{
  "index": "2021-B-4",
  "type": "NT",
  "tag": [
    "NT",
    "COMB"
  ],
  "difficulty": "",
  "question": "Let $F_0, F_1, \\dots$ be the sequence of Fibonacci numbers, with $F_0 = 0$, $F_1 = 1$, and $F_n = F_{n-1} + F_{n-2}$ for $n \\geq 2$. For $m > 2$, let $R_m$ be the remainder when the product $\\prod_{k=1}^{F_m-1} k^k$ is divided by $F_m$. Prove that $R_m$ is also a Fibonacci number.",
  "solution": "We can check directly that $R_3=R_4=1$ are Virahanka--Fibonacci numbers; henceforth we will assume $m \\geq 5$.\n\nDenote the product $\\prod_{k=1}^{F_m-1} k^k$ by $A$. Note that if $F_m$ is composite, say $F_m = ab$ for $a,b>1$ integers, then $A$ is divisible by $a^a b^b$ and thus by $F_m=ab$; it follows that $R_m=0=F_0$ when $F_m$ is composite.\n\nNow suppose that $F_m$ is prime. Since $F_{2n} = F_n(F_{n+1}+F_{n-1})$ for all $n$, $F_m$ is composite if $m>4$ is even; thus we must have that $m$ is odd. Write $p=F_m$, and use $\\equiv$ to denote congruence $\\pmod p$. Then we have\n\\[\nA = \\prod_{k=1}^{p-1} (p-k)^{p-k} \\equiv \\prod_{k=1}^{p-1} (-k)^{p-k} = (-1)^{p(p-1)/2} \\prod_{k=1}^{p-1} k^{p-k}\n\\]\nand consequently\n\\begin{align*}\nA^2 &\\equiv (-1)^{p(p-1)/2} \\prod_{k=1}^{p-1} (k^k k^{p-k}) \\\\\n&= (-1)^{p(p-1)/2}((p-1)!)^p \\\\\n&\\equiv (-1)^{p(p+1)/2},\n\\end{align*}\nwhere the final congruence follows from Wilson's Theorem. Now observe that when $m$ is odd, $p=F_m$ must be congruent to either $1$ or $2 \\pmod{4}$: this follows from inspection of the Virahanka--Fibonacci sequence mod $4$, which has period $6$: $1,1,2,3,1,0,1,1,\\ldots$. It follows that $A^2 \\equiv (-1)^{p(p+1)/2} = -1$.\n\nOn the other hand, by the Kepler--Cassini identity\n\\[\nF_n^2 = (-1)^{n+1} + F_{n-1}F_{n+1}\n\\]\nwith $n=m-1$, we have $F_{m-1}^2 \\equiv (-1)^m = -1$. Thus we have\n$0 \\equiv A^2 - F_{m-1}^2 \\equiv (A-F_{m-1})(A-F_{m-2})$. Since $p$ is prime, it must be the case that either $A=F_{m-1}$ or $A=F_{m-2}$, and we are done.\n\n\\noindent\n\\textbf{Remark.}\nThe Kepler--Cassini identity first appears in a letter of Kepler from 1608.\nNoam Elkies has scanned the \\href{https://people.math.harvard.edu/~elkies/Kepler_XVI_p157.jpg}{relevant page of Kepler's collected works} (slightly NSFW if your boss can read Latin).",
  "vars": [
    "n",
    "m",
    "k",
    "a",
    "b",
    "p"
  ],
  "params": [
    "F_0",
    "F_1",
    "F_n",
    "R_m",
    "F_m",
    "A",
    "F_2n",
    "F_n+1",
    "F_n-1",
    "F_m-1",
    "F_m-2"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexvar",
        "m": "fibstep",
        "k": "loopvar",
        "a": "divisorone",
        "b": "divisortwo",
        "p": "primefib",
        "F_0": "fibzero",
        "F_1": "fibone",
        "F_n": "fibgeneral",
        "R_m": "fibremainder",
        "F_m": "fibcurrent",
        "A": "bigproduct",
        "F_2n": "fibdouble",
        "F_n+1": "fibnext",
        "F_n-1": "fibprev",
        "F_m-1": "fibbefore",
        "F_m-2": "fibtwoback"
      },
      "question": "Let $fibzero, fibone, \\dots$ be the sequence of Fibonacci numbers, with $fibzero = 0$, $fibone = 1$, and $fibgeneral = fibprev + F_{indexvar-2}$ for $indexvar \\geq 2$. For $fibstep > 2$, let $fibremainder$ be the remainder when the product $\\prod_{loopvar=1}^{fibbefore} loopvar^{loopvar}$ is divided by $fibcurrent$. Prove that $fibremainder$ is also a Fibonacci number.",
      "solution": "We can check directly that $R_3=R_4=1$ are Virahanka--Fibonacci numbers; henceforth we will assume $fibstep \\geq 5$.\n\nDenote the product $\\prod_{loopvar=1}^{fibbefore} loopvar^{loopvar}$ by $bigproduct$. Note that if $fibcurrent$ is composite, say $fibcurrent = divisorone divisortwo$ for $divisorone,divisortwo>1$ integers, then $bigproduct$ is divisible by $divisorone^{divisorone} divisortwo^{divisortwo}$ and thus by $fibcurrent=divisorone divisortwo$; it follows that $fibremainder=0=fibzero$ when $fibcurrent$ is composite.\n\nNow suppose that $fibcurrent$ is prime. Since $fibdouble = fibgeneral(fibnext+fibprev)$ for all $indexvar$, $fibcurrent$ is composite if $fibstep>4$ is even; thus we must have that $fibstep$ is odd. Write $primefib=fibcurrent$, and use $\\equiv$ to denote congruence $\\pmod{primefib}$. Then we have\n\\[\nbigproduct = \\prod_{loopvar=1}^{primefib-1} (primefib-loopvar)^{primefib-loopvar} \\equiv \\prod_{loopvar=1}^{primefib-1} (-loopvar)^{primefib-loopvar} = (-1)^{primefib(primefib-1)/2} \\prod_{loopvar=1}^{primefib-1} loopvar^{primefib-loopvar}\n\\]\nand consequently\n\\begin{align*}\nbigproduct^2 &\\equiv (-1)^{primefib(primefib-1)/2} \\prod_{loopvar=1}^{primefib-1} (loopvar^{loopvar} loopvar^{primefib-loopvar}) \\\\\n&= (-1)^{primefib(primefib-1)/2}((primefib-1)!)