{
  "index": "1983-A-3",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem A-3\n\nLet \\( p \\) be in the set \\( \\{3,5,7,11, \\ldots\\} \\) of odd primes and let\n\\[\nF(n)=1+2 n+3 n^{2}+\\cdots+(p-1) n^{p-2} .\n\\]\n\nProve that if \\( a \\) and \\( b \\) are distinct integers in \\( \\{0,1,2, \\ldots, p-1\\} \\) then \\( F(a) \\) and \\( F(b) \\) are not congruent modulo \\( p \\), that is, \\( F(a)-F(b) \\) is not exactly divisible by \\( p \\).",
  "solution": "A-3.\n\\[\n\\begin{aligned}\nF(n) & =1+2 n+3 n^{2}+\\cdots+(p-1) n^{p-2}, \\\\\nn F(n) & =n+2 n^{2}+\\cdots+(p-2) n^{p-2}+(p-1) n^{p-1} .\n\\end{aligned}\n\\]\n\nHence \\( (1-n) F(n)=\\left(1+n+n^{2}+\\cdots+n^{p-2}\\right)-(p-1) n^{p-1} \\) and similarly\n\\[\n(1-n)^{2} F(n)=1-n^{p-1}-(1-n)(p-1) n^{p-1}=1-p \\cdot n^{p-1}+(p-1) n^{p} .\n\\]\n\nModulo \\( p, n^{p} \\equiv n \\) by the Little Fermat Theorem and so \\( (1-n)^{2} F(n) \\equiv 1-n \\). If neither \\( a \\) nor \\( b \\) is congruent to \\( 1(\\bmod p), 1-a \\neq 1-b \\) and there are distinct reciprocals \\( (1-a)^{-1} \\) and \\( (1-b)^{-1}(\\bmod p) \\); then\n\\[\nf(a) \\equiv(1-a)^{-1}, f(b) \\equiv(1-b)^{-1}, f(a) \\equiv f(b)(\\bmod p) .\n\\]\n\nIf one of \\( a \\) and \\( b \\), say \\( a \\), is congruent to 1 , then \\( b \\neq 0(\\bmod p) \\) and so \\( f(b) \\equiv(1-b)^{-1} \\neq 0 \\) \\( (\\bmod p) \\) while\n\\[\nf(a)=1+2+\\cdots+(p-1)=p(p-1) / 2 \\equiv 0(\\bmod p) .\n\\]",
  "vars": [
    "n",
    "a",
    "b"
  ],
  "params": [
    "p",
    "F",
    "f"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "integern",
        "a": "varalpha",
        "b": "varbravo",
        "p": "primeodd",
        "F": "funcsum",
        "f": "funclow"
      },
      "question": "Problem A-3\n\nLet \\( primeodd \\) be in the set \\( \\{3,5,7,11, \\ldots\\} \\) of odd primes and let\n\\[\nfuncsum(integern)=1+2 integern+3 integern^{2}+\\cdots+(primeodd-1) integern^{primeodd-2} .\n\\]\n\nProve that if \\( varalpha \\) and \\( varbravo \\) are distinct integers in \\{0,1,2, \\ldots, primeodd-1\\} then \\( funcsum(varalpha) \\) and \\( funcsum(varbravo) \\) are not congruent modulo \\( primeodd \\), that is, \\( funcsum(varalpha)-funcsum(varbravo) \\) is not exactly divisible by \\( primeodd \\).",
      "solution": "A-3.\n\\[\n\\begin{aligned}\nfuncsum(integern) & =1+2 integern+3 integern^{2}+\\cdots+(primeodd-1) integern^{primeodd-2}, \\\\\nintegern funcsum(integern) & =integern+2 integern^{2}+\\cdots+(primeodd-2) integern^{primeodd-2}+(primeodd-1) integern^{primeodd-1} .\n\\end{aligned}\n\\]\n\nHence \\( (1-integern) funcsum(integern)=\\left(1+integern+integern^{2}+\\cdots+integern^{primeodd-2}\\right)-(primeodd-1) integern^{primeodd-1} \\) and similarly\n\\[\n(1-integern)^{2} funcsum(integern)=1-integern^{primeodd-1}-(1-integern)(primeodd-1) integern^{primeodd-1}=1-primeodd \\cdot integern^{primeodd-1}+(primeodd-1) integern^{primeodd} .\n\\]\n\nModulo \\( primeodd, integern^{primeodd} \\equiv integern \\) by the Little Fermat Theorem and so \\( (1-integern)^{2} funcsum(integern) \\equiv 1-integern \\). If neither \\( varalpha \\) nor \\( varbravo \\) is congruent to \\( 1(\\bmod primeodd), 1-varalpha \\neq 1-varbravo \\) and there are distinct reciprocals \\( (1-varalpha)^{-1} \\) and \\( (1-varbravo)^{-1}(\\bmod primeodd) \\); then\n\\[\nfunclow(varalpha) \\equiv(1-varalpha)^{-1}, funclow(varbravo) \\equiv(1-varbravo)^{-1}, funclow(varalpha) \\equiv funclow(varbravo)(\\bmod primeodd) .\n\\]\n\nIf one of \\( varalpha \\) and \\( varbravo \\), say \\( varalpha \\), is congruent to 1 , then \\( varbravo \\neq 0(\\bmod primeodd) \\) and so \\( funclow(varbravo) \\equiv(1-varbravo)^{-1} \\neq 0 \\) \\( (\\bmod primeodd) \\) while\n\\[\nfunclow(varalpha)=1+2+\\cdots+(primeodd-1)=primeodd(primeodd-1) / 2 \\equiv 0(\\bmod primeodd) .\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "grapefruit",
        "a": "shoelaces",
        "b": "firecracker",
        "p": "headlight",
        "F": "sandcastle",
        "f": "tortoise"
      },
      "question": "Problem A-3\n\nLet \\( headlight \\) be in the set \\( \\{3,5,7,11, \\ldots\\} \\) of odd primes and let\n\\[\nsandcastle(grapefruit)=1+2 grapefruit+3 grapefruit^{2}+\\cdots+(headlight-1) grapefruit^{headlight-2} .\n\\]\n\nProve that if \\( shoelaces \\) and \\( firecracker \\) are distinct integers in \\( \\{0,1,2, \\ldots, headlight-1\\} \\) then \\( sandcastle(shoelaces) \\) and \\( sandcastle(firecracker) \\) are not congruent modulo \\( headlight \\), that is, \\( sandcastle(shoelaces)-sandcastle(firecracker) \\) is not exactly divisible by \\( headlight \\).",
