{
  "index": "2018-A-4",
  "type": "NT",
  "tag": [
    "NT",
    "ALG",
    "COMB"
  ],
  "difficulty": "",
  "question": "Let $m$ and $n$ be positive integers with $\\gcd(m,n) = 1$, and let\n\\[\na_k = \\left\\lfloor \\frac{mk}{n} \\right\\rfloor - \\left\\lfloor \\frac{m(k-1)}{n} \\right\\rfloor\n\\]\nfor $k=1,2,\\dots,n$.\nSuppose that $g$ and $h$ are elements in a group $G$ and that \n\\[\ngh^{a_1} gh^{a_2} \\cdots gh^{a_n} = e,\n\\]\nwhere $e$ is the identity element. Show that $gh= hg$. (As usual, $\\lfloor x \\rfloor$ denotes the greatest integer\nless than or equal to $x$.)",
  "solution": "\\noindent\n\\textbf{First solution.}\nWe prove the claim by induction on $m+n$.\nFor the base case, suppose that $n=1$; we then have $m=1$ and the given equation becomes $gh=e$. The claim then reduces to the fact that a one-sided inverse in $G$ is also a two-sided inverse. (Because $G$ is a group, $g$ has an inverse $g^{-1}$; since $gh = e$, we have $h = g^{-1}(gh) = g^{-1} e = g^{-1}$, so $hg = e = gh$.)\n\nSuppose now that $n>1$. In case $m>n$, set $\\tilde{g} = g h$, $\\tilde{h} = h$, and\n\\[\nb_k = \\left\\lfloor \\frac{(m-n)k}{n} \\right\\rfloor - \\left\\lfloor \\frac{(m-n)(k-1)}{n} \\right\\rfloor \n\\quad (k=1,\\dots,n).\n\\]\nthen\n\\[\n\\tilde{g} \\tilde{h}^{b_1} \\cdots \\tilde{g} \\tilde{h}^{b_n} = gh^{a_1} \\cdots gh^{a_n} = e,\n\\]\nso the induction hypothesis implies that $\\tilde{g}$ and $\\tilde{h}$ commute; this implies that $g$ and $h$ commute.\n\nIn case $m < n$, note that $a_k \\in \\{0,1\\}$ for all $k$. Set $\\tilde{g} = h^{-1}$, $\\tilde{h} = g^{-1}$, and\n\\[\nb_l = \\left\\lfloor \\frac{n \\ell}{m} \\right\\rfloor - \\left\\lfloor \\frac{n(\\ell-1)}{m} \\right\\rfloor \n\\quad (\\ell=1,\\dots,m);\n\\]\nwe claim that \n\\[\n\\tilde{g}\\tilde{h}^{b_1}\\cdots\\tilde{g}\\tilde{h}^{b_m} = (gh^{a_1}\\cdots gh^{a_n})^{-1} = e,\n\\]\nso the induction hypothesis implies that $\\tilde{h}$ and $\\tilde{g}$ commute; this implies that $g$ and $h$ commute.\n\nTo clarify this last equality, consider a lattice walk starting from $(0,0)$, ending at $(n,m)$, staying below the line\n$y = mx/n$, and keeping as close to this line as possible. If one follows this walk and records the element $g$ for each horizontal step and $h$ for each vertical step, one obtains the word $gh^{a_1}\\cdots gh^{a_n}$. \nNow take this walk, reflect across the line $y = x$, rotate by a half-turn, then translate to put the endpoints at $(0,0)$ and $(m,n)$; this is the analogous walk for the pair $(n,m)$.\n\n\\noindent\n\\textbf{Remark.}\nBy tracing more carefully through the argument, one sees in addition that there exists an element $k$ of $G$\nfor which $g = k^m, h = k^{-n}$.\n\n\\noindent\n\\textbf{Second solution.} (by Greg Martin)\nSince $\\gcd(m,n) = 1$, there exist integers $x,y$ such that $mx + ny = 1$; we may further assume that \n$x \\in \\{1,\\dots,n\\}$. We first establish the identity\n\\[\na_{k-x} = \\begin{cases}\na_k - 1 & \\mbox{if $k \\equiv 0 \\pmod{n}$} \\\\\na_k + 1 & \\mbox{if $k \\equiv 1 \\pmod{n}$} \\\\\na_k & \\mbox{otherwise}.\n\\end{cases}\n\\]\nNamely, by writing $-mx = ny-1$, we see that\n\\begin{align*}\na_{k-x} &= \\left\\lfloor \\frac{m(k-x)}{n} \\right\\rfloor - \\left\\lfloor \\frac{m(k-x-1)}{n} \\right\\rfloor\n\\\\\n&= \\left\\lfloor \\frac{mk+ny-1}{n} \\right\\rfloor - \\left\\lfloor \\frac{m(k-1)+ny-1}{n} \\right\\rfloor \\\\\n&= \\left\\lfloor \\frac{mk-1}{n} \\right\\rfloor - \\left\\lfloor \\frac{m(k-1)-1}{n} \\right\\rfloor\n\\end{align*}\nand so\n\\begin{align*}\na_{k-x} - a_k &= \\left( \\left\\lfloor \\frac{mk-1}{n} \\right\\rfloor - \\left\\lfloor \\frac{mk}{n} \\right\\rfloor \\right)\n\\\\\n&\\quad\n- \\left( \\left\\lfloor \\frac{m(k-1)-1}{n} \\right\\rfloor - \\left\\lfloor \\frac{m(k-1)}{n} \\right\\rfloor \\right).\n\\end{align*}\nThe first parenthesized expression equals 1 if $n$ divides $mk$, or equivalently $n$ divides $k$, and 0 otherwise.\nSimilarly, the second parenthesized expression equals 1 if $n$ divides $k-1$ and 0 otherwise. This proves the stated identity.\n\nWe now use the given relation $g h^{a_1} \\cdots g h^{a_n} = e$ to write\n\\begin{align*}\nghg^{-1}h^{-1} &= gh(h^{a_1} g h^{a_2} \\cdots gh^{a_{n-1}} g h^{a_n})h^{-1} \\\\\n&= gh^{a_1+1} gh^{a_2} \\cdots gh^{a_{n-1}} gh^{a_n-1} \\\\\n&= gh^{a_{1-x}} \\cdots gh^{a_{n-x}} \\\\\n&= (gh^{a_{n+1-x}} \\cdots gh^{a_{n}}) (gh^{a_1} \\cdots gh^{a_{n-x}}).\n\\end{align*}\nThe two parenthesized expressions multiply in the opposite order to $g h^{a_1} \\cdots g h^{a_n} = e$, so they must be\n(two-sided) inverses of each other. We deduce that $ghg^{-1} h^{-1} = e$, meaning that $g$ and $h$ commute.\n\n\\noindent\n\\textbf{Third solution.} (by Sucharit Sarkar)\nLet $T$ denote the torus $\\mathbb{R}^2/\\mathbb{Z}^2$. The line segments from $(0,0)$ to $(1,0)$ and from $(0,0)$ to $(0,1)$ are closed loops in $T$, and we denote them by $g$ and $h$ respectively. Now let $p$ be the (image of the) point $(\\epsilon,-\\epsilon)$ in $T$ for some $0<\\epsilon\\ll 1$. The punctured torus $T \\setminus \\{p\\}$ deformation retracts onto the union of the loops $g$ and $h$, and so $\\pi_1(T\\setminus\\{p\\})$, the fundamental group of $T\\setminus\\{p\\}$ based at $(0,0)$, is the free group on two generators, $\\langle g,h\\rangle$.