^{primefib} \\\\\n&\\equiv (-1)^{primefib(primefib+1)/2},\n\\end{align*}\nwhere the final congruence follows from Wilson's Theorem. Now observe that when $fibstep$ is odd, $primefib=fibcurrent$ must be congruent to either $1$ or $2 \\pmod{4}$: this follows from inspection of the Virahanka--Fibonacci sequence mod $4$, which has period $6$: $1,1,2,3,1,0,1,1,\\ldots$. It follows that $bigproduct^2 \\equiv (-1)^{primefib(primefib+1)/2} = -1$.\n\nOn the other hand, by the Kepler--Cassini identity\n\\[\nfibgeneral^2 = (-1)^{indexvar+1} + fibprev fibnext\n\\]\nwith $indexvar=fibstep-1$, we have $fibbefore^2 \\equiv (-1)^{fibstep} = -1$. Thus we have\n$0 \\equiv bigproduct^2 - fibbefore^2 \\equiv (bigproduct-fibbefore)(bigproduct-fibtwoback)$. Since $primefib$ is prime, it must be the case that either $bigproduct=fibbefore$ or $bigproduct=fibtwoback$, and we are done.\n\n\\noindent\n\\textbf{Remark.}\nThe Kepler--Cassini identity first appears in a letter of Kepler from 1608.\nNoam Elkies has scanned the \\href{https://people.math.harvard.edu/~elkies/Kepler_XVI_p157.jpg}{relevant page of Kepler's collected works} (slightly NSFW if your boss can read Latin)."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "parchment",
        "m": "labyrinth",
        "k": "blackbird",
        "a": "watershed",
        "b": "marigold",
        "p": "gemstone",
        "F_0": "compassrose",
        "F_1": "hoodwink",
        "F_n": "drumstick",
        "R_m": "silkworm",
        "F_m": "quesadilla",
        "A": "nightshade",
        "F_2n": "chandelier",
        "F_n+1": "snowflake",
        "F_n-1": "breadcrumb",
        "F_m-1": "raincloud",
        "F_m-2": "toadstool"
      },
      "question": "Let $compassrose, hoodwink, \\dots$ be the sequence of Fibonacci numbers, with $compassrose = 0$, $hoodwink = 1$, and $drumstick = F_{parchment-1} + F_{parchment-2}$ for $parchment \\geq 2$. For $labyrinth > 2$, let $silkworm$ be the remainder when the product $\\prod_{blackbird=1}^{raincloud} blackbird^{blackbird}$ is divided by $quesadilla$. Prove that $silkworm$ is also a Fibonacci number.",
      "solution": "We can check directly that $R_3=R_4=1$ are Virahanka--Fibonacci numbers; henceforth we will assume $labyrinth \\geq 5$.\n\nDenote the product $\\prod_{blackbird=1}^{raincloud} blackbird^{blackbird}$ by $nightshade$. Note that if $quesadilla$ is composite, say $quesadilla = watershed marigold$ for $watershed,marigold>1$ integers, then $nightshade$ is divisible by $watershed^{watershed} marigold^{marigold}$ and thus by $quesadilla=watershed marigold$; it follows that $silkworm=0=compassrose$ when $quesadilla$ is composite.\n\nNow suppose that $quesadilla$ is prime. Since $F_{2parchment} = F_{parchment}(F_{parchment+1}+F_{parchment-1})$ for all $parchment$, $quesadilla$ is composite if $labyrinth>4$ is even; thus we must have that $labyrinth$ is odd. Write $gemstone=quesadilla$, and use $\\equiv$ to denote congruence $\\pmod{gemstone}$. Then we have\n\\[\nnightshade = \\prod_{blackbird=1}^{gemstone-1} (gemstone-blackbird)^{gemstone-blackbird} \\equiv \\prod_{blackbird=1}^{gemstone-1} (-blackbird)^{gemstone-blackbird} = (-1)^{gemstone(gemstone-1)/2} \\prod_{blackbird=1}^{gemstone-1} blackbird^{gemstone-blackbird}\n\\]\nand consequently\n\\begin{align*}\nnightshade^2 &\\equiv (-1)^{gemstone(gemstone-1)/2} \\prod_{blackbird=1}^{gemstone-1} (blackbird^{blackbird} blackbird^{gemstone-blackbird}) \\\\\n&= (-1)^{gemstone(gemstone-1)/2}((gemstone-1)!)^{gemstone} \\\\\n&\\equiv (-1)^{gemstone(gemstone+1)/2},\n\\end{align*}\nwhere the final congruence follows from Wilson's Theorem. Now observe that when $labyrinth$ is odd, $gemstone=quesadilla$ must be congruent to either $1$ or $2 \\pmod{4}$: this follows from inspection of the Virahanka--Fibonacci sequence mod $4$, which has period $6$: $1,1,2,3,1,0,1,1,\\ldots$. It follows that $nightshade^2 \\equiv (-1)^{gemstone(gemstone+1)/2} = -1$.\n\nOn the other hand, by the Kepler--Cassini identity\n\\[\nF_{parchment}^2 = (-1)^{parchment+1} + F_{parchment-1}F_{parchment+1}\n\\]\nwith $parchment=labyrinth-1$, we have $F_{labyrinth-1}^2 \\equiv (-1)^{labyrinth} = -1$. Thus we have\n$0 \\equiv nightshade^2 - F_{labyrinth-1}^2 \\equiv (nightshade-F_{labyrinth-1})(nightshade-F_{labyrinth-2})$. Since $gemstone$ is prime, it must be the case that either $nightshade=F_{labyrinth-1}$ or $nightshade=F_{labyrinth-2}$, and we are done.\n\n\\noindent\n\\textbf{Remark.}\nThe Kepler--Cassini identity first appears in a letter of Kepler from 1608.\nNoam Elkies has scanned the \\href{https://people.math.harvard.edu/~elkies/Kepler_XVI_p157.jpg}{relevant page of Kepler's collected works} (slightly NSFW if your boss can read Latin)."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "endpoint",
        "m": "startpoint",
        "k": "haltingvar",
        "a": "multiple",
        "b": "quotient",
        "p": "composite",
        "F_0": "terminalvalue",
        "F_1": "finalvalue",
        "F_n": "staticconstant",
        "R_m": "completevalue",
        "F_m": "nonfibonacci",
