\]\n\nLet  \n\\[\nr_{1}=1<r_{2}=2<r_{3}=3<r_{4}=4\n\\]\nbe the roots of $P$.  \nBecause $(r_{j}^{\\,k})_{0\\le k,j\\le 3}$ is a nonsingular Vandermonde\nmatrix, for each fixed $x$ the solution of (1) has the classical form  \n\\[\na_{n}(x)=\\sum_{j=1}^{4}c_{j}(x)\\,r_{j}^{\\,n}\\qquad(n\\ge 0),\\tag{2}\n\\]\nwith uniquely determined real coefficients $c_{j}(x)$ depending\n\\emph{continuously} on $x$.\n\n------------------------------------------------------------------\n2.  A unique dominant root and its positivity\n------------------------------------------------------------------\nFor $x>0$ define  \n\\[\n\\kappa(x):=\\max\\{\\,r_{j}\\mid c_{j}(x)\\neq 0\\},\\qquad\nC(x):=c_{\\kappa(x)}(x).\n\\]\n\nBecause $r_{1},\\dots ,r_{4}$ are positive, \\eqref{2} together with $a_{n}(x)>0$\nimplies $C(x)\\neq 0$.  Moreover  \n\\[\n\\lim_{n\\to\\infty}\\frac{a_{n+1}(x)}{a_{n}(x)}=\\kappa(x),\\tag{3}\n\\]\nso $\\kappa(x)$ is the growth rate of the orbit.\n\nPositivity of every $a_{n}(x)$ enforces $C(x)>0$:\nif $C(x)<0$ the right–hand side of \\eqref{2} becomes negative for all\nlarge $n$ because $r_{j}^{\\,n}/\\kappa(x)^{\\,n}\\to0$ when $r_{j}<\\kappa(x)$.\n\nLemma 2.1 (local constancy of $\\kappa$).  \nThe map $\\kappa:(0,\\infty)\\to\\{1,2,3,4\\}$ is locally constant, hence\nconstant.\n\nProof.  \nFix $x_{0}>0$ and let $\\kappa_{0}:=\\kappa(x_{0})$.  \nBecause $C(x_{0})>0$ and $C$ is continuous, there exists $\\delta>0$\nsuch that $C(x)\\ge C(x_{0})/2>0$ whenever $\\lvert x-x_{0}\\rvert<\\delta$.\nFor such $x$ write \\eqref{2} as  \n\\[\na_{n}(x)=C(x)\\,\\kappa_{0}^{\\,n}+R_{n}(x),\\qquad\nR_{n}(x):=\\sum_{r_{j}<\\kappa_{0}}c_{j}(x)\\,r_{j}^{\\,n}.\n\\]\nSince $r_{j}<\\kappa_{0}$, the ratio $R_{n}(x)/\\kappa_{0}^{\\,n}$ tends to\n$0$ uniformly in $x$ from the above $\\delta$–neighbourhood.  Hence for\nlarge $n$ the term $C(x)\\kappa_{0}^{\\,n}$ dominates and forces\n$\\kappa(x)=\\kappa_{0}$.  Thus $\\kappa$ is constant on\n$(x_{0}-\\delta,x_{0}+\\delta)$, proving the lemma. ∎\n\nConsequently there is a single  \n\\[\n\\kappa\\in\\{1,2,3,4\\}\\quad\\text{such that}\\quad\\kappa(x)\\equiv\\kappa.\\tag{4}\n\\]\n\n------------------------------------------------------------------\n3.  Normalising the subordinate coefficients\n------------------------------------------------------------------\nFor $r\\in\\{2,3\\}$ with $r<\\kappa$ define  \n\\[\n\\gamma_{r}(x):=\n\\begin{cases}\n\\dfrac{c_{r}(x)}{C(x)},&\\text{if }r<\\kappa,\\\\[6pt]\n0,&\\text{if }r\\ge\\kappa.\n\\end{cases}\n\\]\nBy the same domination argument as in Lemma 2.1 one obtains that every\n$\\gamma_{r}(x)$ is locally constant, hence constant on $(0,\\infty)$.  \nWe drop the argument of $\\gamma_{r}$ and write simply $\\gamma_{r}$.  \nSet  \n\\[\n\\lambda:=1+\\gamma_{2}+\\gamma_{3}\\quad(\\lambda>0).