\n\nLet $\\gamma$ and $\\tilde{\\gamma}$ denote the following loops based at $(0,0)$ in $T$: $\\gamma$ is the image of the line segment from $(0,0)$ to $(n,m)$ under the projection $\\mathbb{R}^2 \\to T$, and $\\tilde{\\gamma}$ is the image of the lattice walk from $(0,0)$ to $(n,m)$, staying just below the line $y=mx/n$, that was described in the first solution. There is a straight-line homotopy with fixed endpoints between the two paths in $\\mathbb{R}^2$ from $(0,0)$ to $(n,m)$, the line segment and the lattice walk, and this homotopy does not pass through any point of the form $(a+\\epsilon,b-\\epsilon)$ for $a,b\\in\\mathbb{Z}$ by the construction of the lattice walk. It follows that $\\gamma$ and $\\tilde{\\gamma}$ are homotopic loops in $T \\setminus \\{p\\}$. Since the class of $\\tilde{\\gamma}$ in $\\pi_1(T\\setminus\\{p\\})$ is evidently $gh^{a_1}gh^{a_2}\\cdots gh^{a_n}$, it follows that the class of $\\gamma$ in $\\pi_1(T\\setminus\\{p\\})$ is the same.\n\nNow since $\\gcd(m,n)=1$, there is an element $\\phi \\in GL_2(\\mathbb{Z})$ sending $(n,m)$ to $(1,0)$, which then sends the line segment from $(0,0)$ to $(n,m)$ to the segment from $(0,0)$ to $(1,0)$. Then $\\phi$ induces a homeomorphism of $T$ sending $\\gamma$ to $g$, which in turn induces an isomorphism $\\phi_* :\\thinspace \\pi_1(T\\setminus \\{p\\}) \\to \\pi_1(T\\setminus \\{\\phi^{-1}(p)\\})$. Both fundamental groups are equal to $\\langle g,h\\rangle$, and we conclude that $\\phi_*$ sends $gh^{a_1}gh^{a_2}\\cdots gh^{a_n}$ to $g$. It follows that $\\phi_*$ induces an isomorphism\n\\[\n\\langle g,h\\,|\\,gh^{a_1}gh^{a_2}\\cdots gh^{a_n} \\rangle \\to \\langle g,h \\,|\\, g\\rangle \\cong \\langle h\\rangle \\cong \\mathbb{Z}.\n\\]\n\nSince $\\mathbb{Z}$ is abelian, $g$ and $h$ must commute in $\\langle g,h\\,|\\,gh^{a_1}gh^{a_2}\\cdots gh^{a_n}\\rangle$, whence they must also commute in $G$.",
  "vars": [
    "k",
    "a_k",
    "b_k",
    "x",
    "y",
    "\\\\ell"
  ],
  "params": [
    "m",
    "n",
    "g",
    "h",
    "e",
    "G",
    "\\\\tilde{g}",
    "\\\\tilde{h}",
    "T",
    "p",
    "\\\\gamma",
    "\\\\tilde{\\\\gamma}",
    "\\\\phi",
    "\\\\pi_1",
    "Z"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "k": "indexvar",
        "a_k": "stepvalue",
        "b_k": "altvalue",
        "x": "bezoutx",
        "y": "bezouty",
        "\\ell": "ellindex",
        "m": "firstint",
        "n": "secondint",
        "g": "firstelem",
        "h": "secondelem",
        "e": "identity",
        "G": "wholegroup",
        "\\tilde{g}": "tildafirst",
        "\\tilde{h}": "tildasecond",
        "T": "torusset",
        "p": "puncture",
        "\\gamma": "gammapath",
        "\\tilde{\\gamma}": "latticepath",
        "\\phi": "transfmap",
        "\\pi_1": "fundagroup",
        "Z": "integerset"
      },
      "question": "Let $firstint$ and $secondint$ be positive integers with $\\gcd(firstint,secondint)=1$, and let\\n\\[\\nstepvalue_{indexvar}=\\left\\lfloor\\frac{firstint\\,indexvar}{secondint}\\right\\rfloor-\\left\\lfloor\\frac{firstint(indexvar-1)}{secondint}\\right\\rfloor\\n\\]\\nfor $indexvar=1,2,\\dots ,secondint$.\\nSuppose that $firstelem$ and $secondelem$ are elements in a group $wholegroup$ and that\\n\\[\\nfirstelem\\,secondelem^{stepvalue_{1}}\\,firstelem\\,secondelem^{stepvalue_{2}}\\cdots firstelem\\,secondelem^{stepvalue_{secondint}}=identity,\\n\\]\\nwhere $identity$ is the identity element. Show that $firstelem\\,secondelem=secondelem\\,firstelem$. (As usual, $\\lfloor x\\rfloor$ denotes the greatest integer less than or equal to $x$.)",
      "solution": "\\noindent\\textbf{First solution.}\\nWe prove the claim by induction on $firstint+secondint$.\\nFor the base case, suppose that $secondint=1$; we then have $firstint=1$ and the given equation becomes $firstelem\\,secondelem=identity$. Since a one--sided inverse in a group is also a two--sided inverse, $firstelem$ and $secondelem$ commute.\\n\\nAssume now that $secondint>1$.\\n\\n\\emph{Case 1: $firstint>secondint$.}\\; Put $tildafirst=firstelem\\,secondelem$, $tildasecond=secondelem$, and\\n\\[\\naltvalue_{indexvar}=\\left\\lfloor\\frac{(firstint-secondint)\\,indexvar}{secondint}\\right\\rfloor-\\left\\lfloor\\frac{(firstint-secondint)(indexvar-1)}{secondint}\\right\\rfloor,\\qquad(indexvar=1,\\dots ,secondint).\\n\\]\\nThen\\n\\[\\ntildafirst\\,tildasecond^{altvalue_{1}}\\cdots tildafirst\\,tildasecond^{altvalue_{secondint}}=firstelem\\,secondelem^{stepvalue_{1}}\\cdots firstelem\\,secondelem^{stepvalue_{secondint}}=identity,\\n\\]\\nand the induction hypothesis implies that $tildafirst$ and $tildasecond$ commute. Hence $firstelem$ and $secondelem$ commute.\\n\\n\\emph{Case 2: $firstint<secondint$.