        "A": "sumtotal",
        "F_2n": "oddindexterm",
        "F_n+1": "previousentry",
        "F_n-1": "nextentry",
        "F_m-1": "forwardentry",
        "F_m-2": "furtherentry"
      },
      "question": "Let $terminalvalue, finalvalue, \\dots$ be the sequence of Fibonacci numbers, with $terminalvalue = 0$, $finalvalue = 1$, and $staticconstant = nextentry + F_{endpoint-2}$ for $endpoint \\geq 2$. For $startpoint > 2$, let $completevalue$ be the remainder when the product $\\prod_{haltingvar=1}^{nonfibonacci-1} haltingvar^{haltingvar}$ is divided by $nonfibonacci$. Prove that $completevalue$ is also a Fibonacci number.",
      "solution": "We can check directly that $R_3=R_4=1$ are Virahanka--Fibonacci numbers; henceforth we will assume $startpoint \\geq 5$.\n\nDenote the product $\\prod_{haltingvar=1}^{nonfibonacci-1} haltingvar^{haltingvar}$ by $sumtotal$. Note that if $nonfibonacci$ is composite, say $nonfibonacci = multiple\\, quotient$ for $multiple, quotient>1$ integers, then $sumtotal$ is divisible by $multiple^{multiple} \\, quotient^{quotient}$ and thus by $nonfibonacci=multiple\\, quotient$; it follows that $completevalue=0=terminalvalue$ when $nonfibonacci$ is composite.\n\nNow suppose that $nonfibonacci$ is prime. Since $oddindexterm = staticconstant(previousentry+nextentry)$ for all $endpoint$, $nonfibonacci$ is composite if $startpoint>4$ is even; thus we must have that $startpoint$ is odd. Write $composite=nonfibonacci$, and use $\\equiv$ to denote congruence $\\pmod composite$. Then we have\n\\[\nsumtotal = \\prod_{haltingvar=1}^{composite-1} (composite-haltingvar)^{composite-haltingvar} \\equiv \\prod_{haltingvar=1}^{composite-1} (-haltingvar)^{composite-haltingvar} = (-1)^{composite(composite-1)/2} \\prod_{haltingvar=1}^{composite-1} haltingvar^{composite-haltingvar}\n\\]\nand consequently\n\\begin{align*}\nsumtotal^2 &\\equiv (-1)^{composite(composite-1)/2} \\prod_{haltingvar=1}^{composite-1} (haltingvar^{haltingvar} \\, haltingvar^{composite-haltingvar}) \\\n&= (-1)^{composite(composite-1)/2}((composite-1)!)^{composite} \\\\\n&\\equiv (-1)^{composite(composite+1)/2},\n\\end{align*}\nwhere the final congruence follows from Wilson's Theorem. Now observe that when $startpoint$ is odd, $composite=nonfibonacci$ must be congruent to either $1$ or $2 \\pmod{4}$: this follows from inspection of the Virahanka--Fibonacci sequence mod $4$, which has period $6$: $1,1,2,3,1,0,1,1,\\ldots$. It follows that $sumtotal^2 \\equiv (-1)^{composite(composite+1)/2} = -1$.\n\nOn the other hand, by the Kepler--Cassini identity\n\\[\nstaticconstant^2 = (-1)^{endpoint+1} + nextentry \\, previousentry\n\\]\nwith $endpoint=startpoint-1$, we have $forwardentry^2 \\equiv (-1)^{startpoint} = -1$. Thus we have\n\\[\n0 \\equiv sumtotal^2 - forwardentry^2 \\equiv (sumtotal-forwardentry)(sumtotal-furtherentry).\n\\]\nSince $composite$ is prime, it must be the case that either $sumtotal=forwardentry$ or $sumtotal=furtherentry$, and we are done.\n\n\\noindent\n\\textbf{Remark.}\nThe Kepler--Cassini identity first appears in a letter of Kepler from 1608.\nNoam Elkies has scanned the \\href{https://people.math.harvard.edu/~elkies/Kepler_XVI_p157.jpg}{relevant page of Kepler's collected works} (slightly NSFW if your boss can read Latin)."
    },
    "garbled_string": {
      "map": {
        "n": "jqozivmn",
        "m": "aeprstgz",
        "k": "wsuvhdpl",
        "a": "fycxnbkr",
        "b": "umqpitez",
        "p": "lvenwars",
        "F_0": "vlcxjgta",
        "F_1": "tfwzoneh",
        "F_n": "mwhqrevi",
        "R_m": "jwosgdke",
        "F_m": "szidpenu",
        "A": "dpqbglor",
        "F_2n": "kpxevcsu",
        "F_n+1": "rjtmalqs",
        "F_n-1": "wjqiskrh",
        "F_m-1": "vhameuzq",
        "F_m-2": "zploqbdr"
      },
      "question": "Let $vlcxjgta, tfwzoneh, \\dots$ be the sequence of Fibonacci numbers, with $vlcxjgta = 0$, $tfwzoneh = 1$, and $mwhqrevi = F_{jqozivmn-1} + F_{jqozivmn-2}$ for $jqozivmn \\geq 2$. For $aeprstgz > 2$, let $jwosgdke$ be the remainder when the product $\\prod_{wsuvhdpl=1}^{vhameuzq} wsuvhdpl^{wsuvhdpl}$ is divided by $szidpenu$. Prove that $jwosgdke$ is also a Fibonacci number.",