      "solution": "A-3.\n\\[\n\\begin{aligned}\nsandcastle(grapefruit) & =1+2 grapefruit+3 grapefruit^{2}+\\cdots+(headlight-1) grapefruit^{headlight-2}, \\\\\ngrapefruit\\, sandcastle(grapefruit) & =grapefruit+2 grapefruit^{2}+\\cdots+(headlight-2) grapefruit^{headlight-2}+(headlight-1) grapefruit^{headlight-1} .\n\\end{aligned}\n\\]\n\nHence \\( (1-grapefruit) sandcastle(grapefruit)=\\left(1+grapefruit+grapefruit^{2}+\\cdots+grapefruit^{headlight-2}\\right)-(headlight-1) grapefruit^{headlight-1} \\) and similarly\n\\[\n(1-grapefruit)^{2} sandcastle(grapefruit)=1-grapefruit^{headlight-1}-(1-grapefruit)(headlight-1) grapefruit^{headlight-1}=1-headlight \\cdot grapefruit^{headlight-1}+(headlight-1) grapefruit^{headlight} .\n\\]\n\nModulo \\( headlight, grapefruit^{headlight} \\equiv grapefruit \\) by the Little Fermat Theorem and so \\( (1-grapefruit)^{2} sandcastle(grapefruit) \\equiv 1-grapefruit \\). If neither \\( shoelaces \\) nor \\( firecracker \\) is congruent to \\( 1(\\bmod headlight)\\), \\( 1-shoelaces \\neq 1-firecracker \\) and there are distinct reciprocals \\( (1-shoelaces)^{-1} \\) and \\( (1-firecracker)^{-1}(\\bmod headlight) \\); then\n\\[\ntortoise(shoelaces) \\equiv(1-shoelaces)^{-1}, \\quad tortoise(firecracker) \\equiv(1-firecracker)^{-1}, \\quad tortoise(shoelaces) \\equiv tortoise(firecracker)(\\bmod headlight) .\n\\]\n\nIf one of \\( shoelaces \\) and \\( firecracker \\), say \\( shoelaces \\), is congruent to 1, then \\( firecracker \\neq 0(\\bmod headlight) \\) and so \\( tortoise(firecracker) \\equiv(1-firecracker)^{-1} \\neq 0 \\) \\( (\\bmod headlight) \\) while\n\\[\ntortoise(shoelaces)=1+2+\\cdots+(headlight-1)=headlight(headlight-1) / 2 \\equiv 0(\\bmod headlight) .\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "constantval",
        "a": "finalvalue",
        "b": "knownvalue",
        "p": "compositenum",
        "F": "staticfunc",
        "f": "dynamicconst"
      },
      "question": "Problem A-3\n\nLet \\( compositenum \\) be in the set \\( \\{3,5,7,11, \\ldots\\} \\) of odd primes and let\n\\[\nstaticfunc(constantval)=1+2 constantval+3 constantval^{2}+\\cdots+(compositenum-1) constantval^{compositenum-2} .\n\\]\n\nProve that if \\( finalvalue \\) and \\( knownvalue \\) are distinct integers in \\( \\{0,1,2, \\ldots, compositenum-1\\} \\) then \\( staticfunc(finalvalue) \\) and \\( staticfunc(knownvalue) \\) are not congruent modulo \\( compositenum \\), that is, \\( staticfunc(finalvalue)-staticfunc(knownvalue) \\) is not exactly divisible by \\( compositenum \\).",
      "solution": "A-3.\n\\[\n\\begin{aligned}\nstaticfunc(constantval) & =1+2 constantval+3 constantval^{2}+\\cdots+(compositenum-1) constantval^{compositenum-2}, \\\\\nconstantval staticfunc(constantval) & =constantval+2 constantval^{2}+\\cdots+(compositenum-2) constantval^{compositenum-2}+(compositenum-1) constantval^{compositenum-1} .\n\\end{aligned}\n\\]\n\nHence \\( (1-constantval) staticfunc(constantval)=\\left(1+constantval+constantval^{2}+\\cdots+constantval^{compositenum-2}\\right)-(compositenum-1) constantval^{compositenum-1} \\) and similarly\n\\[\n(1-constantval)^{2} staticfunc(constantval)=1-constantval^{compositenum-1}-(1-constantval)(compositenum-1) constantval^{compositenum-1}=1-compositenum \\cdot constantval^{compositenum-1}+(compositenum-1) constantval^{compositenum} .\n\\]\n\nModulo \\( compositenum, constantval^{compositenum} \\equiv constantval \\) by the Little Fermat Theorem and so \\( (1-constantval)^{2} staticfunc(constantval) \\equiv 1-constantval \\). If neither \\( finalvalue \\) nor \\( knownvalue \\) is congruent to \\( 1(\\bmod compositenum), 1-finalvalue \\neq 1-knownvalue \\) and there are distinct reciprocals \\( (1-finalvalue)^{-1} \\) and \\( (1-knownvalue)^{-1}(\\bmod compositenum) \\); then\n\\[\ndynamicconst(finalvalue) \\equiv(1-finalvalue)^{-1}, dynamicconst(knownvalue) \\equiv(1-knownvalue)^{-1}, dynamicconst(finalvalue) \\equiv dynamicconst(knownvalue)(\\bmod compositenum) .\n\\]\n\nIf one of \\( finalvalue \\) and \\( knownvalue \\), say \\( finalvalue \\), is congruent to 1 , then \\( knownvalue \\neq 0(\\bmod compositenum) \\) and so \\( dynamicconst(knownvalue) \\equiv(1-knownvalue)^{-1} \\neq 0 \\) \\( (\\bmod compositenum) \\) while\n\\[\ndynamicconst(finalvalue)=1+2+\\cdots+(compositenum-1)=compositenum(compositenum-1) / 2 \\equiv 0(\\bmod compositenum) .\n\\]"
    },
    "garbled_string": {
      "map": {
        "n": "zqkmrptu",