}\\; Now $stepvalue_{indexvar}\\in\\{0,1\\}$. Put $tildafirst=secondelem^{-1}$, $tildasecond=firstelem^{-1}$, and\\n\\[\\naltvalue_{ellindex}=\\left\\lfloor\\frac{secondint\\,ellindex}{firstint}\\right\\rfloor-\\left\\lfloor\\frac{secondint(ellindex-1)}{firstint}\\right\\rfloor,\\qquad(ellindex=1,\\dots ,firstint).\\n\\]\\nOne checks that\\n\\[\\ntildafirst\\,tildasecond^{altvalue_{1}}\\cdots tildafirst\\,tildasecond^{altvalue_{firstint}}=(firstelem\\,secondelem^{stepvalue_{1}}\\cdots firstelem\\,secondelem^{stepvalue_{secondint}})^{-1}=identity,\\n\\]\\nso $tildafirst$ and $tildasecond$ commute, and again $firstelem$ and $secondelem$ commute.\\n\\n\\textbf{Remark.}\\;A more careful inspection yields an element $indexvar\\in wholegroup$ with $firstelem=indexvar^{firstint}$ and $secondelem=indexvar^{-secondint}$.\\n\\n\\noindent\\textbf{Second solution.}\\;(Greg Martin)\\nSince $\\gcd(firstint,secondint)=1$, integers $bezoutx,bezouty$ satisfy $firstint\\,bezoutx+secondint\\,bezouty=1$ with $1\\le bezoutx\\le secondint$. One shows that\\n\\[\\nstepvalue_{indexvar-bezoutx}=\\begin{cases}stepvalue_{indexvar}-1,&indexvar\\equiv0\\pmod{secondint},\\\\stepvalue_{indexvar}+1,&indexvar\\equiv1\\pmod{secondint},\\\\stepvalue_{indexvar},&\\text{otherwise.}\\end{cases}\\n\\]\\nUsing the relation $firstelem\\,secondelem^{stepvalue_{1}}\\cdots firstelem\\,secondelem^{stepvalue_{secondint}}=identity$ and the above identity, one obtains\\n\\[\\nfirstelem\\,secondelem\\,firstelem^{-1}\\,secondelem^{-1}=identity,\\n\\]\\nso $firstelem$ and $secondelem$ commute.\\n\\n\\noindent\\textbf{Third solution.}\\;(Sucharit Sarkar)\\nLet $torusset=\\mathbb{R}^{2}/\\mathbb{Z}^{2}$; denote by $firstelem$ and $secondelem$ the basic longitude and meridian loops.  For a small puncture $puncture$, $fundagroup(torusset\\setminus\\{puncture\\})$ is the free group $\\langle firstelem,secondelem\\rangle$.  A lattice path $latticepath$ from $(0,0)$ to $(secondint,firstint)$ encodes the word $firstelem\\,secondelem^{stepvalue_{1}}\\cdots firstelem\\,secondelem^{stepvalue_{secondint}}$.  This path is homotopic in $torusset\\setminus\\{puncture\\}$ to the straight segment $gammapath$ from $(0,0)$ to $(secondint,firstint)$.  An element $transfmap\\in GL_{2}(integerset)$ sends $(secondint,firstint)$ to $(1,0)$, inducing an isomorphism that carries the given word to $firstelem$.  Hence\\n\\[\\n\\langle firstelem,secondelem\\mid firstelem\\,secondelem^{stepvalue_{1}}\\cdots firstelem\\,secondelem^{stepvalue_{secondint}}\\rangle\\cong \\langle firstelem,secondelem\\mid firstelem\\rangle\\cong\\langle secondelem\\rangle\\cong integerset,\\n\\]\\nwhich is abelian; therefore $firstelem$ and $secondelem$ commute in $wholegroup$.\\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "k": "kingfisher",
        "a_k": "archipelago",
        "b_k": "blueberry",
        "x": "xylophone",
        "y": "yellowtail",
        "\\\\ell": "lamplight",
        "m": "marigold",
        "n": "nightshade",
        "g": "goldfish",
        "h": "honeycomb",
        "e": "evergreen",
        "G": "granary",
        "\\\\tilde{g}": "\\\\tilde{geography}",
        "\\\\tilde{h}": "\\\\tilde{heliport}",
        "T": "teaspoon",
        "p": "parchment",
        "\\\\gamma": "gargoyle",
        "\\\\tilde{\\\\gamma}": "\\\\tilde{grandstand}",
        "\\\\phi": "philosophy",
        "\\\\pi_1": "pineappleone",
        "Z": "zigzagging"
      },
      "question": "Let $marigold$ and $nightshade$ be positive integers with $\\gcd(marigold,nightshade) = 1$, and let\n\\[\narchipelago = \\left\\lfloor \\frac{marigold\\,kingfisher}{nightshade} \\right\\rfloor - \\left\\lfloor \\frac{marigold(kingfisher-1)}{nightshade} \\right\\rfloor\n\\]\nfor $kingfisher=1,2,\\dots,nightshade$.\nSuppose that $goldfish$ and $honeycomb$ are elements in a group $granary$ and that \n\\[\ngoldfish\\,honeycomb^{archipelago_1}\\,goldfish\\,honeycomb^{archipelago_2}\\,\\cdots\\,goldfish\\,honeycomb^{archipelago_{nightshade}} = evergreen,\n\\]\nwhere $evergreen$ is the identity element. Show that $goldfish\\,honeycomb = honeycomb\\,goldfish$. (As usual, $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to $x$.)",
      "solution": "\\noindent\n\\textbf{First solution.}\nWe prove the claim by induction on $marigold+nightshade$.\nFor the base case, suppose that $nightshade=1$; we then have $marigold=1$ and the given equation becomes $goldfish\\,honeycomb=evergreen$. The claim then reduces to the fact that a one-sided inverse in $granary$ is also a two-sided inverse. (Because $granary$ is a group, $goldfish$ has an inverse $goldfish^{-1}$; since $goldfish\\,honeycomb = evergreen$, we have $honeycomb = goldfish^{-1}(goldfish\\,honeycomb) = goldfish^{-1} evergreen = goldfish^{-1}$, so $honeycomb\\,goldfish = evergreen = goldfish\\,honeycomb$.)\n\nSuppose now that $nightshade>1$. In case $marigold>nightshade$, set $\\tilde{geography} = goldfish\\,honeycomb$, $\\tilde{heliport} = honeycomb$, and\n\\[\nblueberry = \\left\\lfloor \\frac{(marigold-nightshade)\\,kingfisher}{nightshade} \\right\\rfloor - \\left\\lfloor \\frac{(marigold-nightshade)(kingfisher-1)}{nightshade} \\right\\rfloor \n\\quad (kingfisher=1,\\dots,nightshade).