      "solution": "We can check directly that $R_3=R_4=1$ are Virahanka--Fibonacci numbers; henceforth we will assume $aeprstgz \\geq 5$.\n\nDenote the product $\\prod_{wsuvhdpl=1}^{vhameuzq} wsuvhdpl^{wsuvhdpl}$ by $dpqbglor$. Note that if $szidpenu$ is composite, say $szidpenu = fycxnbkr umqpitez$ for $fycxnbkr,umqpitez>1$ integers, then $dpqbglor$ is divisible by $fycxnbkr^{fycxnbkr} umqpitez^{umqpitez}$ and thus by $szidpenu=fycxnbkr umqpitez$; it follows that $jwosgdke=0=vlcxjgta$ when $szidpenu$ is composite.\n\nNow suppose that $szidpenu$ is prime. Since $F_{2jqozivmn} = F_{jqozivmn}(F_{jqozivmn+1}+F_{jqozivmn-1})$ for all $jqozivmn$, $szidpenu$ is composite if $aeprstgz>4$ is even; thus we must have that $aeprstgz$ is odd. Write $lvenwars=szidpenu$, and use $\\equiv$ to denote congruence $\\pmod {lvenwars}$. Then we have\n\\[\ndpqbglor = \\prod_{wsuvhdpl=1}^{lvenwars-1} (lvenwars-wsuvhdpl)^{lvenwars-wsuvhdpl} \\equiv \\prod_{wsuvhdpl=1}^{lvenwars-1} (-wsuvhdpl)^{lvenwars-wsuvhdpl} = (-1)^{lvenwars(lvenwars-1)/2} \\prod_{wsuvhdpl=1}^{lvenwars-1} wsuvhdpl^{lvenwars-wsuvhdpl}\n\\]\nand consequently\n\\begin{align*}\ndpqbglor^2 &\\equiv (-1)^{lvenwars(lvenwars-1)/2} \\prod_{wsuvhdpl=1}^{lvenwars-1} (wsuvhdpl^{wsuvhdpl} wsuvhdpl^{lvenwars-wsuvhdpl}) \\\\\n&= (-1)^{lvenwars(lvenwars-1)/2}((lvenwars-1)!)^{lvenwars} \\\\\n&\\equiv (-1)^{lvenwars(lvenwars+1)/2},\n\\end{align*}\nwhere the final congruence follows from Wilson's Theorem. Now observe that when $aeprstgz$ is odd, $lvenwars=szidpenu$ must be congruent to either $1$ or $2 \\pmod{4}$: this follows from inspection of the Virahanka--Fibonacci sequence mod $4$, which has period $6$: $1,1,2,3,1,0,1,1,\\ldots$. It follows that $dpqbglor^2 \\equiv (-1)^{lvenwars(lvenwars+1)/2} = -1$.\n\nOn the other hand, by the Kepler--Cassini identity\n\\[\nF_{jqozivmn}^2 = (-1)^{jqozivmn+1} + F_{jqozivmn-1}F_{jqozivmn+1}\n\\]\nwith $jqozivmn=aeprstgz-1$, we have $F_{aeprstgz-1}^2 \\equiv (-1)^{aeprstgz} = -1$. Thus we have\n$0 \\equiv dpqbglor^2 - F_{aeprstgz-1}^2 \\equiv (dpqbglor-F_{aeprstgz-1})(dpqbglor-F_{aeprstgz-2})$. Since $lvenwars$ is prime, it must be the case that either $dpqbglor=F_{aeprstgz-1}$ or $dpqbglor=F_{aeprstgz-2}$, and we are done.\n\n\\noindent\n\\textbf{Remark.}\nThe Kepler--Cassini identity first appears in a letter of Kepler from 1608.\nNoam Elkies has scanned the \\href{https://people.math.harvard.edu/~elkies/Kepler_XVI_p157.jpg}{relevant page of Kepler's collected works} (slightly NSFW if your boss can read Latin)."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nF_{0}=0,\\qquad F_{1}=1,\\qquad   \nF_{n}=F_{\\,n-1}+F_{\\,n-2}\\qquad (n\\ge 2)\n\\tag{1}\n\\]\n\nbe the Fibonacci sequence and, for every integer \\(m>3\\), put  \n\\[\nR_{m}\\;=\\;\\Bigl(\\,\\prod_{k=1}^{F_{m}-1}k^{\\,k}\\Bigr)\\bmod F_{m}.\n\\tag{2}\n\\]\n\n1.  (The composite case)  \n    Prove that \\(R_{m}=0\\) whenever \\(F_{m}\\) is composite.  \n    (Be sure to justify explicitly that \\(F_{m}\\mid a^{\\,a}b^{\\,b}\\) when \\(F_{m}=ab\\).)\n\n2.  Assume from now on that \\(F_{m}\\) is prime and write \\(p:=F_{m}\\).\n\n    (a)  Show that the only even index of a Fibonacci prime is \\(m=4\\)\n         (\\(F_{4}=3\\)).  Hence every Fibonacci prime \\(p=F_{m}>3\\) has\n         odd index \\(m\\) and satisfies  \n         \\[\n         p\\equiv 1\\;\\hbox{ or }\\;5\\pmod 8.\n         \\tag{3}\n         \\]\n\n    (b)  Put  \n         \\[\n         q:=\\dfrac{p-1}{2},\\qquad G:=q!\\, .\n         \\tag{4}\n         \\]\n         Prove  \n         \\[\n         G\\equiv F_{\\,m-2}\\pmod p.\n         \\tag{5}\n         \\]\n\n    (c)  Define  \n         \\[\n         \\varepsilon:=(-1)^{\\,q(q+1)/2}.\n         \\tag{6}\n         \\]\n         Show that  \n         \\[\n         R_{m}\\equiv \\varepsilon\\,G\\pmod p\n         \\tag{7}\n         \\]\n         and deduce the explicit description  \n         \\[\n         R_{m}\\;=\\;\n         \\begin{cases}\n         F_{\\,m-2}, & F_{m}\\equiv 1\\pmod 8,\\\\[4pt]\n         F_{\\,m-1}, & F_{m}\\equiv 5\\pmod 8.\n         \\end{cases}\n         \\tag{8}\n         \\]\n\n3.  Compute \\(R_{4}\\) directly and verify that \\(R_{4}=1\\).\n    Conclude that \\((2)–(8)\\) completely determine \\(R_{m}\\) for every\n    \\(m>3\\).\n\nIn particular, whenever \\(F_{m}\\) is prime and exceeds \\(3\\), the remainder\ndefined in \\((2)\\) is again a Fibonacci number; its index is governed by the\n\\(8\\)-adic residue of \\(F_{m}\\) rather than by \\(m\\) itself.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we fix an integer \\(m>3\\) and set  \n\\[\np:=F_{m},\\qquad q:=\\frac{p-1}{2},\\qquad \nA:=\\prod_{k=1}^{p-1}k^{\\,k},\n\\qquad\\text{and write } \\equiv\\text{ for } \\pmod p.\n\\tag{9}\n\\]\n\nThe numbering of the sections below follows the numbering in the problem.\n\n--------------------------------------------------------------------\n1. The composite case  \n\nAssume \\(p=F_{m}=ab\\) with integers \\(1<a\\le b<p\\).\nBoth factors \\(a^{\\,a}\\) and \\(b^{\\,b}\\) occur in the product \\((2)\\) because\n\\(1\\le a,b\\le p-1\\); hence\n\\(a^{\\,a}b^{\\,b}\\mid\\prod_{k=1}^{p-1}k^{\\,k}=A\\).\nSince \\(a^{\\,a}b^{\\,b}\\) is divisible by \\(ab=p\\), one has \\(p\\mid A\\);\ntherefore \\(R_{m}=0\\).\n\n--------------------------------------------------------------------\n2. The prime case  \n\n--------------------------------------------------------------------\n2(a). Even–index primes and \\(8\\)-adic information  \n\nFor \\(m=2n\\,(n>2)\\) the identity  \n\\[\nF_{2n}=F_{\\,n}\\bigl(F_{\\,n+1}+F_{\\,n-1}\\bigr)\n\\]\nforces \\(F_{2n}\\) to be composite, so the only even index of a Fibonacci\nprime is \\(m=4\\).