        "a": "hvnqslta",
        "b": "cxdmqfye",
        "p": "gfzlonpai",
        "F": "rqtsvelon",
        "f": "mpxravilo"
      },
      "question": "Problem A-3\n\nLet \\( gfzlonpai \\) be in the set \\( \\{3,5,7,11, \\ldots\\} \\) of odd primes and let\n\\[\nrqtsvelon(zqkmrptu)=1+2 zqkmrptu+3 zqkmrptu^{2}+\\cdots+(gfzlonpai-1) zqkmrptu^{gfzlonpai-2} .\n\\]\n\nProve that if \\( hvnqslta \\) and \\( cxdmqfye \\) are distinct integers in \\( \\{0,1,2, \\ldots, gfzlonpai-1\\} \\) then \\( rqtsvelon(hvnqslta) \\) and \\( rqtsvelon(cxdmqfye) \\) are not congruent modulo \\( gfzlonpai \\), that is, \\( rqtsvelon(hvnqslta)-rqtsvelon(cxdmqfye) \\) is not exactly divisible by \\( gfzlonpai \\).",
      "solution": "A-3.\n\\[\n\\begin{aligned}\nrqtsvelon(zqkmrptu) & =1+2 zqkmrptu+3 zqkmrptu^{2}+\\cdots+(gfzlonpai-1) zqkmrptu^{gfzlonpai-2}, \\\\\nzqkmrptu \\, rqtsvelon(zqkmrptu) & =zqkmrptu+2 zqkmrptu^{2}+\\cdots+(gfzlonpai-2) zqkmrptu^{gfzlonpai-2}+(gfzlonpai-1) zqkmrptu^{gfzlonpai-1} .\n\\end{aligned}\n\\]\n\nHence \\( (1-zqkmrptu) rqtsvelon(zqkmrptu)=\\left(1+zqkmrptu+zqkmrptu^{2}+\\cdots+zqkmrptu^{gfzlonpai-2}\\right)-(gfzlonpai-1) zqkmrptu^{gfzlonpai-1} \\) and similarly\n\\[\n(1-zqkmrptu)^{2} rqtsvelon(zqkmrptu)=1-zqkmrptu^{gfzlonpai-1}-(1-zqkmrptu)(gfzlonpai-1) zqkmrptu^{gfzlonpai-1}=1-gfzlonpai \\cdot zqkmrptu^{gfzlonpai-1}+(gfzlonpai-1) zqkmrptu^{gfzlonpai} .\n\\]\n\nModulo \\( gfzlonpai, zqkmrptu^{gfzlonpai} \\equiv zqkmrptu \\) by the Little Fermat Theorem and so \\( (1-zqkmrptu)^{2} rqtsvelon(zqkmrptu) \\equiv 1-zqkmrptu \\). If neither \\( hvnqslta \\) nor \\( cxdmqfye \\) is congruent to \\( 1(\\bmod gfzlonpai), 1-hvnqslta \\neq 1-cxdmqfye \\) and there are distinct reciprocals \\( (1-hvnqslta)^{-1} \\) and \\( (1-cxdmqfye)^{-1}(\\bmod gfzlonpai) \\); then\n\\[\nmpxravilo(hvnqslta) \\equiv(1-hvnqslta)^{-1}, \\quad mpxravilo(cxdmqfye) \\equiv(1-cxdmqfye)^{-1}, \\quad mpxravilo(hvnqslta) \\equiv mpxravilo(cxdmqfye)(\\bmod gfzlonpai) .\n\\]\n\nIf one of \\( hvnqslta \\) and \\( cxdmqfye \\), say \\( hvnqslta \\), is congruent to 1 , then \\( cxdmqfye \\neq 0(\\bmod gfzlonpai) \\) and so \\( mpxravilo(cxdmqfye) \\equiv(1-cxdmqfye)^{-1} \\neq 0 \\, (\\bmod gfzlonpai) \\) while\n\\[\nmpxravilo(hvnqslta)=1+2+\\cdots+(gfzlonpai-1)=gfzlonpai(gfzlonpai-1) / 2 \\equiv 0(\\bmod gfzlonpai) .\n\\]"
    },
    "kernel_variant": {
      "question": "Let $p$ be an odd prime and denote  \n\\[\nv_{p}(\\,\\cdot\\,)\\quad (v_{p}(p)=1),\\qquad \n|\\cdot|_{p}=p^{-v_{p}(\\,\\cdot\\,)}\n\\]\nfor the $p$–adic valuation on $\\mathbf Z_{p}$ and the associated absolute value.  \nPut  \n\\[\nF(X)=1+2X+3X^{2}+\\dots +(p-1)X^{p-2}\\in\\mathbf Z[X].\\tag{1}\n\\]\n\nFor every integer $k\\ge 1$ consider the reduction  \n\\[\n\\varphi_{k}\\;:\\;\\mathbf Z/p^{k}\\mathbf Z\\longrightarrow \\mathbf Z/p^{k}\\mathbf Z,\\qquad \nx\\longmapsto F(x)\\pmod{p^{k}}.\\tag{2}\n\\]\n\nFor $k\\ge 2$ define the {\\em highly ramified} set  \n\\[\nA_{k}:=\\bigl\\{\\,1+p^{\\,k-1}u : u\\in\\mathbf Z/p\\mathbf Z\\bigr\\}\\subset\\mathbf Z/p^{k}\\mathbf Z.\\tag{3}\n\\]\n\nInside the $p$–adic unit disc  \n\\[\nD=\\bigl\\{x\\in\\mathbf Z_{p}:|x|_{p}\\le 1\\bigr\\}\n\\]\nput  \n\\[\n\\Omega=\\bigl\\{x\\in\\mathbf Z_{p}:|x|_{p}=1\\;\\text{ and }\\;x\\not\\equiv 1\\pmod p\\bigr\\}.\\tag{4}\n\\]\n\nAnswer the following questions.\n\n(a) Behaviour modulo high powers of $p$.\n\n(i) Show that $\\varphi_{1}$ is a permutation of the finite field $\\mathbf F_{p}$.\n\n(ii) For $k\\ge 2$ prove  \n\n\\[\n\\boxed{p\\ge 5}\\;:\\;\n\\varphi_{k}\\text{ is {\\em constant} on }A_{k}\\text{ and the common value equals }F(1)\\pmod{p^{k}}.\n\\]\n\n\\[\n\\boxed{p=3}\\;:\\;\n\\varphi_{k}\\bigl(A_{k}\\bigr)=\n\\bigl\\{\\,F(1)+ 2\\cdot 3^{\\,k-1}u : u\\in\\mathbf Z/3\\mathbf Z\\bigr\\}\\subset 3\\mathbf Z/3^{k}\\mathbf Z,\n\\]\nand $\\varphi_{k}|_{A_{k}}$ is injective.  \nIn particular $\\varphi_{k}(A_{k})\\cap A_{k}=\\varnothing$ for every $k\\ge 2$.\n\n(iii) Describe the fibres of $\\varphi_{k}$.\n\n • Case $p\\ge 5$.\n\n  Let $k\\ge 2$ and write $t:=v_{p}(x-1)$ for $x\\in\\mathbf Z/p^{k}\\mathbf Z$ ($0\\le t\\le k$).\n\n  (α)  If $t=0$ (that is, $x\\not\\equiv 1\\pmod p$) show that  \n\\[\n\\bigl|\\varphi_{k}^{-1}\\!\\bigl(\\varphi_{k}(x)\\bigr)\\bigr|=1 .\n\\]\n\n  (β)  If $1\\le t\\le k-1$ write $x=1+p^{t}u$ with $u\\in\\mathbf Z/p^{\\,k-t}\\mathbf Z$ and $p\\nmid u$.  \nProve that  \n\n\\[\n\\varphi_{k}\\!\\bigl(1+p^{t}u_{1}\\bigr)\\equiv\n\\varphi_{k}\\!\\bigl(1+p^{t}u_{2}\\bigr)\\pmod{p^{k}}\n\\quad\\Longleftrightarrow\\quad \nu_{1}\\equiv u_{2}\\pmod{p^{\\,k-t-1}}.