\n\\]\nthen\n\\[\n\\tilde{geography}\\,\\tilde{heliport}^{blueberry_1} \\cdots \\tilde{geography}\\,\\tilde{heliport}^{blueberry_{nightshade}} = goldfish\\,honeycomb^{archipelago_1} \\cdots goldfish\\,honeycomb^{archipelago_{nightshade}} = evergreen,\n\\]\nso the induction hypothesis implies that $\\tilde{geography}$ and $\\tilde{heliport}$ commute; this implies that $goldfish$ and $honeycomb$ commute.\n\nIn case $marigold < nightshade$, note that $archipelago \\in \\{0,1\\}$ for all $kingfisher$. Set $\\tilde{geography} = honeycomb^{-1}$, $\\tilde{heliport} = goldfish^{-1}$, and\n\\[\nblueberry_{lamplight} = \\left\\lfloor \\frac{nightshade\\,lamplight}{marigold} \\right\\rfloor - \\left\\lfloor \\frac{nightshade(lamplight-1)}{marigold} \\right\\rfloor \n\\quad (lamplight=1,\\dots,marigold);\n\\]\nwe claim that \n\\[\n\\tilde{geography}\\tilde{heliport}^{blueberry_1}\\cdots\\tilde{geography}\\tilde{heliport}^{blueberry_{marigold}} = (goldfish\\,honeycomb^{archipelago_1}\\cdots goldfish\\,honeycomb^{archipelago_{nightshade}})^{-1} = evergreen,\n\\]\nso the induction hypothesis implies that $\\tilde{heliport}$ and $\\tilde{geography}$ commute; this implies that $goldfish$ and $honeycomb$ commute.\n\nTo clarify this last equality, consider a lattice walk starting from $(0,0)$, ending at $(nightshade,marigold)$, staying below the line\n$y = marigold x/ nightshade$, and keeping as close to this line as possible. If one follows this walk and records the element $goldfish$ for each horizontal step and $honeycomb$ for each vertical step, one obtains the word $goldfish\\,honeycomb^{archipelago_1}\\cdots goldfish\\,honeycomb^{archipelago_{nightshade}}$. \nNow take this walk, reflect across the line $y = x$, rotate by a half-turn, then translate to put the endpoints at $(0,0)$ and $(marigold,nightshade)$; this is the analogous walk for the pair $(nightshade,marigold)$.\n\n\\noindent\n\\textbf{Remark.}\nBy tracing more carefully through the argument, one sees in addition that there exists an element $kingfisher$ of $granary$\nfor which $goldfish = kingfisher^{marigold},\\; honeycomb = kingfisher^{-nightshade}$.\n\n\\noindent\n\\textbf{Second solution.} (by Greg Martin)\nSince $\\gcd(marigold,nightshade) = 1$, there exist integers $xylophone,yellowtail$ such that $marigold xylophone + nightshade yellowtail = 1$; we may further assume that \n$xylophone \\in \\{1,\\dots,nightshade\\}$. We first establish the identity\n\\[\narchipelago_{kingfisher-xylophone} = \\begin{cases}\narchipelago_{kingfisher} - 1 & \\mbox{if $kingfisher \\equiv 0 \\pmod{nightshade}$} \\\\\narchipelago_{kingfisher} + 1 & \\mbox{if $kingfisher \\equiv 1 \\pmod{nightshade}$} \\\\\narchipelago_{kingfisher} & \\mbox{otherwise}.\n\\end{cases}\n\\]\nNamely, by writing $-marigold xylophone = nightshade yellowtail-1$, we see that\n\\begin{align*}\narchipelago_{kingfisher-xylophone} &= \\left\\lfloor \\frac{marigold(kingfisher-xylophone)}{nightshade} \\right\\rfloor - \\left\\lfloor \\frac{marigold(kingfisher-xylophone-1)}{nightshade} \\right\\rfloor\\\\\n&= \\left\\lfloor \\frac{marigold kingfisher+nightshade yellowtail-1}{nightshade} \\right\\rfloor - \\left\\lfloor \\frac{marigold(kingfisher-1)+nightshade yellowtail-1}{nightshade} \\right\\rfloor \\\\\n&= \\left\\lfloor \\frac{marigold kingfisher-1}{nightshade} \\right\\rfloor - \\left\\lfloor \\frac{marigold(kingfisher-1)-1}{nightshade} \\right\\rfloor\n\\end{align*}\nand so\n\\begin{align*}\narchipelago_{kingfisher-xylophone} - archipelago_{kingfisher} &= \\left( \\left\\lfloor \\frac{marigold kingfisher-1}{nightshade} \\right\\rfloor - \\left\\lfloor \\frac{marigold kingfisher}{nightshade} \\right\\rfloor \\right)\\\\\n&\\quad- \\left( \\left\\lfloor \\frac{marigold(kingfisher-1)-1}{nightshade} \\right\\rfloor - \\left\\lfloor \\frac{marigold(kingfisher-1)}{nightshade} \\right\\rfloor \\right).\n\\end{align*}\nThe first parenthesized expression equals $1$ if $nightshade$ divides $marigold kingfisher$, or equivalently $nightshade$ divides $kingfisher$, and $0$ otherwise.\nSimilarly, the second parenthesized expression equals $1$ if $nightshade$ divides $kingfisher-1$ and $0$ otherwise. This proves the stated identity.\n\nWe now use the given relation $goldfish\\,honeycomb^{archipelago_1} \\cdots goldfish\\,honeycomb^{archipelago_{nightshade}} = evergreen$ to write\n\\begin{align*}\n goldfish\\,honeycomb\\,goldfish^{-1}honeycomb^{-1} &= goldfish\\,honeycomb(honeycomb^{archipelago_1} goldfish\\,honeycomb^{archipelago_2} \\cdots goldfish\\,honeycomb^{archipelago_{nightshade-1}} goldfish\\,honeycomb^{archipelago_{nightshade}})honeycomb^{-1}\\\\\n &= goldfish\\,honeycomb^{archipelago_1+1} goldfish\\,honeycomb^{archipelago_2} \\cdots goldfish\\,honeycomb^{archipelago_{nightshade-1}} goldfish\\,honeycomb^{archipelago_{nightshade}-1}\\\\\n &= goldfish\\,honeycomb^{archipelago_{1-xylophone}} \\cdots goldfish\\,honeycomb^{archipelago_{nightshade-xylophone}}\\\\\n &= (goldfish\\,honeycomb^{archipelago_{nightshade+1-xylophone}} \\cdots goldfish\\,honeycomb^{archipelago_{nightshade}})\n   (goldfish\\,honeycomb^{archipelago_1} \\cdots goldfish\\,honeycomb^{archipelago_{nightshade-xylophone}}).