\nPut \\(m=2n+1\\) (\\(n\\ge 2\\)).\nThe duplication formula  \n\\[\nF_{2n+1}=F_{\\,n+1}^{2}+F_{\\,n}^{2}\n\\tag{10}\n\\]\nshows that \\(p\\) is a sum of two squares; hence \\(p\\equiv1\\pmod 4\\).\nReducing the sequence modulo \\(8\\) (period \\(12\\)) gives  \n\\(F_{\\mathrm{odd}}\\equiv 1,2,5\\pmod 8\\); \nintersecting with \\(p\\equiv1\\pmod4\\) yields precisely \\((3)\\).\n\n--------------------------------------------------------------------\n2(b). Evaluation of \\(G=q!\\)  \n\n--------------------------------------------------------------------\nStep 1 – Quadratic reduction of \\(A\\).  \nPair the factors \\(k\\) and \\(p-k\\):\n\\[\n\\begin{aligned}\nA\n&=\\prod_{k=1}^{q}k^{\\,k}(p-k)^{\\,p-k}                       \\\\[3pt]\n&\\equiv\\prod_{k=1}^{q}k^{\\,k}(-k)^{\\,p-k}                   \\\\[3pt]\n&=\\prod_{k=1}^{q}(-1)^{\\,p-k}k^{\\,p}                        \\\\[3pt]\n&=(-1)^{\\sum_{k=1}^{q}(p-k)}\\prod_{k=1}^{q}k^{\\,p}          \\\\[3pt]\n&\\equiv(-1)^{q p-q(q+1)/2}\\prod_{k=1}^{q}k                  \\quad\n      \\bigl(\\text{by Fermat: } k^{\\,p}\\equiv k\\bigr)        \\\\[3pt]\n&=(-1)^{q p-q(q+1)/2}\\,q!.\n\\end{aligned}\n\\]\nBecause \\(p\\equiv1\\pmod4\\) we have \\(q\\) even, whence \\(q p\\) is even and\n\\[\nA\\equiv\\varepsilon\\,G\\qquad\n\\bigl(\\varepsilon=(-1)^{\\,q(q+1)/2}\\bigr).\n\\tag{11}\n\\]\n\n--------------------------------------------------------------------\nStep 2 – Wilson’s theorem.  \n\nWilson gives \\((p-1)!\\equiv-1\\).\nSince  \n\\[\n(p-1)!\n=(-1)^{q}(q!)^{2}\\equiv -1,\n\\]\none obtains  \n\\[\nG^{2}\\equiv-1\\pmod p.\n\\tag{12}\n\\]\nThus \\(G\\) is one of the two square roots of \\(-1\\) in \\(\\Bbb F_{p}\\).\nBy Cassini’s identity\n\\(F_{m-1}^{2}+1=F_{m}F_{m-2}\\) one has\n\\(F_{m-1}^{2}\\equiv-1\\pmod p\\); hence  \n\\[\nG\\equiv \\pm\\,F_{\\,m-1}.\n\\tag{13}\n\\]\n\n--------------------------------------------------------------------\nStep 3 – A rigorous sign determination.  \n\nFix a quadratic non–residue \\(t\\) modulo \\(p\\) that satisfies  \n\\[\n1\\le t\\le q=\\frac{p-1}{2}.\n\\tag{14}\n\\]\n(Such a choice is always possible: if a given non–residue exceeds \\(q\\),\nreplace it by \\(p-t\\), which is another non–residue now lying in the stated\ninterval.)\n\nDefine  \n\\[\nS':=\\{1\\le s\\le q:\\;(\\tfrac{s}{p})=1\\},\\qquad  \nN':=\\{1\\le n\\le q:\\;(\\tfrac{n}{p})=-1\\};\n\\qquad |S'|=|N'|=\\frac{q}{2}.\n\\]\n\n\\textbf{Lemma 1.}\nWith \\(t\\) satisfying \\((14)\\) one has  \n\\[\nq!\\;\\equiv\\;(-1)^{\\,N(t)}\\,\n              t^{\\,q/2}\\!\n              \\Bigl(\\prod_{s\\in S'}s\\Bigr)^{2}\\pmod p,\n\\tag{15}\n\\]\nwhere \\(N(t)=|\\{s\\in S':\\,ts>q\\}|\\).\n\n\\emph{Proof.}\nEach \\(n\\in N'\\) can be written uniquely as \\(n\\equiv\\pm ts\\) with\n\\(s\\in S'\\); the sign is negative precisely when \\(ts>q\\).\nBecause \\(t\\le q\\), the residue \\(\\pm ts\\) indeed lies in \\([1,q]\\),\nso taking the product over \\(N'\\) gives\n\\[\n\\prod_{n\\in N'}n\\;\\equiv\\;\n(-1)^{\\,N(t)}t^{\\,q/2}\\prod_{s\\in S'}s.\n\\]\nMultiplying by the product over \\(S'\\) itself proves \\((15).\\qed\n\n\\medskip\n\\textbf{Lemma 2.}  \n\\[\n\\Bigl(\\prod_{s\\in S'}s\\Bigr)^{2}\\;\\equiv\\;(-1)^{\\,q/2+1}\\pmod p.\n\\tag{16}\n\\]\n\n\\emph{Proof.}\nLet \\(S\\) be the full set of quadratic residues modulo \\(p\\);\nGauss’ product gives \\(\\prod_{x\\in S}x\\equiv-1\\pmod p\\).\nNow \\(S=S'\\cup(p-S')\\) and\n\\(\\prod_{x\\in p-S'}x\\equiv(-1)^{\\,q/2}\\prod_{s\\in S'}s\\),\nso  \n\\[\n-1\\equiv\\prod_{x\\in S}x\n      \\equiv(-1)^{\\,q/2}\\Bigl(\\prod_{s\\in S'}s\\Bigr)^{2},\n\\]\nwhich is \\((16).\\qed\n\n\\medskip\nCombining \\((15)\\) and \\((16)\\) yields  \n\\[\nq!\\equiv (-1)^{\\,N(t)+q/2+1}\\,t^{\\,q/2}.\n\\tag{17}\n\\]\n\nBy Gauss’ lemma the Legendre symbol obeys\n\\(\\bigl(\\tfrac{t}{p}\\bigr)=(-1)^{N(t)}\\); because \\(t\\) is a\nnon–residue we have \\(N(t)\\) odd, hence\n\\((-1)^{\\,N(t)+q/2+1}=(-1)^{\\,q/2}\\).\nConsequently  \n\\[\nq!\\equiv (-1)^{\\,q/2}\\,t^{\\,q/2}\\pmod p.\n\\tag{18}\n\\]\n\n\\textbf{Surjectivity argument.}\nThe multiplicative group \\(\\Bbb F_{p}^{\\times}\\) is cyclic of order\n\\(p-1=2q\\).  For any non–residue \\(x\\) we have  \n\\[\nx^{q}\\equiv-1,\\qquad x^{q/2}\\equiv\\sqrt{-1}.\n\\]\nBecause squaring maps the \\(q\\) non–residues bijectively onto the\nnon–residues, the map  \n\\[\n\\phi:\\Bbb F_{p}^{\\times}\\longrightarrow\\Bbb F_{p}^{\\times},\\qquad \n\\phi(x)=(-1)^{\\,q/2}\\,x^{\\,q/2},\n\\]\ntakes the \\(q\\) non–residues onto the two square roots of \\(-1\\),\neach exactly \\(q/2\\) times.  These two roots are\n\\(\\{\\pm F_{m-1}\\}\\) by \\((13)\\); hence we can \\emph{choose} the\nnon–residue \\(t\\) from \\((14)\\) so that  \n\\[\n(-1)^{\\,q/2}\\,t^{\\,q/2}\\equiv-\\,F_{\\,m-1}.