\\tag{5}\n\\]\n\n------------------------------------------------------------------\n4.  How the coefficients propagate under $f$\n------------------------------------------------------------------\nSince $a_{n+1}(x)=a_{n}\\bigl(f(x)\\bigr)$, relation \\eqref{2} with\n$x\\mapsto f(x)$ gives  \n\\[\n\\sum_{j=1}^{4}c_{j}\\bigl(f(x)\\bigr)\\,r_{j}^{\\,n}\n      =\\sum_{j=1}^{4}c_{j}(x)\\,r_{j}^{\\,n+1}\\qquad(\\forall n\\ge 0).\n\\]\nLinear independence of the sequences $r_{j}^{\\,n}$ yields  \n\\[\nc_{j}\\bigl(f(x)\\bigr)=r_{j}\\,c_{j}(x)\\qquad(j=1,2,3,4).\\tag{6}\n\\]\nIn particular  \n\\[\nC\\bigl(f(x)\\bigr)=\\kappa\\,C(x),\\qquad\nc_{1}\\bigl(f(x)\\bigr)=c_{1}(x).\\tag{7}\n\\]\n\n------------------------------------------------------------------\n5.  Constancy of $c_{1}$ and an affine formula for $C$\n------------------------------------------------------------------\n\\emph{Case $\\kappa>1$.}  \nThen by \\eqref{3} all orbits are unbounded, hence\n$\\mathcal O(x)$ is unbounded for each $x$.  Equation \\eqref{7} shows that\n$c_{1}$ is constant on $\\mathcal O(x)$, and by continuity it is constant\non $\\overline{\\mathcal O(x)}=[x,\\infty)$.  Varying $x$ therefore forces  \n\\[\nc_{1}(x)\\equiv A\\quad(A\\in\\mathbb R).\\tag{8}\n\\]\n\nWith $n=0$ in \\eqref{2} we obtain  \n\\[\nx=A+\\lambda\\,C(x)\\quad\\Longrightarrow\\quad\nC(x)=\\frac{x-A}{\\lambda}\\qquad(x>0).\\tag{9}\n\\]\nAs $C$ is positive and increasing, $\\lambda>0$ and $A\\le 0$.  Writing  \n\\[\nB:=-\\frac{A}{\\lambda}\\ge 0\n\\]\nwe may rewrite \\eqref{9} as  \n\\[\nC(x)=\\frac{x}{\\lambda}+B\\quad(x>0).\\tag{10}\n\\]\n\n\\emph{Case $\\kappa=1$.}  \nHere \\eqref{2} becomes $a_{n}(x)=c_{1}(x)$ for all $n$.  \nSince $a_{0}(x)=x$, we get $c_{1}(x)=x$.  Then  \n\\[\na_{1}(x)=f(x)=c_{1}(x)=x,\n\\]\nso \\emph{necessarily} $f(x)=x$ and no other coefficients occur.\nConsequently $c_{1}$ is automatically constant along every orbit.  \nThis treats simultaneously the constancy issue for $c_{1}$ and settles\nthe whole case $\\kappa=1$.\n\nFrom now on we assume $\\kappa>1$.\n\n------------------------------------------------------------------\n6.  Vanishing of the subordinate coefficients\n------------------------------------------------------------------\nFor $r\\in\\{2,3\\}$ with $r<\\kappa$ we have, by \\eqref{6} and \\eqref{10},  \n\\[\n\\gamma_{r}\\,C\\bigl(f(x)\\bigr)\n   =c_{r}\\bigl(f(x)\\bigr)\n   =r\\,c_{r}(x)\n   =r\\,\\gamma_{r}\\,C(x)\n   =r\\,\\gamma_{r}\\Bigl(\\tfrac{x}{\\lambda}+B\\Bigr).\n\\]\nOn the other hand, since $C\\bigl(f(x)\\bigr)=\\kappa\\,C(x)$,\n\\[\n\\gamma_{r}\\,\\kappa\\Bigl(\\tfrac{x}{\\lambda}+B\\Bigr)\n   =r\\,\\gamma_{r}\\Bigl(\\tfrac{x}{\\lambda}+B\\Bigr)\\qquad(x>0).\n\\]\nBecause the bracket is strictly positive, the factor\n$(\\kappa-r)$ forces $\\gamma_{r}=0$.  Consequently  \n\\[\n\\gamma_{2}=\\gamma_{3}=0,\\qquad\\lambda=1\\quad\\text{by}\\;(\\ref{5}).