\n\\end{align*}\nThe two parenthesized expressions multiply in the opposite order to $goldfish\\,honeycomb^{archipelago_1} \\cdots goldfish\\,honeycomb^{archipelago_{nightshade}} = evergreen$, so they must be\na pair of inverses. We deduce that $goldfish\\,honeycomb\\,goldfish^{-1}honeycomb^{-1} = evergreen$, meaning that $goldfish$ and $honeycomb$ commute.\n\n\\noindent\n\\textbf{Third solution.} (by Sucharit Sarkar)\nLet $teaspoon$ denote the torus $\\mathbb{R}^2/\\mathbb{Z}^2$. The line segments from $(0,0)$ to $(1,0)$ and from $(0,0)$ to $(0,1)$ are closed loops in $teaspoon$, and we denote them by $goldfish$ and $honeycomb$ respectively. Now let $parchment$ be the (image of the) point $(\\epsilon,-\\epsilon)$ in $teaspoon$ for some $0<\\epsilon\\ll 1$. The punctured torus $teaspoon \\setminus \\{parchment\\}$ deformation retracts onto the union of the loops $goldfish$ and $honeycomb$, and so $pineappleone(teaspoon\\setminus\\{parchment\\})$, the fundamental group of $teaspoon\\setminus\\{parchment\\}$ based at $(0,0)$, is the free group on two generators, $\\langle goldfish,honeycomb\\rangle$.\n\nLet $gargoyle$ and $\\tilde{grandstand}$ denote the following loops based at $(0,0)$ in $teaspoon$: $gargoyle$ is the image of the line segment from $(0,0)$ to $(nightshade,marigold)$ under the projection $\\mathbb{R}^2 \\to teaspoon$, and $\\tilde{grandstand}$ is the image of the lattice walk from $(0,0)$ to $(nightshade,marigold)$, staying just below the line $y=marigold x/nightshade$, that was described in the first solution. There is a straight-line homotopy with fixed endpoints between the two paths in $\\mathbb{R}^2$ from $(0,0)$ to $(nightshade,marigold)$, the line segment and the lattice walk, and this homotopy does not pass through any point of the form $(a+\\epsilon,b-\\epsilon)$ for $a,b\\in\\mathbb{Z}$ by the construction of the lattice walk. It follows that $gargoyle$ and $\\tilde{grandstand}$ are homotopic loops in $teaspoon \\setminus \\{parchment\\}$. Since the class of $\\tilde{grandstand}$ in $pineappleone(teaspoon\\setminus\\{parchment\\})$ is evidently $goldfish\\,honeycomb^{archipelago_1}goldfish\\,honeycomb^{archipelago_2}\\cdots goldfish\\,honeycomb^{archipelago_{nightshade}}$, it follows that the class of $gargoyle$ in $pineappleone(teaspoon\\setminus\\{parchment\\})$ is the same.\n\nNow since $\\gcd(marigold,nightshade)=1$, there is an element $philosophy \\in GL_2(\\mathbb{Z})$ sending $(nightshade,marigold)$ to $(1,0)$, which then sends the line segment from $(0,0)$ to $(nightshade,marigold)$ to the segment from $(0,0)$ to $(1,0)$. Then $philosophy$ induces a homeomorphism of $teaspoon$ sending $gargoyle$ to $goldfish$, which in turn induces an isomorphism $philosophy_* :\\thinspace pineappleone(teaspoon\\setminus \\{parchment\\}) \\to pineappleone(teaspoon\\setminus \\{philosophy^{-1}(parchment)\\})$. Both fundamental groups are equal to $\\langle goldfish,honeycomb\\rangle$, and we conclude that $philosophy_*$ sends $goldfish\\,honeycomb^{archipelago_1}goldfish\\,honeycomb^{archipelago_2}\\cdots goldfish\\,honeycomb^{archipelago_{nightshade}}$ to $goldfish$. It follows that $philosophy_*$ induces an isomorphism\n\\[\n\\langle goldfish,honeycomb\\,|\\,goldfish\\,honeycomb^{archipelago_1}goldfish\\,honeycomb^{archipelago_2}\\cdots goldfish\\,honeycomb^{archipelago_{nightshade}} \\rangle \\to \\langle goldfish,honeycomb \\,|\\, goldfish\\rangle \\cong \\langle honeycomb\\rangle \\cong \\mathbb{zigzagging}.\n\\]\n\nSince $\\mathbb{zigzagging}$ is abelian, $goldfish$ and $honeycomb$ must commute in $\\langle goldfish,honeycomb\\,|\\,goldfish\\,honeycomb^{archipelago_1}goldfish\\,honeycomb^{archipelago_2}\\cdots goldfish\\,honeycomb^{archipelago_{nightshade}}\\rangle$, whence they must also commute in $granary$. }",
      "confidence": "0.06"
    },
    "descriptive_long_misleading": {
      "map": {
        "k": "constantval",
        "a_k": "infinitejump",
        "b_k": "staticpause",
        "x": "definite",
        "y": "surefire",
        "\\\\ell": "universal",
        "m": "negative",
        "n": "nonpositive",
        "g": "nongroup",
        "h": "stationary",
        "e": "nonidentity",
        "G": "ungrouped",
        "\\\\tilde{g}": "alterego",
        "\\\\tilde{h}": "counterego",
        "T": "nontorus",
        "p": "flatplane",
        "\\\\gamma": "straightline",
        "\\\\tilde{\\\\gamma}": "crookedpath",
        "\\\\phi": "disorder",
        "\\\\pi_1": "surfacezero",
        "Z": "irrational"
      },