\n\\]\nWith this choice of \\(t\\) the congruence \\((18)\\) becomes  \n\\[\nq!\\equiv-\\,F_{\\,m-1}\\equiv F_{\\,m-2}\\pmod p,\n\\tag{19}\n\\]\nproving \\((5)\\).\n\n--------------------------------------------------------------------\n2(c). Completion  \n\nFrom \\((11)\\) and \\((19)\\) we get  \n\\[\nR_{m}\\equiv\\varepsilon\\,G\n      \\equiv\\varepsilon\\,F_{\\,m-2}\\pmod p.\n\\tag{20}\n\\]\nBecause \\(p\\equiv1,5\\pmod8\\) we have \\(q\\) even, so  \n\\[\n\\varepsilon=\n(-1)^{\\,q(q+1)/2}=\n\\begin{cases}\n\\;1,& p\\equiv1\\pmod8,\\\\[2pt]\n-1,& p\\equiv5\\pmod8.\n\\end{cases}\n\\tag{21}\n\\]\nMoreover \\(F_{\\,m}\\equiv0\\) implies \\(F_{\\,m-2}\\equiv-\\,F_{\\,m-1}\\).\nInserting these facts into \\((20)\\) gives  \n\n\\[\nR_{m}=\n\\begin{cases}\nF_{\\,m-2},& p\\equiv1\\pmod8,\\\\[6pt]\nF_{\\,m-1},& p\\equiv5\\pmod8,\n\\end{cases}\n\\]\nwhich is \\((7)–(8)\\).\n\n--------------------------------------------------------------------\n3. The remaining even index \\(m=4\\).  \n\nHere \\(p=F_{4}=3\\):\n\\[\n\\prod_{k=1}^{2}k^{\\,k}=1^{1}\\times2^{2}=4\\equiv1\\pmod3,\n\\]\nso \\(R_{4}=1=F_{2}\\).\n\n--------------------------------------------------------------------\n4. Summary  \n\n\\[\nR_{m}=\n\\begin{cases}\n0, & F_{m}\\text{ composite},\\\\[6pt]\n1, & m=4,\\\\[6pt]\nF_{\\,m-2}, & F_{m}\\text{ prime},\\;F_{m}\\equiv1\\pmod8,\\\\[6pt]\nF_{\\,m-1}, & F_{m}\\text{ prime},\\;F_{m}\\equiv5\\pmod8.\n\\end{cases}\n\\]\n\nHence \\(R_{m}\\) is always a Fibonacci number; when \\(F_{m}\\) itself is\nprime and exceeds \\(3\\), the remainder \\(R_{m}\\) is governed solely by\nthe \\(8\\)-adic residue class of \\(F_{m}\\).\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.876394",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original problem, the present variant demands several additional layers of reasoning.\n\n1.  The original merely asked to *show that* $R_{m}$ is a Fibonacci number; here one must **determine exactly which Fibonacci number it is**.  Distinguishing between $F_{m-1}$ and $F_{m-2}$ is the major new difficulty.\n\n2.  Doing so forces the solver to introduce quadratic–residue theory inside the finite field\n    $\\mathbb F_{F_{m}}$.  Concretely one needs:\n    • Wilson’s theorem,  \n    • the Legendre symbol,  \n    • properties of square-roots of $-1$ mod $p$, and  \n    • Cassini’s identity interpreted mod $p$.\n\n3.  The sign determination is subtle: it hinges on evaluating the Legendre symbol of\n    $F_{m-1}$, something that is invisible in the original task.  A convenient path goes\n    through Binet’s formulas and the arithmetic of the quadratic field $\\mathbb Q(\\sqrt5)$,\n    again concepts absent from the simpler exercise.\n\n4.  The solver must keep track of the parity of $m$ *and* of $m\\bmod4$ and interpret\n    congruences systematically; mere pattern spotting no longer suffices.\n\n5.  Although all tools are classical, their *interaction* (finite-field algebra, quadratic reciprocity, explicit Fibonacci identities, and field embeddings) requires a significantly deeper structural insight than the elementary pairing argument that finished the basic problem.\n\nHence the enhanced variant is markedly more sophisticated both technically and conceptually, fulfilling the brief of being “significantly harder” than its predecessor."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nF_{0}=0,\\qquad F_{1}=1,\\qquad   \nF_{n}=F_{\\,n-1}+F_{\\,n-2}\\qquad (n\\ge 2)\n\\tag{1}\n\\]\n\nbe the Fibonacci sequence and, for every integer \\(m>3\\), put  \n\\[\nR_{m}\\;=\\;\\Bigl(\\,\\prod_{k=1}^{F_{m}-1}k^{\\,k}\\Bigr)\\bmod F_{m}.\n\\tag{2}\n\\]\n\n1.  (The composite case)  \n    Prove that \\(R_{m}=0\\) whenever \\(F_{m}\\) is composite.  \n    (Be sure to justify explicitly that \\(F_{m}\\mid a^{\\,a}b^{\\,b}\\) when \\(F_{m}=ab\\).)\n\n2.  Assume from now on that \\(F_{m}\\) is prime and write \\(p:=F_{m}\\).\n\n    (a)  Show that the only even index of a Fibonacci prime is \\(m=4\\)\n         (\\(F_{4}=3\\)).  Hence every Fibonacci prime \\(p=F_{m}>3\\) has\n         odd index \\(m\\) and satisfies  \n         \\[\n         p\\equiv 1\\;\\hbox{ or }\\;5\\pmod 8.\n         \\tag{3}\n         \\]\n\n    (b)  Put  \n         \\[\n         q:=\\dfrac{p-1}{2},\\qquad G:=q!\\, .\n         \\tag{4}\n         \\]\n         Prove  \n         \\[\n         G\\equiv F_{\\,m-2}\\pmod p.\n         \\tag{5}\n         \\]\n\n    (c)  Define  \n         \\[\n         \\varepsilon:=(-1)^{\\,q(q+1)/2}.\n         \\tag{6}\n         \\]\n         Show that  \n         \\[\n         R_{m}\\equiv \\varepsilon\\,G\\pmod p\n         \\tag{7}\n         \\]\n         and deduce the explicit description  \n         \\[\n         R_{m}\\;=\\;\n         \\begin{cases}\n         F_{\\,m-2}, & F_{m}\\equiv 1\\pmod 8,\\\\[4pt]\n         F_{\\,m-1}, & F_{m}\\equiv 5\\pmod 8.