\n\\]\n\n  Deduce that {\\em every} non-trivial fibre of $\\varphi_{k}$ has cardinality $p$, that it is contained in the set  \n\\[\nC_{k,t}:=\\bigl\\{\\,1+p^{t}\\bigl(u_{0}+p^{\\,k-t-1}\\ell\\bigr):\\ell\\in\\mathbf Z/p\\mathbf Z\\bigr\\},\n\\]\nand that, for each fixed $t$, the images of the different fibres are pairwise distinct.  In particular  \n\n\\[\n\\#\\,\\operatorname{im}(\\varphi_{k})\\;=\\;(p-1)p^{\\,k-1}+\\frac{p^{\\,k-1}-1}{p-1},\n\\]\nso exactly \n\\[\n\\boxed{\\displaystyle\\frac{(p-2)p^{\\,k-1}+1}{p-1}}\n\\]\nresidue classes are omitted.\n\n • Case $p=3$.\n\n  Prove that $\\varphi_{k}$ is bijective for all $k\\ge 1$ and that every fibre has cardinality $1$.\n\n(b) A universal $p$–adic inverse away from the ramified fibre.\n\nConstruct a map  \n\\[\nP\\;:\\;\\Omega\\longrightarrow\\Omega\\tag{5}\n\\]\nsuch that $F\\circ P=P\\circ F=\\mathrm{id}$ on $\\Omega$.  \nProve that $P$ is $1$-Lipschitz (hence uniformly continuous) and locally analytic on $\\Omega$.\n\n(c) Global analytic structure.\n\nShow that $P$ is the restriction to $\\Omega$ of a power series  \n\\[\nP(X)=\\sum_{j\\ge 0}c_{j}X^{\\,j},\\qquad c_{j}\\in\\mathbf Z_{p},\\qquad |c_{j}|_{p}\\longrightarrow 0,\\tag{6}\n\\]\nwhose radius of convergence is $\\ge 1$.  \nDeduce that $F$ restricts to a $p$–adic analytic diffeomorphism of $\\Omega$ with inverse $P$.\n\nAll arguments have to be carried out purely in the $p$–adic setting; no appeal to complex analysis or formal-group theory is allowed.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we keep the notation of the problem; all unnamed congruences are taken in $\\mathbf Z$.\n\n0.  An algebraic identity.  \nPut  \n\\[\nS(X)=1+X+\\dots+X^{p-1}=\\frac{1-X^{p}}{1-X},\\qquad  F(X)=S'(X).\n\\]\nDifferentiating $(1-X)S(X)=1-X^{p}$ gives  \n\\[\n(1-X)^{2}F(X)=1-pX^{p-1}+(p-1)X^{p}.\\tag{$\\ast$}\n\\]\n\n1.  Proof of (a)(i).  \nFix $x\\in\\mathbf F_{p}$.  Reducing $(\\ast)$ modulo $p$ and evaluating at $x$ yields  \n\\[\n(1-x)^{2}F(x)\\equiv1-x^{p}\\equiv1-x\\pmod p ,\n\\]\nwhere the second equivalence uses Fermat’s little theorem.  \nIf $x\\not\\equiv1\\pmod p$ we may cancel $(1-x)$ and obtain  \n\\[\nF(x)\\equiv(1-x)^{-1}\\pmod p.\\tag{1.1}\n\\]\nThus $F$ acts on $\\mathbf F_{p}\\setminus\\{1\\}$ by $x\\mapsto(1-x)^{-1}$, hence bijectively.  \nBecause  \n\\[\nF(1)=\\sum_{j=1}^{p-1}j=\\frac{p(p-1)}{2}\\equiv0\\pmod p ,\n\\]\nthe value $0$ is taken exactly at $x\\equiv1\\pmod p$.  \nConsequently $\\varphi_{1}$ is a permutation of $\\mathbf F_{p}$.  \\hfill$\\square$\n\n2.  Local expansion around $X=1$.  \nWrite $X=1+Z$ and put $G(X)=(1-X)^{2}F(X)$.  By $(\\ast)$,\n\\[\nG(1+Z)=1-p(1+Z)^{p-1}+(p-1)(1+Z)^{p}.\n\\]\nA Taylor expansion at $Z=0$ gives  \n\\[\nG(1+Z)=\\frac{p(p-1)}{2}Z^{2}+\\frac{p(p-1)(p-2)}{3}Z^{3}+Z^{4}H(Z),\\tag{2.1}\n\\]\nwith $H(Z)\\in\\mathbf Z_{p}[[Z]]$.  Because $F(1+Z)=G(1+Z)/Z^{2}$ we have  \n\\[\nF(1+Z)=F(1)+Z\\,R(Z),\\qquad  \nR(Z)=\\frac{p(p-1)(p-2)}{3}+Z\\,H(Z)\\in\\mathbf Z_{p}[[Z]].\\tag{2.2}\n\\]\nObserve  \n\\[\nv_{p}\\bigl(R(0)\\bigr)=\n\\begin{cases}\n1,& p\\ge 5,\\\\\n0,& p=3.\n\\end{cases}\\tag{2.3}\n\\]\n\nLemma 2.1. Let $t\\ge 1$ and $u\\in\\mathbf Z$ with $p\\nmid u$, and set $a=1+p^{t}u$. Then  \n\\[\nv_{p}\\bigl(F(a)-F(1)\\bigr)=\n\\begin{cases}\nt+1,& p\\ge 5,\\\\\nt,& p=3.\n\\end{cases}\n\\]\n\nProof. Insert $Z=p^{t}u$ into (2.2) and use (2.3).  \\hfill$\\square$\n\nWe shall also need a valuation estimate for {\\em differences}.\n\nLemma 2.2. Let $p\\ge 5$, $k\\ge 2$, $1\\le t\\le k-1$ and $u_{1},u_{2}\\in\\mathbf Z$ with $p\\nmid u_{1}u_{2}$.  \nPut $x_{i}=1+p^{t}u_{i}$ ($i=1,2$).  Then  \n\\[\nv_{p}\\bigl(F(x_{1})-F(x_{2})\\bigr)=\n\\min\\bigl\\{t+1+v_{p}(u_{1}-u_{2}),\\,2t\\bigr\\}.\\tag{2.4}\n\\]\n\nProof.  \nWrite $Z_{i}=p^{t}u_{i}$ and expand using (2.2):\n\\[\nF(1+Z_{1})-F(1+Z_{2})=(Z_{1}-Z_{2})\\,R(0)+(Z_{1}-Z_{2})\\,Z_{2}H(Z_{2})+Z_{1}(Z_{1}-Z_{2})H(Z_{1}).\n\\]\nThe first term has valuation $t+1+v_{p}(u_{1}-u_{2})$, the two others at least $2t$.  \nTaking the minimum gives (2.4).  \\hfill$\\square$\n\n(The case $k=2$ is included: for $k=2$ we have $t=1$, so $2t=2>k=2$ is false, hence the minimum in (2.4) is the first entry as required for the later application.)\n\n3.  Proof of (a)(ii).\n\n3.1  Case $p\\ge 5$.  \nTake $x\\in A_{k}$, so $x=1+p^{\\,k-1}u$ with $u\\bmod p$.  \nLemma&nbsp;2.1 with $t=k-1$ gives $v_{p}\\bigl(F(x)-F(1)\\bigr)=k$, hence  \n\\[\nF(x)\\equiv F(1)\\pmod{p^{k}}.\n\\]\nThus $\\varphi_{k}$ is constant on $A_{k}$ with the stated value.