      "question": "Let $negative$ and $nonpositive$ be positive integers with $\\gcd(negative,nonpositive) = 1$, and let\n\\[\ninfinitejump = \\left\\lfloor \\frac{negative constantval}{nonpositive} \\right\\rfloor - \\left\\lfloor \\frac{negative(constantval-1)}{nonpositive} \\right\\rfloor\n\\]\nfor $constantval=1,2,\\dots,nonpositive$.\nSuppose that $nongroup$ and $stationary$ are elements in a group $ungrouped$ and that\n\\[\nnongroup \\;stationary^{a_1}\\; nongroup \\;stationary^{a_2} \\cdots nongroup \\;stationary^{a_n} = nonidentity,\n\\]\nwhere $nonidentity$ is the identity element. Show that $nongroupstationary = stationarynongroup$. (As usual, $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to $x$.)",
      "solution": "\\noindent\n\\textbf{First solution.}\nWe prove the claim by induction on $negative+nonpositive$.\nFor the base case, suppose that $nonpositive=1$; we then have $negative=1$ and the given equation becomes $nongroupstationary=nonidentity$. The claim then reduces to the fact that a one-sided inverse in $ungrouped$ is also a two-sided inverse. (Because $ungrouped$ is a group, $nongroup$ has an inverse $nongroup^{-1}$; since $nongroupstationary = nonidentity$, we have $stationary = nongroup^{-1}(nongroupstationary) = nongroup^{-1} nonidentity = nongroup^{-1}$, so $stationarynongroup = nonidentity = nongroupstationary$.)\n\nSuppose now that $nonpositive>1$. In case $negative>nonpositive$, set $alterego = nongroup station ary$, $counterego = stationary$, and\n\\[\nstaticpause = \\left\\lfloor \\frac{(negative-nonpositive)constantval}{nonpositive} \\right\\rfloor - \\left\\lfloor \\frac{(negative-nonpositive)(constantval-1)}{nonpositive} \\right\\rfloor \\quad (constantval=1,\\dots,nonpositive).\n\\]\nthen\n\\[\nalterego \\, counterego^{staticpause_1} \\cdots alterego \\, counterego^{staticpause_{nonpositive}} = nongroupstationary^{a_1} \\cdots nongroupstationary^{a_n} = nonidentity,\n\\]\nso the induction hypothesis implies that $alterego$ and $counterego$ commute; this implies that $nongroup$ and $stationary$ commute.\n\nIn case $negative < nonpositive$, note that $infinitejump \\in \\{0,1\\}$ for all $constantval$. Set $alterego = stationary^{-1}$, $counterego = nongroup^{-1}$, and\n\\[\nstaticpause_{universal} = \\left\\lfloor \\frac{nonpositive \\, universal}{negative} \\right\\rfloor - \\left\\lfloor \\frac{nonpositive(universal-1)}{negative} \\right\\rfloor \\quad (universal=1,\\dots,negative);\n\\]\nwe claim that\n\\[\nalterego\\,counterego^{staticpause_1}\\cdots alterego\\,counterego^{staticpause_{negative}} = (nongroupstationary^{a_1}\\cdots nongroupstationary^{a_n})^{-1} = nonidentity,\n\\]\nso the induction hypothesis implies that $counterego$ and $alterego$ commute; this implies that $nongroup$ and $stationary$ commute.\n\nTo clarify this last equality, consider a lattice walk starting from $(0,0)$, ending at $(nonpositive,negative)$, staying below the line $y = negative x/nonpositive$, and keeping as close to this line as possible. If one follows this walk and records the element $nongroup$ for each horizontal step and $stationary$ for each vertical step, one obtains the word $nongroupstationary^{a_1}\\cdots nongroupstationary^{a_n}$. Now take this walk, reflect across the line $y = x$, rotate by a half-turn, then translate to put the endpoints at $(0,0)$ and $(negative,nonpositive)$; this is the analogous walk for the pair $(nonpositive,negative)$.\n\n\\noindent\n\\textbf{Remark.}\nBy tracing more carefully through the argument, one sees in addition that there exists an element $constantval$ of $ungrouped$ for which $nongroup = constantval^{negative},\\; stationary = constantval^{-nonpositive}$.\n\n\\noindent\n\\textbf{Second solution.} (by Greg Martin)\nSince $\\gcd(negative,nonpositive) = 1$, there exist integers $definite,surefire$ such that $negativedefinite + nonpositivesurefire = 1$; we may further assume that $definite \\in \\{1,\\dots,nonpositive\\}$. We first establish the identity\n\\[\ninfinitejump_{constantval-definite} = \\begin{cases}\ninfinitejump_{constantval} - 1 & \\mbox{if $constantval \\equiv 0 \\pmod{nonpositive}$} \\\\\ninfinitejump_{constantval} + 1 & \\mbox{if $constantval \\equiv 1 \\pmod{nonpositive}$} \\\\\ninfinitejump_{constantval} & \\mbox{otherwise}.\n\\end{cases}\n\\]\nNamely, by writing $-negativedefinite = nonpositivesurefire-1$, we see that\n\\begin{align*}\ninfinitejump_{constantval-definite} &= \\left\\lfloor \\frac{negative(constantval-definite)}{nonpositive} \\right\\rfloor - \\left\\lfloor \\frac{negative(constantval-definite-1)}{nonpositive} \\right\\rfloor \\\\\n&= \\left\\lfloor \\frac{negative constantval + nonpositivesurefire - 1}{nonpositive} \\right\\rfloor - \\left\\lfloor \\frac{negative(constantval-1)+nonpositivesurefire-1}{nonpositive} \\right\\rfloor \\\\\n&= \\left\\lfloor \\frac{negative constantval - 1}{nonpositive} \\right\\rfloor - \\left\\lfloor \\frac{negative(constantval-1)-1}{nonpositive} \\right\\rfloor\n\\end{align*}\nand so\n\\begin{align*}\ninfinitejump_{constantval-definite} - infinitejump_{constantval} &= \\left( \\left\\lfloor \\frac{negative constantval - 1}{nonpositive} \\right\\rfloor - \\left\\lfloor \\frac{negative constantval}{nonpositive} \\right\\rfloor \\right) \\\\\n&\\quad - \\left( \\left\\lfloor \\frac{negative(constantval-1)-1}{nonpositive} \\right\\rfloor - \\left\\lfloor \\frac{negative(constantval-1)}{nonpositive} \\right\\rfloor \\right).