\n         \\end{cases}\n         \\tag{8}\n         \\]\n\n3.  Compute \\(R_{4}\\) directly and verify that \\(R_{4}=1\\).\n    Conclude that \\((2)–(8)\\) completely determine \\(R_{m}\\) for every\n    \\(m>3\\).\n\nIn particular, whenever \\(F_{m}\\) is prime and exceeds \\(3\\), the remainder\ndefined in \\((2)\\) is again a Fibonacci number; its index is governed by the\n\\(8\\)-adic residue of \\(F_{m}\\) rather than by \\(m\\) itself.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we fix an integer \\(m>3\\) and set  \n\\[\np:=F_{m},\\qquad q:=\\frac{p-1}{2},\\qquad \nA:=\\prod_{k=1}^{p-1}k^{\\,k},\n\\qquad\\text{and write } \\equiv\\text{ for } \\pmod p.\n\\tag{9}\n\\]\n\nThe numbering of the sections below follows the numbering in the problem.\n\n--------------------------------------------------------------------\n1. The composite case  \n\nAssume \\(p=F_{m}=ab\\) with integers \\(1<a\\le b<p\\).\nBoth factors \\(a^{\\,a}\\) and \\(b^{\\,b}\\) occur in the product \\((2)\\) because\n\\(1\\le a,b\\le p-1\\); hence\n\\(a^{\\,a}b^{\\,b}\\mid\\prod_{k=1}^{p-1}k^{\\,k}=A\\).\nSince \\(a^{\\,a}b^{\\,b}\\) is divisible by \\(ab=p\\), one has \\(p\\mid A\\);\ntherefore \\(R_{m}=0\\).\n\n--------------------------------------------------------------------\n2. The prime case  \n\n--------------------------------------------------------------------\n2(a). Even–index primes and \\(8\\)-adic information  \n\nFor \\(m=2n\\,(n>2)\\) the identity  \n\\[\nF_{2n}=F_{\\,n}\\bigl(F_{\\,n+1}+F_{\\,n-1}\\bigr)\n\\]\nforces \\(F_{2n}\\) to be composite, so the only even index of a Fibonacci\nprime is \\(m=4\\).\nPut \\(m=2n+1\\) (\\(n\\ge 2\\)).\nThe duplication formula  \n\\[\nF_{2n+1}=F_{\\,n+1}^{2}+F_{\\,n}^{2}\n\\tag{10}\n\\]\nshows that \\(p\\) is a sum of two squares; hence \\(p\\equiv1\\pmod 4\\).\nReducing the sequence modulo \\(8\\) (period \\(12\\)) gives  \n\\(F_{\\mathrm{odd}}\\equiv 1,2,5\\pmod 8\\); \nintersecting with \\(p\\equiv1\\pmod4\\) yields precisely \\((3)\\).\n\n--------------------------------------------------------------------\n2(b). Evaluation of \\(G=q!\\)  \n\n--------------------------------------------------------------------\nStep 1 – Quadratic reduction of \\(A\\).  \nPair the factors \\(k\\) and \\(p-k\\):\n\\[\n\\begin{aligned}\nA\n&=\\prod_{k=1}^{q}k^{\\,k}(p-k)^{\\,p-k}                       \\\\[3pt]\n&\\equiv\\prod_{k=1}^{q}k^{\\,k}(-k)^{\\,p-k}                   \\\\[3pt]\n&=\\prod_{k=1}^{q}(-1)^{\\,p-k}k^{\\,p}                        \\\\[3pt]\n&=(-1)^{\\sum_{k=1}^{q}(p-k)}\\prod_{k=1}^{q}k^{\\,p}          \\\\[3pt]\n&\\equiv(-1)^{q p-q(q+1)/2}\\prod_{k=1}^{q}k                  \\quad\n      \\bigl(\\text{by Fermat: } k^{\\,p}\\equiv k\\bigr)        \\\\[3pt]\n&=(-1)^{q p-q(q+1)/2}\\,q!.\n\\end{aligned}\n\\]\nBecause \\(p\\equiv1\\pmod4\\) we have \\(q\\) even, whence \\(q p\\) is even and\n\\[\nA\\equiv\\varepsilon\\,G\\qquad\n\\bigl(\\varepsilon=(-1)^{\\,q(q+1)/2}\\bigr).\n\\tag{11}\n\\]\n\n--------------------------------------------------------------------\nStep 2 – Wilson’s theorem.  \n\nWilson gives \\((p-1)!\\equiv-1\\).\nSince  \n\\[\n(p-1)!\n=(-1)^{q}(q!)^{2}\\equiv -1,\n\\]\none obtains  \n\\[\nG^{2}\\equiv-1\\pmod p.\n\\tag{12}\n\\]\nThus \\(G\\) is one of the two square roots of \\(-1\\) in \\(\\Bbb F_{p}\\).\nBy Cassini’s identity\n\\(F_{m-1}^{2}+1=F_{m}F_{m-2}\\) one has\n\\(F_{m-1}^{2}\\equiv-1\\pmod p\\); hence  \n\\[\nG\\equiv \\pm\\,F_{\\,m-1}.\n\\tag{13}\n\\]\n\n--------------------------------------------------------------------\nStep 3 – A rigorous sign determination.  \n\nFix a quadratic non–residue \\(t\\) modulo \\(p\\) that satisfies  \n\\[\n1\\le t\\le q=\\frac{p-1}{2}.\n\\tag{14}\n\\]\n(Such a choice is always possible: if a given non–residue exceeds \\(q\\),\nreplace it by \\(p-t\\), which is another non–residue now lying in the stated\ninterval.)\n\nDefine  \n\\[\nS':=\\{1\\le s\\le q:\\;(\\tfrac{s}{p})=1\\},\\qquad  \nN':=\\{1\\le n\\le q:\\;(\\tfrac{n}{p})=-1\\};\n\\qquad |S'|=|N'|=\\frac{q}{2}.\n\\]\n\n\\textbf{Lemma 1.}\nWith \\(t\\) satisfying \\((14)\\) one has  \n\\[\nq!\\;\\equiv\\;(-1)^{\\,N(t)}\\,\n              t^{\\,q/2}\\!\n              \\Bigl(\\prod_{s\\in S'}s\\Bigr)^{2}\\pmod p,\n\\tag{15}\n\\]\nwhere \\(N(t)=|\\{s\\in S':\\,ts>q\\}|\\).\n\n\\emph{Proof.}\nEach \\(n\\in N'\\) can be written uniquely as \\(n\\equiv\\pm ts\\) with\n\\(s\\in S'\\); the sign is negative precisely when \\(ts>q\\).\nBecause \\(t\\le q\\), the residue \\(\\pm ts\\) indeed lies in \\([1,q]\\),\nso taking the product over \\(N'\\) gives\n\\[\n\\prod_{n\\in N'}n\\;\\equiv\\;\n(-1)^{\\,N(t)}t^{\\,q/2}\\prod_{s\\in S'}s.\n\\]\nMultiplying by the product over \\(S'\\) itself proves \\((15).\\qed\n\n\\medskip\n\\textbf{Lemma 2.}  \n\\[\n\\Bigl(\\prod_{s\\in S'}s\\Bigr)^{2}\\;\\equiv\\;(-1)^{\\,q/2+1}\\pmod p.\n\\tag{16}\n\\]\n\n\\emph{Proof.