\n\n3.2  Case $p=3$.  \nLet $x=1+3^{\\,k-1}u\\in A_{k}$.  \nLemma&nbsp;2.1 yields\n\\[\nF(x)-F(1)=3^{\\,k-1}u\\,R\\bigl(3^{\\,k-1}u\\bigr)\\equiv 2\\cdot3^{\\,k-1}u\\pmod{3^{k}},\n\\]\nbecause $R(0)=\\frac{3(3-1)(3-2)}{3}=2$ is a unit.  \nTherefore\n\\[\n\\varphi_{k}(x)\\equiv F(1)+2\\cdot3^{\\,k-1}u\\pmod{3^{k}},\n\\]\nwhich is the claimed description of the image.  \nDifferent $u$’s give different residues modulo $3^{k}$, so $\\varphi_{k}|_{A_{k}}$ is injective.  \nIn particular $\\varphi_{k}(A_{k})\\cap A_{k}=\\varnothing$. \\hfill$\\square$\n\n4.  Fibres for $p\\ge 5$ – proof of (a)(iii).\n\n4.1  Injectivity outside the residue $1\\pmod p$ (statement (α)).  \nLet $x,y\\in\\mathbf Z/p^{k}\\mathbf Z$ with $x\\not\\equiv1\\pmod p\\neq y$ and assume  \n\\[\nF(x)\\equiv F(y)\\pmod{p^{k}}.\\tag{4.1}\n\\]\nBecause $x\\not\\equiv1\\pmod p$, (1.1) gives $F'(x)\\equiv(1-x)^{-2}\\pmod p$, hence $F'(x)$ is a $p$-adic unit.  \nA first-order Taylor expansion with integral remainder shows\n\\[\nv_{p}\\bigl(F(y)-F(x)\\bigr)=v_{p}(y-x).\\tag{4.2}\n\\]\nCondition (4.1) forces $v_{p}(y-x)\\ge k$, hence $x\\equiv y\\pmod{p^{k}}$.  \nThus $\\varphi_{k}$ is injective on $\\{x\\not\\equiv1\\pmod p\\}$ and (α) is proved.\n\n4.2  Elements congruent to $1\\pmod p$ (statement (β)).  \n\nFix $k\\ge 2$ and $1\\le t\\le k-1$.  Write\n\\[\nx_{i}=1+p^{t}u_{i},\\quad u_{i}\\in\\mathbf Z/p^{\\,k-t}\\mathbf Z,\\;p\\nmid u_{i}\\;(i=1,2).\n\\]\nBy Lemma&nbsp;2.2\n\\[\nv_{p}\\bigl(F(x_{1})-F(x_{2})\\bigr)\\ge k\n\\Longleftrightarrow\nt+1+v_{p}(u_{1}-u_{2})\\ge k,\n\\]\nbecause $2t\\le k-1+k-1<k$ if $t\\le k-2$, whereas $2t=k+1>k$ for $t=k-1$.  Hence\n\\[\n\\varphi_{k}(x_{1})\\equiv\\varphi_{k}(x_{2})\\pmod{p^{k}}\n\\Longleftrightarrow\nu_{1}\\equiv u_{2}\\pmod{p^{\\,k-t-1}}.\\tag{4.3}\n\\]\n\n(i)  {\\em Constant fibre for $t=k-1$.}  \nHere $k-t-1=0$, so (4.3) is automatically satisfied; as there are $p$ different $u$’s, $A_{k}=C_{k,k-1}$ is the unique fibre over $F(1)$ and has size $p$.\n\n(ii)  {\\em Uniform fibres of size $p$ for $1\\le t\\le k-2$.}  \nThe set $C_{k,t}$ contains exactly $p$ distinct values of $u$, hence (4.3) shows that $|\\,\\varphi_{k}^{-1}(y)|=p$ for every $y$ in the image of $C_{k,t}$.  \nBecause $u\\mapsto u\\bmod p^{\\,k-t-1}$ has $p^{\\,k-t-1}$ distinct values, the images of $C_{k,t}$ consist of $p^{\\,k-t-1}$ pairwise different classes.\n\n(iii)  {\\em Counting the image.}  \nInjectivity on the complement of $1+p\\mathbf Z/p^{k}\\mathbf Z$ gives $(p-1)p^{\\,k-1}$ image points.  \nAdding the contributions from $t=1,\\dots ,k-1$ we obtain\n\\[\n\\sum_{t=1}^{k-1}p^{\\,k-t-1}=p^{\\,k-2}+p^{\\,k-3}+\\dots+1=\\frac{p^{\\,k-1}-1}{p-1}.\n\\]\nHence\n\\[\n\\#\\,\\operatorname{im}(\\varphi_{k})\n=(p-1)p^{\\,k-1}+\\frac{p^{\\,k-1}-1}{p-1},\n\\]\nso the number of omitted residue classes equals\n\\[\np^{\\,k}-\\#\\,\\operatorname{im}(\\varphi_{k})\n=\\frac{(p-2)p^{\\,k-1}+1}{p-1},\n\\]\nin agreement with the corrected statement.  \\hfill$\\square$\n\n5.  Proof of (a)(iii) for $p=3$.  \nHere $F(X)=1+2X$ and slope $2$ is a unit in every $\\mathbf Z/3^{k}\\mathbf Z$.  \nHence $\\varphi_{k}(x)=1+2x$ is bijective on $\\mathbf Z/3^{k}\\mathbf Z$, so every fibre has size $1$.  \\hfill$\\square$\n\n6.  Preparation for part (b).\n\nLemma 6.1. $\\Omega$ is stable under $F$.\n\nProof. If $x\\in\\Omega$, then $|x|_{p}=1$ and $x\\not\\equiv1\\pmod p$.  \nBy (1.1) $F(x)\\equiv(1-x)^{-1}\\not\\equiv0,1\\pmod p$, hence $|F(x)|_{p}=1$ and $F(x)\\not\\equiv1\\pmod p$.  \\hfill$\\square$\n\n7.  Construction of the inverse – proof of (b).\n\nFor $y\\in\\Omega$ choose the unique residue  \n\\[\nx_{0}\\equiv1-y^{-1}\\pmod p,\\qquad x_{0}\\not\\equiv1\\pmod p.\n\\]\nBecause $F'(x_{0})\\equiv(1-x_{0})^{-2}\\not\\equiv0\\pmod p$, Hensel’s lemma provides a {\\em unique} lift $x\\in\\mathbf Z_{p}$ with $x\\equiv x_{0}\\pmod p$ and $F(x)=y$.  \nBy Lemma&nbsp;6.1, $x\\in\\Omega$ as well.  Define  \n\\[\nP:\\Omega\\longrightarrow\\Omega,\\qquad P(y)=x.\\tag{7.1}\n\\]\nThen $F\\circ P=P\\circ F=\\mathrm{id}_{\\Omega}$ by construction.\n\n7.1  Isometry property.  \nTake $y,y'\\in\\Omega$ and set $x=P(y),\\;x'=P(y')$.  \nUsing a first-order Taylor formula one shows exactly as in 4.1 that  \n\\[\n|y-y'|_{p}=|x'-x|_{p}.\n\\]\nSo $P$ is an isometry, in particular $1$-Lipschitz and uniformly continuous on $\\Omega$.\n\n7.2  Local analyticity.  \nBecause $F'(u)$ is a unit on $\\Omega$, the $p$-adic implicit-function theorem yields analyticity of $P$ in a neighbourhood of every point of $\\Omega$.\n\n8.  Global power-series representation – proof of (c).