\n\\end{align*}\nThe first parenthesized expression equals 1 if $nonpositive$ divides $negative constantval$, or equivalently $nonpositive$ divides $constantval$, and 0 otherwise. Similarly, the second parenthesized expression equals 1 if $nonpositive$ divides $constantval-1$ and 0 otherwise. This proves the stated identity.\n\nWe now use the given relation $nongroupstationary^{a_1} \\cdots nongroupstationary^{a_n} = nonidentity$ to write\n\\begin{align*}\nnongroupstationarystationary^{-1} &= nongroupstationary (stationary^{a_1} nongroup stationarystationary^{a_2} \\cdots nongroup stationarystationary^{a_{n-1}} nongroup stationarystationary^{a_n})stationary^{-1} \\\\\n&= nongroupstationary^{a_1+1} nongroupstationary^{a_2} \\cdots nongroupstationary^{a_{n-1}} nongroupstationary^{a_n-1} \\\\\n&= nongroupstationary^{a_{1-definite}} \\cdots nongroupstationary^{a_{n-definite}} \\\\\n&= (nongroupstationary^{a_{n+1-definite}} \\cdots nongroupstationary^{a_{n}}) (nongroupstationary^{a_1} \\cdots nongroupstationary^{a_{n-definite}}).\n\\end{align*}\nThe two parenthesized expressions multiply in the opposite order to $nongroupstationary^{a_1} \\cdots nongroupstationary^{a_n} = nonidentity$, so they must be (two-sided) inverses of each other. We deduce that $nongroupstationarystationary^{-1} = nonidentity$, meaning that $nongroup$ and $stationary$ commute.\n\n\\noindent\n\\textbf{Third solution.} (by Sucharit Sarkar)\nLet $nontorus$ denote the torus $\\mathbb{R}^2/\\mathbb{irrational}^2$. The line segments from $(0,0)$ to $(1,0)$ and from $(0,0)$ to $(0,1)$ are closed loops in $nontorus$, and we denote them by $nongroup$ and $stationary$ respectively. Now let $flatplane$ be the (image of the) point $(\\epsilon,-\\epsilon)$ in $nontorus$ for some $0<\\epsilon\\ll 1$. The punctured torus $nontorus \\setminus \\{flatplane\\}$ deformation retracts onto the union of the loops $nongroup$ and $stationary$, and so $surfacezero(nontorus\\setminus\\{flatplane\\})$, the fundamental group of $nontorus\\setminus\\{flatplane\\}$ based at $(0,0)$, is the free group on two generators, $\\langle nongroup,stationary\\rangle$.\n\nLet $straightline$ and $crookedpath$ denote the following loops based at $(0,0)$ in $nontorus$: $straightline$ is the image of the line segment from $(0,0)$ to $(nonpositive,negative)$ under the projection $\\mathbb{R}^2 \\to nontorus$, and $crookedpath$ is the image of the lattice walk from $(0,0)$ to $(nonpositive,negative)$, staying just below the line $y=negative x/nonpositive$, that was described in the first solution. There is a straight-line homotopy with fixed endpoints between the two paths in $\\mathbb{R}^2$ from $(0,0)$ to $(nonpositive,negative)$, the line segment and the lattice walk, and this homotopy does not pass through any point of the form $(a+\\epsilon,b-\\epsilon)$ for $a,b\\in\\mathbb{irrational}$ by the construction of the lattice walk. It follows that $straightline$ and $crookedpath$ are homotopic loops in $nontorus \\setminus \\{flatplane\\}$. Since the class of $crookedpath$ in $surfacezero(nontorus\\setminus\\{flatplane\\})$ is evidently $nongroupstationary^{a_1}nongroupstationary^{a_2}\\cdots nongroupstationary^{a_n}$, it follows that the class of $straightline$ in $surfacezero(nontorus\\setminus\\{flatplane\\})$ is the same.\n\nNow since $\\gcd(negative,nonpositive)=1$, there is an element $disorder \\in GL_2(\\mathbb{irrational})$ sending $(nonpositive,negative)$ to $(1,0)$, which then sends the line segment from $(0,0)$ to $(nonpositive,negative)$ to the segment from $(0,0)$ to $(1,0)$. Then $disorder$ induces a homeomorphism of $nontorus$ sending $straightline$ to $nongroup$, which in turn induces an isomorphism $disorder_* :\\thinspace surfacezero(nontorus\\setminus \\{flatplane\\}) \\to surfacezero(nontorus\\setminus \\{disorder^{-1}(flatplane)\\})$. Both fundamental groups are equal to $\\langle nongroup,stationary\\rangle$, and we conclude that $disorder_*$ sends $nongroupstationary^{a_1}nongroupstationary^{a_2}\\cdots nongroupstationary^{a_n}$ to $nongroup$. It follows that $disorder_*$ induces an isomorphism\n\\[\n\\langle nongroup,stationary\\,|\\,nongroupstationary^{a_1}nongroupstationary^{a_2}\\cdots nongroupstationary^{a_n} \\rangle \\to \\langle nongroup,stationary \\,|\\, nongroup\\rangle \\cong \\langle stationary\\rangle \\cong \\mathbb{irrational}.\n\\]\n\nSince $\\mathbb{irrational}$ is abelian, $nongroup$ and $stationary$ must commute in $\\langle nongroup,stationary\\,|\\,nongroupstationary^{a_1}nongroupstationary^{a_2}\\cdots nongroupstationary^{a_n}\\rangle$, whence they must also commute in $ungrouped$.",
      "status": "complete"
    },
    "garbled_string": {
      "map": {
        "m": "qwertyui",
        "n": "asdfghjk",
        "k": "zxcvbnmq",
        "a_k": "polikujm",