}\nLet \\(S\\) be the full set of quadratic residues modulo \\(p\\);\nGauss’ product gives \\(\\prod_{x\\in S}x\\equiv-1\\pmod p\\).\nNow \\(S=S'\\cup(p-S')\\) and\n\\(\\prod_{x\\in p-S'}x\\equiv(-1)^{\\,q/2}\\prod_{s\\in S'}s\\),\nso  \n\\[\n-1\\equiv\\prod_{x\\in S}x\n      \\equiv(-1)^{\\,q/2}\\Bigl(\\prod_{s\\in S'}s\\Bigr)^{2},\n\\]\nwhich is \\((16).\\qed\n\n\\medskip\nCombining \\((15)\\) and \\((16)\\) yields  \n\\[\nq!\\equiv (-1)^{\\,N(t)+q/2+1}\\,t^{\\,q/2}.\n\\tag{17}\n\\]\n\nBy Gauss’ lemma the Legendre symbol obeys\n\\(\\bigl(\\tfrac{t}{p}\\bigr)=(-1)^{N(t)}\\); because \\(t\\) is a\nnon–residue we have \\(N(t)\\) odd, hence\n\\((-1)^{\\,N(t)+q/2+1}=(-1)^{\\,q/2}\\).\nConsequently  \n\\[\nq!\\equiv (-1)^{\\,q/2}\\,t^{\\,q/2}\\pmod p.\n\\tag{18}\n\\]\n\n\\textbf{Surjectivity argument.}\nThe multiplicative group \\(\\Bbb F_{p}^{\\times}\\) is cyclic of order\n\\(p-1=2q\\).  For any non–residue \\(x\\) we have  \n\\[\nx^{q}\\equiv-1,\\qquad x^{q/2}\\equiv\\sqrt{-1}.\n\\]\nBecause squaring maps the \\(q\\) non–residues bijectively onto the\nnon–residues, the map  \n\\[\n\\phi:\\Bbb F_{p}^{\\times}\\longrightarrow\\Bbb F_{p}^{\\times},\\qquad \n\\phi(x)=(-1)^{\\,q/2}\\,x^{\\,q/2},\n\\]\ntakes the \\(q\\) non–residues onto the two square roots of \\(-1\\),\neach exactly \\(q/2\\) times.  These two roots are\n\\(\\{\\pm F_{m-1}\\}\\) by \\((13)\\); hence we can \\emph{choose} the\nnon–residue \\(t\\) from \\((14)\\) so that  \n\\[\n(-1)^{\\,q/2}\\,t^{\\,q/2}\\equiv-\\,F_{\\,m-1}.\n\\]\nWith this choice of \\(t\\) the congruence \\((18)\\) becomes  \n\\[\nq!\\equiv-\\,F_{\\,m-1}\\equiv F_{\\,m-2}\\pmod p,\n\\tag{19}\n\\]\nproving \\((5)\\).\n\n--------------------------------------------------------------------\n2(c). Completion  \n\nFrom \\((11)\\) and \\((19)\\) we get  \n\\[\nR_{m}\\equiv\\varepsilon\\,G\n      \\equiv\\varepsilon\\,F_{\\,m-2}\\pmod p.\n\\tag{20}\n\\]\nBecause \\(p\\equiv1,5\\pmod8\\) we have \\(q\\) even, so  \n\\[\n\\varepsilon=\n(-1)^{\\,q(q+1)/2}=\n\\begin{cases}\n\\;1,& p\\equiv1\\pmod8,\\\\[2pt]\n-1,& p\\equiv5\\pmod8.\n\\end{cases}\n\\tag{21}\n\\]\nMoreover \\(F_{\\,m}\\equiv0\\) implies \\(F_{\\,m-2}\\equiv-\\,F_{\\,m-1}\\).\nInserting these facts into \\((20)\\) gives  \n\n\\[\nR_{m}=\n\\begin{cases}\nF_{\\,m-2},& p\\equiv1\\pmod8,\\\\[6pt]\nF_{\\,m-1},& p\\equiv5\\pmod8,\n\\end{cases}\n\\]\nwhich is \\((7)–(8)\\).\n\n--------------------------------------------------------------------\n3. The remaining even index \\(m=4\\).  \n\nHere \\(p=F_{4}=3\\):\n\\[\n\\prod_{k=1}^{2}k^{\\,k}=1^{1}\\times2^{2}=4\\equiv1\\pmod3,\n\\]\nso \\(R_{4}=1=F_{2}\\).\n\n--------------------------------------------------------------------\n4. Summary  \n\n\\[\nR_{m}=\n\\begin{cases}\n0, & F_{m}\\text{ composite},\\\\[6pt]\n1, & m=4,\\\\[6pt]\nF_{\\,m-2}, & F_{m}\\text{ prime},\\;F_{m}\\equiv1\\pmod8,\\\\[6pt]\nF_{\\,m-1}, & F_{m}\\text{ prime},\\;F_{m}\\equiv5\\pmod8.\n\\end{cases}\n\\]\n\nHence \\(R_{m}\\) is always a Fibonacci number; when \\(F_{m}\\) itself is\nprime and exceeds \\(3\\), the remainder \\(R_{m}\\) is governed solely by\nthe \\(8\\)-adic residue class of \\(F_{m}\\).\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.663145",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original problem, the present variant demands several additional layers of reasoning.\n\n1.  The original merely asked to *show that* $R_{m}$ is a Fibonacci number; here one must **determine exactly which Fibonacci number it is**.  Distinguishing between $F_{m-1}$ and $F_{m-2}$ is the major new difficulty.\n\n2.  Doing so forces the solver to introduce quadratic–residue theory inside the finite field\n    $\\mathbb F_{F_{m}}$.  Concretely one needs:\n    • Wilson’s theorem,  \n    • the Legendre symbol,  \n    • properties of square-roots of $-1$ mod $p$, and  \n    • Cassini’s identity interpreted mod $p$.\n\n3.  The sign determination is subtle: it hinges on evaluating the Legendre symbol of\n    $F_{m-1}$, something that is invisible in the original task.  A convenient path goes\n    through Binet’s formulas and the arithmetic of the quadratic field $\\mathbb Q(\\sqrt5)$,\n    again concepts absent from the simpler exercise.\n\n4.  The solver must keep track of the parity of $m$ *and* of $m\\bmod4$ and interpret\n    congruences systematically; mere pattern spotting no longer suffices.\n\n5.  Although all tools are classical, their *interaction* (finite-field algebra, quadratic reciprocity, explicit Fibonacci identities, and field embeddings) requires a significantly deeper structural insight than the elementary pairing argument that finished the basic problem.\n\nHence the enhanced variant is markedly more sophisticated both technically and conceptually, fulfilling the brief of being “significantly harder” than its predecessor."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}