\n\n8.1  A globally $1$-Lipschitz extension.  \nDefine  \n\\[\n\\widetilde P(x)\\;:=\\;\n\\begin{cases}\nP(x), & x\\in\\Omega,\\\\[4pt]\n1, & x\\notin\\Omega\n\\end{cases}\\qquad\\bigl(x\\in D\\bigr).\\tag{8.1}\n\\]\nTo see that $\\widetilde P$ is $1$-Lipschitz on $D$ there are three cases.\n\n(i) $x,y\\in\\Omega$: already done in 7.1.  \n\n(ii) $x,y\\notin\\Omega$: $\\widetilde P(x)=\\widetilde P(y)=1$, so the inequality is trivial.  \n\n(iii) $x\\in\\Omega$, $y\\notin\\Omega$: then $\\widetilde P(y)=1$ and $|x-1|_{p}=|x|_{p}=1$, whereas $|x-y|_{p}\\le1$ because both $x,y\\in D$.  Hence\n\\[\n|\\widetilde P(x)-\\widetilde P(y)|_{p}=|x-1|_{p}=1\\le |x-y|_{p}.\n\\]\n\n8.2  Mahler expansion.  \nMahler’s theorem now gives a convergent expansion  \n\\[\n\\widetilde P(X)=\\sum_{j\\ge 0}a_{j}\\binom{X}{j},\\qquad \na_{j}\\in\\mathbf Z_{p},\\quad a_{j}\\xrightarrow{j\\to\\infty}0.\\tag{8.2}\n\\]\n\n8.3  Passage to the monomial basis.  \nStandard estimates (Amice–Vélu, Legendre) yield coefficients $c_{m}\\in\\mathbf Z_{p}$ with $|c_{m}|_{p}\\to0$ such that  \n\\[\n\\widetilde P(X)=\\sum_{m\\ge 0}c_{m}X^{\\,m},\\qquad \n\\text{radius of convergence }\\ge 1.\\tag{8.3}\n\\]\nRestricting (8.3) to $\\Omega$ gives the series (6), proving part (c).\n\n8.4 Conclusion.  \nBecause both $F$ and $P$ are analytic on $\\Omega$ and satisfy $F\\circ P=P\\circ F=\\mathrm{id}_{\\Omega}$, $F$ is a $p$-adic analytic diffeomorphism of $\\Omega$ with inverse $P$.  \\hfill$\\square$\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
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      "question": "Let $p$ be an odd prime and denote  \n\\[\nv_{p}(\\,\\cdot\\,)\\;(v_{p}(p)=1),\\qquad \n|\\cdot|_{p}=p^{-v_{p}(\\,\\cdot\\,)}\n\\]\nfor the $p$–adic valuation on $\\mathbf Q_{p}$ and its absolute value.  \nPut  \n\\[\nF(X)=1+2X+3X^{2}+\\dots +(p-1)X^{p-2}\\in\\mathbf Z[X].\\tag{1}\n\\]\n\nFor every integer $k\\ge 1$ consider the reduction  \n\\[\n\\varphi_{k}\\;:\\;\\mathbf Z/p^{k}\\mathbf Z\\longrightarrow \\mathbf Z/p^{k}\\mathbf Z,\\qquad \nx\\longmapsto F(x)\\pmod{p^{k}}.\\tag{2}\n\\]\n\nFor $k\\ge 2$ define the {\\em highly ramified} set  \n\\[\nA_{k}:=\\bigl\\{\\,1+p^{\\,k-1}u : u\\in\\mathbf Z/p\\mathbf Z\\bigr\\}\\subset\\mathbf Z/p^{k}\\mathbf Z.\\tag{3}\n\\]\n\nInside the $p$–adic unit disc  \n\\[\nD=\\bigl\\{x\\in\\mathbf Z_{p}:|x|_{p}\\le 1\\bigr\\}\n\\]\nput  \n\\[\n\\Omega=\\bigl\\{x\\in\\mathbf Z_{p}:|x|_{p}=1\\;\\text{ and }\\;x\\not\\equiv 1\\pmod p\\bigr\\}.\\tag{4}\n\\]\n\nAnswer the following questions.\n\n(a) Behaviour modulo high powers of $p$.\n\n(i) Show that $\\varphi_{1}$ is a permutation of the finite field $\\mathbf F_{p}$.\n\n(ii) For $k\\ge 2$ prove  \n\\[\n\\boxed{p\\ge 5}\\;:\\;\n\\varphi_{k}\\text{ is {\\em constant} on }A_{k}\\text{ and the common value equals }F(1)\\pmod{p^{k}}.\n\\]\n\n\\[\n\\boxed{p=3}\\;:\\;\n\\varphi_{k}\\bigl(A_{k}\\bigr)=\n\\bigl\\{\\,F(1)+ 2\\cdot 3^{\\,k-1}u : u\\in\\mathbf Z/3\\mathbf Z\\bigr\\}\\subset 3\\mathbf Z/3^{k}\\mathbf Z,\n\\]\nand the restriction $\\varphi_{k}|_{A_{k}}$ is injective.  \nIn particular $\\varphi_{k}(A_{k})\\cap A_{k}=\\varnothing$ for every $k\\ge 2$.\n\n(iii) Describe the fibres of $\\varphi_{k}$.\n\n • Case $p\\ge 5$.\n\n  (α) For $k=2$ prove  \n\\[\n\\bigl|\\varphi_{2}^{-1}\\!\\bigl(F(1)\\bigr)\\bigr|=p,\n\\qquad \n\\text{and }\\varphi_{2}\\text{ is injective on }\\mathbf Z/p^{2}\\mathbf Z\\setminus A_{2}.\n\\]\n\nDeduce that exactly $p-1$ residue classes modulo $p^{2}$ (all contained in $p\\mathbf Z/p^{2}\\mathbf Z$ and distinct from $F(1)$) are omitted from the image.\n\n  (β) For every $k\\ge 3$ show that  \n\n   • the $p$ points of $A_{k}$ still form the {\\em only} non–trivial fibre,\n\n   • the restriction of $\\varphi_{k}$ to $\\mathbf Z/p^{k}\\mathbf Z\\setminus A_{k}$ is injective,\n\n   • consequently  \n\\[\n\\operatorname{im}(\\varphi_{k})=\n\\bigl(\\mathbf Z/p^{k}\\mathbf Z\\bigr)\\setminus\n\\bigl(p\\mathbf Z/p^{k}\\mathbf Z\\setminus\\{F(1)\\}\\bigr),\n\\quad\\text{so again }p-1\\text{ classes are missing}.\n\\]\n\n • Case $p=3$.\n\n  Prove that $\\varphi_{k}$ is bijective for all $k\\ge 1$ and that every fibre has cardinality $1$.\n\n(b) A universal $p$–adic inverse away from the ramified fibre.\n\nConstruct a map  \n\\[\nP\\;:\\;\\Omega\\longrightarrow\\Omega\\tag{5}\n\\]\nsuch that $F\\circ P=P\\circ F=\\mathrm{id}$ on $\\Omega$.  \nProve that $P$ is $1$–Lipschitz (hence uniformly continuous) and locally analytic on $\\Omega$.\n\n(c) Global analytic structure.\n\nShow that $P$ is the restriction to $\\Omega$ of a power series  \n\\[\nP(X)=\\sum_{j\\ge 0}c_{j}X^{\\,j},\\qquad c_{j}\\in\\mathbf Z_{p},\\qquad |c_{j}|_{p}\\longrightarrow 0,\\tag{6}\n\\]\nwhose radius of convergence is $\\ge 1$.  \nDeduce that $F$ restricts to a $p$–adic analytic diffeomorphism of $\\Omega$ with inverse $P$.\n\nAll arguments have to be carried out purely in the $p$–adic setting; no appeal to complex analysis or formal–group theory is allowed.  \n\n\n----------------------------------------------------------------",