\\tag{11}\n\\]\nNow \\eqref{10} simplifies to  \n\\[\nC(x)=x+B\\qquad(x>0).\\tag{12}\n\\]\n\n------------------------------------------------------------------\n7.  Determining $f$\n------------------------------------------------------------------\nInsert $n=1$ in \\eqref{2}, use \\eqref{8} and \\eqref{12}:\n\\[\nf(x)=\\sum_{j=1}^{4}r_{j}\\,c_{j}(x)\n     =x+(\\kappa-1)\\,C(x)\n     =x+(\\kappa-1)(x+B)\n     =\\kappa x+(\\kappa-1)B.\\tag{13}\n\\]\nSet  \n\\[\nc:=(\\kappa-1)B\\ge 0.\n\\]\nEquation \\eqref{13} gives $f(x)=\\kappa x+c$ when $\\kappa>1$, while the\ncase $\\kappa=1$ supplied $f(x)=x$.  We have thus derived the complete\nlist announced in the problem statement.\n\n------------------------------------------------------------------\n8.  Verification and completeness\n------------------------------------------------------------------\nFor an affine map $f(x)=\\kappa x+c$ one has  \n\\[\nf^{\\circ n}(x)=\\kappa^{n}x+c\\frac{\\kappa^{n}-1}{\\kappa-1}\\qquad(n\\ge 1).\n\\]\nSubstituting in $(\\ast)$ and using $P(\\kappa)=0$ gives identically $0$,\nand continuity/monotonicity are obvious for $c\\ge 0$\n(with $c=0$ when $\\kappa=1$).  \nConversely, the necessity has been established above, so the solution set\nis exactly the four families  \n\\[\nf_{1,0}(x)=x,\\qquad f_{2,c}(x)=2x+c,\\qquad f_{3,c}(x)=3x+c,\\qquad\nf_{4,c}(x)=4x+c\\quad(c\\ge 0).\\qquad\\Box\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.549817",
        "was_fixed": false,
        "difficulty_analysis": "• Higher-order composition: The relation involves the 4-fold iterate of f, turning the\n  two‐term second-order recursion of the original into a five-term fourth-order one.\n• Spectral theory:  Solving the problem now needs linear-algebraic machinery (companion\n  matrices, eigenvalues, Perron–Frobenius theorem) instead of an elementary quadratic\n  characteristic polynomial.\n• Multiple positive eigenvalues:  The roots 1,2,3,4 are all positive, so the\n  sign-alternation trick that dispatches the −3 root in the original problem no longer\n  works; one must analyse long-term growth and use limit arguments.\n• Dynamical-systems viewpoint:  Step 5 converts the functional equation into a statement\n  about the ω-limit set of a discrete dynamical system and shows that all orbits share\n  the same Lyapunov quotient.\n• Uniqueness via closure of quotient set:  Establishing that g(x)=f(x)/x is constant\n  requires a delicate combination of monotonicity, orbit analysis and Perron–Frobenius\n  asymptotics.\n\nThese layers of linear-algebraic and dynamical techniques render the enhanced\nvariant markedly more intricate than both the original and the former kernel\nversion, which were solvable with a single elementary recursion plus a sign\nargument."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}