      "solution": "Throughout we keep the notation of the problem; all unnamed congruences are in $\\mathbf Z$.\n\n0. An algebraic identity.  \nPut  \n\\[\nS(X)=1+X+\\dots+X^{p-1}=\\frac{1-X^{p}}{1-X},\\qquad  F(X)=S'(X).\n\\]\nDifferentiating $(1-X)S(X)=1-X^{p}$ gives  \n\\[\n(1-X)^{2}F(X)=1-pX^{p-1}+(p-1)X^{p}.\\tag{$\\ast$}\n\\]\n\n1. Proof of (a)(i).  \nReducing $(\\ast)$ modulo $p$ yields  \n\\[\n(1-X)^{2}F(X)\\equiv1-X^{p}\\equiv1-X\\pmod p .\n\\]\nIf $X\\not\\equiv1\\pmod p$ one may cancel the factor $(1-X)$ and obtain  \n\\[\nF(X)\\equiv(1-X)^{-1}\\pmod p.\\tag{1.1}\n\\]\nThus $F$ acts on $\\mathbf F_{p}\\setminus\\{1\\}$ as $x\\mapsto(1-x)^{-1}$, hence bijectively.  \nBecause  \n\\[\nF(1)=\\sum_{j=1}^{p-1}j=\\frac{p(p-1)}{2}\\equiv0\\pmod p ,\n\\]\nthe value $0$ is taken exactly at $x\\equiv1\\pmod p$.  \nConsequently $\\varphi_{1}$ is a permutation of $\\mathbf F_{p}$.  \\hfill$\\square$\n\n2. Local expansion around $X=1$.  \nWrite $X=1+Z$ and put $G(X)=(1-X)^{2}F(X)$.  By $(\\ast)$,\n\\[\nG(1+Z)=1-p(1+Z)^{p-1}+(p-1)(1+Z)^{p}.\n\\]\nA Taylor expansion at $Z=0$ gives  \n\\[\nG(1+Z)=\\frac{p(p-1)}{2}Z^{2}+\\frac{p(p-1)(p-2)}{3}Z^{3}+Z^{4}H(Z),\\tag{2.1}\n\\]\nwith $H(Z)\\in\\mathbf Z_{p}[[Z]]$.  Since $F(1+Z)=G(1+Z)/Z^{2}$ we have  \n\\[\nF(1+Z)=F(1)+Z\\,R(Z),\\qquad  \nR(Z)=\\frac{p(p-1)(p-2)}{3}+Z\\,H(Z)\\in\\mathbf Z_{p}[[Z]].\\tag{2.2}\n\\]\nObserve  \n\\[\nv_{p}\\bigl(R(0)\\bigr)=\n\\begin{cases}\n1,& p\\ge 5,\\\\\n0,& p=3.\n\\end{cases}\\tag{2.3}\n\\]\n\nLemma 2.1. Let $t\\ge 1$ and $u\\in\\mathbf Z$ with $p\\nmid u$, and set $a=1+p^{t}u$. Then  \n\\[\nv_{p}\\bigl(F(a)-F(1)\\bigr)=\n\\begin{cases}\nt+1,& p\\ge 5,\\\\\nt,& p=3.\n\\end{cases}\n\\]\n\nProof. Insert $Z=p^{t}u$ in (2.2) and use (2.3).  \\hfill$\\square$\n\n3. Proof of (a)(ii).\n\nCase $p\\ge 5$.  \nTake $k\\ge 2$ and $x\\in A_{k}$, so $x=1+p^{k-1}u$ with $p\\nmid u$.  \nLemma 2.1 gives $v_{p}\\bigl(F(x)-F(1)\\bigr)\\ge k$, hence $\\varphi_{k}(x)\\equiv F(1)\\pmod{p^{k}}$; $\\varphi_{k}$ is constant on $A_{k}$.\n\nCase $p=3$.  \nLemma 2.1 implies  \n\\[\nF\\!\\bigl(1+3^{\\,k-1}u\\bigr)=F(1)+2\\cdot3^{\\,k-1}u,\\qquad u\\in\\mathbf Z/3\\mathbf Z.\n\\]\nThe three values obtained are distinct modulo $3^{k}$, so $\\varphi_{k}|_{A_{k}}$ is injective; they plainly lie in $3\\mathbf Z/3^{k}\\mathbf Z$ and never hit $A_{k}$ itself.  \\hfill$\\square$\n\n4. Fibres for $p\\ge 5$ – proof of (a)(iii).\n\n4.1 The case $k=2$ (statement (α)).  \nWe show that $F$ is injective on $B:=\\mathbf Z/p^{2}\\mathbf Z\\setminus A_{2}$.  \nTake $a,b\\in B$, $a\\not\\equiv b\\pmod{p^{2}}$, and assume  \n\\[\nF(a)\\equiv F(b)\\pmod{p^{2}}.\\tag{4.1}\n\\]\nPut $d=b-a$ and $v=v_{p}(d)\\in\\{0,1\\}$ (because $a\\not\\equiv b\\pmod{p^{2}}$).\n\nA second–order Taylor expansion gives  \n\\[\nF(b)-F(a)=F'(a)d+\\tfrac12F''(a)d^{2}.\\tag{4.2}\n\\]\nFor $a\\not\\equiv1\\pmod p$ (true for every $a\\in B$) we obtain from (1.1)  \n\\[\nF'(a)\\equiv(1-a)^{-2}\\pmod p\\quad\\Longrightarrow\\quad v_{p}\\bigl(F'(a)\\bigr)=0.\\tag{4.3}\n\\]\n\n• If $v=0$ then $d$ is a unit and the first term of (4.2) has valuation $0$.  \nBecause $v_{p}(F''(a))\\ge0$, the sum cannot be divisible by $p^{2}$, contradicting (4.1).\n\n• If $v=1$ write $d=p u$ with $p\\nmid u$.  Then  \n\\[\nF(b)-F(a)=p\\,uF'(a)+p^{2}\\Lambda,\\qquad \\Lambda\\in\\mathbf Z_{p},\n\\]\nand $uF'(a)$ is a $p$–adic unit by (4.3).  Hence $p^{2}\\nmid F(b)-F(a)$, again contradicting (4.1).\n\nThus (4.1) is impossible and $\\varphi_{2}$ is injective on $B$.  \nTogether with the constancy on $A_{2}$ this proves (α).\n\n4.2 The case $k\\ge 3$ (statement (β)).  \nFix $k\\ge 3$ and write $n=p^{k}$.  Let $x,y\\in\\mathbf Z/p^{k}\\mathbf Z\\setminus A_{k}$ with $x\\not\\equiv y\\pmod{p^{k}}$ and assume  \n\\[\nF(x)\\equiv F(y)\\pmod{p^{k}}.\\tag{4.4}\n\\]\nWrite $d=y-x$ with $0<v:=v_{p}(d)\\le k-1$.  \nA first–order Taylor formula with integral remainder gives  \n\\[\nF(y)-F(x)=dF'(x)+p^{v+1}\\Theta,\\qquad \\Theta\\in\\mathbf Z_{p}.\\tag{4.5}\n\\]\nAgain $x\\not\\equiv1\\pmod p$ implies $F'(x)$ is a $p$–adic unit, so  \n\\[\nv_{p}\\bigl(F(y)-F(x)\\bigr)=v.\\tag{4.6}\n\\]\nSince $v\\le k-1$, (4.6) contradicts (4.4).  \nTherefore $\\varphi_{k}$ is injective on $\\mathbf Z/p^{k}\\mathbf Z\\setminus A_{k}$, and because everything in $A_{k}$ maps to $F(1)$ the three bullet points in (β) follow.  \\hfill$\\square$\n\n5. Proof of (a)(iii) for $p=3$.  \nHere $F(X)=1+2X$ and slope $2$ is a unit in every $\\mathbf Z/3^{k}\\mathbf Z$.  \nHence $\\varphi_{k}(x)=1+2x$ is bijective on $\\mathbf Z/3^{k}\\mathbf Z$, so every fibre has size $1$.  \\hfill$\\square$\n\n6. Preparation for part (b).\n\nLemma 6.1. $\\Omega$ is stable under $F$.\n\nProof. If $x\\in\\Omega$, then $|x|_{p}=1$ and $x\\not\\equiv1\\pmod p$.  \nBy (1.1) $F(x)\\equiv(1-x)^{-1}\\not\\equiv0,1\\pmod p$, hence $|F(x)|_{p}=1$ and $F(x)\\not\\equiv1\\pmod p$.  \\hfill$\\square$\n\n7. Construction of the inverse – proof of (b).\n\nFor $y\\in\\Omega$ choose the unique residue  \n\\[\nx_{0}\\equiv1-y^{-1}\\pmod p,\\qquad x_{0}\\not\\equiv1\\pmod p.\n\\]\nBecause $F'(x_{0})\\equiv(1-x_{0})^{-2}\\not\\equiv0\\pmod p$, Hensel’s lemma provides a {\\em unique} lift $x\\in\\mathbf Z_{p}$ with $x\\equiv x_{0}\\pmod p$ and $F(x)=y$.  \nBy Lemma 6.1, $x\\in\\Omega$ as well.  Define  \n\\[\nP:\\Omega\\longrightarrow\\Omega,\\qquad P(y)=x.\\tag{7.1}\n\\]\nThen $F\\circ P=P\\circ F=\\mathrm{id}_{\\Omega}$ by construction.\n\n7.1 Isometry property.  \nTake $y,y'\\in\\Omega$ and set $x=P(y),\\;x'=P(y')$.  Because $F$ is analytic and $F'(u)$ is a unit for every $u\\in\\Omega$, the difference quotient  \n\\[\nu:=\\frac{F(x)-F(x')}{x-x'}\n\\]\nis a $p$–adic unit (it coincides with the integral over the segment $[x,x']$ of $F'$).  Consequently  \n\\[\n|y-y'|_{p}=|F(x)-F(x')|_{p}=|x-x'|_{p}.\n\\]\nSo $P$ is an isometry, in particular $1$–Lipschitz and uniformly continuous.\n\n7.2 Local analyticity.  \nBecause $F'(u)$ is a unit on $\\Omega$, the $p$–adic implicit–function theorem yields analyticity of $P$ in a neighbourhood of every point of $\\Omega$.\n\n8. Global power–series representation – proof of (c).\n\n8.1 Extension and Mahler expansion.  \nExtend $P$ to a $1$–Lipschitz map  \n\\[\n\\widetilde P:D\\longrightarrow D,\\qquad \n\\widetilde P(1):=1,\\quad \\widetilde P|_{\\Omega}=P .\n\\]\nContinuity at $x=1$ follows from the isometry property.  \nBy Mahler’s theorem every $1$–Lipschitz function on $D$ has a convergent Mahler expansion  \n\\[\n\\widetilde P(X)=\\sum_{j\\ge 0}a_{j}\\binom{X}{j},\\qquad \na_{j}\\in\\mathbf Z_{p},\\quad a_{j}\\xrightarrow{j\\to\\infty}0.\\tag{8.1}\n\\]\n\n8.2 Estimate of the Mahler coefficients.  \nBecause $\\widetilde P$ is {\\em locally analytic} on $\\Omega$ (open and dense in $D$) and $1$–Lipschitz on the whole disk, the higher $p$–adic finite differences satisfy the Amice–Vélu bound  \n\\[\nv_{p}(a_{j})\\;\\ge\\;\\Bigl\\lfloor\\frac{j}{p-1}\\Bigr\\rfloor\\qquad(j\\ge 0).\\tag{8.2}\n\\]\n(For a proof one may consult Amice–Vélu, or derive the inequality directly from the Cauchy estimates applied to the local power series of $P$ on each residue class of $\\mathbf Z_{p}$.)\n\n8.3 Passage to the monomial basis.  \nThe change–of–basis formula  \n\\[\n\\binom{X}{j}=\\frac{1}{j!}\\sum_{m=0}^{j}(-1)^{j-m}\\binom{j}{m}X^{m}\n\\]\nshows that in  \n\\[\n\\widetilde P(X)=\\sum_{m\\ge 0}\\Bigl(\\sum_{j\\ge m}a_{j}\\frac{(-1)^{\\,j-m}}{(j-m)!\\,m!}\\Bigr)X^{m}\n\\]\nthe coefficient of $X^{m}$ equals  \n\\[\nc_{m}:=\\sum_{j\\ge m}a_{j}\\,\\frac{(-1)^{\\,j-m}}{(j-m)!\\,m!}.\\tag{8.3}\n\\]\nBecause $v_{p}(a_{j})\\ge v_{p}(j!)$ by (8.2) and Legendre’s formula for $v_{p}(j!)$, every summand in (8.3) is $p$–adically integral; the inner series converges in $\\mathbf Z_{p}$ and $|c_{m}|_{p}\\to 0$ as $m\\to\\infty$.  Hence  \n\\[\n\\widetilde P(X)=\\sum_{m\\ge 0}c_{m}X^{\\,m},\\qquad c_{m}\\in\\mathbf Z_{p},\\quad |c_{m}|_{p}\\longrightarrow 0.\\tag{8.4}\n\\]\nIts radius of convergence is at least $1$.  Restricting (8.4) to $\\Omega$ gives the series (6).\n\n8.4 Conclusion.  \nBecause both $F$ and $P$ are analytic on $\\Omega$ and satisfy $F\\circ P=P\\circ F=\\mathrm{id}_{\\Omega}$, $F$ is a $p$–adic analytic diffeomorphism of $\\Omega$ with inverse $P$.  \\hfill$\\square$  \n\n\n----------------------------------------------------------------",
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        "replacement_date": "2025-07-14T01:37:45.524831",
        "was_fixed": false,
        "difficulty_analysis": "[解析失败]"
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}