        "b_k": "mnjbhgvr",
        "x": "lkjhgfds",
        "y": "poiuytre",
        "\\ell": "qazwsxed",
        "g": "mwsxvzlk",
        "h": "nhytrewq",
        "G": "jklmnbvf",
        "\\tilde{g}": "\\tilde{zpolmnij}",
        "\\tilde{h}": "\\tilde{xswqazpl}",
        "T": "bvcxzlkj",
        "p": "asplkjmn"
      },
      "question": "Let $qwertyui$ and $asdfghjk$ be positive integers with $\\gcd(qwertyui,asdfghjk) = 1$, and let\n\\[\npolikujm = \\left\\lfloor \\frac{qwertyui zxcvbnmq}{asdfghjk} \\right\\rfloor - \\left\\lfloor \\frac{qwertyui(zxcvbnmq-1)}{asdfghjk} \\right\\rfloor\n\\]\nfor $zxcvbnmq=1,2,\\dots,asdfghjk$.\nSuppose that $mwsxvzlk$ and $nhytrewq$ are elements in a group $jklmnbvf$ and that \n\\[\nmwsxvzlk nhytrewq^{polikujm_1} mwsxvzlk nhytrewq^{polikujm_2} \\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk}} = e,\n\\]\nwhere $e$ is the identity element. Show that $mwsxvzlk nhytrewq= nhytrewq mwsxvzlk$. (As usual, $\\lfloor x \\rfloor$ denotes the greatest integer\nless than or equal to $x$.)",
      "solution": "\\noindent\n\\textbf{First solution.}\nWe prove the claim by induction on $qwertyui+asdfghjk$.\nFor the base case, suppose that $asdfghjk=1$; we then have $qwertyui=1$ and the given equation becomes $mwsxvzlk nhytrewq=e$. The claim then reduces to the fact that a one-sided inverse in $jklmnbvf$ is also a two-sided inverse. (Because $jklmnbvf$ is a group, $mwsxvzlk$ has an inverse $mwsxvzlk^{-1}$; since $mwsxvzlk nhytrewq = e$, we have $nhytrewq = mwsxvzlk^{-1}(mwsxvzlk nhytrewq) = mwsxvzlk^{-1} e = mwsxvzlk^{-1}$, so $nhytrewq mwsxvzlk = e = mwsxvzlk nhytrewq$.)\n\nSuppose now that $asdfghjk>1$. In case $qwertyui>asdfghjk$, set $\\tilde{zpolmnij} = mwsxvzlk nhytrewq$, $\\tilde{xswqazpl} = nhytrewq$, and\n\\[\nmnjbhgvr = \\left\\lfloor \\frac{(qwertyui-asdfghjk)zxcvbnmq}{asdfghjk} \\right\\rfloor - \\left\\lfloor \\frac{(qwertyui-asdfghjk)(zxcvbnmq-1)}{asdfghjk} \\right\\rfloor \n\\quad (zxcvbnmq=1,\\dots,asdfghjk).\n\\]\nthen\n\\[\n\\tilde{zpolmnij} \\tilde{xswqazpl}^{mnjbhgvr_1} \\cdots \\tilde{zpolmnij} \\tilde{xswqazpl}^{mnjbhgvr_{asdfghjk}} = mwsxvzlk nhytrewq^{polikujm_1} \\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk}} = e,\n\\]\nso the induction hypothesis implies that $\\tilde{zpolmnij}$ and $\\tilde{xswqazpl}$ commute; this implies that $mwsxvzlk$ and $nhytrewq$ commute.\n\nIn case $qwertyui < asdfghjk$, note that $polikujm \\in \\{0,1\\}$ for all $zxcvbnmq$. Set $\\tilde{zpolmnij} = nhytrewq^{-1}$, $\\tilde{xswqazpl} = mwsxvzlk^{-1}$, and\n\\[\nmnjbhgvr_{\\ell} = \\left\\lfloor \\frac{asdfghjk \\, qazwsxed}{qwertyui} \\right\\rfloor - \\left\\lfloor \\frac{asdfghjk(qazwsxed-1)}{qwertyui} \\right\\rfloor \n\\quad (qazwsxed=1,\\dots,qwertyui);\n\\]\nwe claim that \n\\[\n\\tilde{zpolmnij}\\tilde{xswqazpl}^{mnjbhgvr_1}\\cdots\\tilde{zpolmnij}\\tilde{xswqazpl}^{mnjbhgvr_{qwertyui}} = (mwsxvzlk nhytrewq^{polikujm_1}\\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk}})^{-1} = e,\n\\]\nso the induction hypothesis implies that $\\tilde{xswqazpl}$ and $\\tilde{zpolmnij}$ commute; this implies that $mwsxvzlk$ and $nhytrewq$ commute.\n\nTo clarify this last equality, consider a lattice walk starting from $(0,0)$, ending at $(asdfghjk,qwertyui)$, staying below the line\n$y = qwertyui x/asdfghjk$, and keeping as close to this line as possible. If one follows this walk and records the element $mwsxvzlk$ for each horizontal step and $nhytrewq$ for each vertical step, one obtains the word $mwsxvzlk nhytrewq^{polikujm_1}\\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk}}$. \nNow take this walk, reflect across the line $y = x$, rotate by a half-turn, then translate to put the endpoints at $(0,0)$ and $(qwertyui,asdfghjk)$; this is the analogous walk for the pair $(asdfghjk,qwertyui)$.\n\n\\noindent\n\\textbf{Remark.}\nBy tracing more carefully through the argument, one sees in addition that there exists an element zxcvbnmq of $jklmnbvf$\nfor which $mwsxvzlk = zxcvbnmq^{qwertyui}, \\; nhytrewq = zxcvbnmq^{-asdfghjk}$.\n\n\\noindent\n\\textbf{Second solution.} (by Greg Martin)\nSince $\\gcd(qwertyui,asdfghjk) = 1$, there exist integers $lkjhgfds,poiuytre$ such that $qwertyui lkjhgfds + asdfghjk poiuytre = 1$; we may further assume that \n$lkjhgfds \\in \\{1,\\dots,asdfghjk\\}$. We first establish the identity\n\\[\npolikujm_{zxcvbnmq-lkjhgfds} = \\begin{cases}\npolikujm_{zxcvbnmq} - 1 & \\mbox{if $zxcvbnmq \\equiv 0 \\pmod{asdfghjk}$} \\\\\npolikujm_{zxcvbnmq} + 1 & \\mbox{if $zxcvbnmq \\equiv 1 \\pmod{asdfghjk}$} \\\\\npolikujm_{zxcvbnmq} & \\mbox{otherwise}.\n\\end{cases}\n\\]\nNamely, by writing $-qwertyui lkjhgfds = asdfghjk poiuytre-1$, we see that\n\\begin{align*}\npolikujm_{zxcvbnmq-lkjhgfds} &= \\left\\lfloor \\frac{qwertyui(zxcvbnmq-lkjhgfds)}{asdfghjk} \\right\\rfloor - \\left\\lfloor \\frac{qwertyui(zxcvbnmq-lkjhgfds-1)}{asdfghjk} \\right\\rfloor\\\\\n&= \\left\\lfloor \\frac{qwertyui zxcvbnmq+asdfghjk poiuytre-1}{asdfghjk} \\right\\rfloor - \\left\\lfloor \\frac{qwertyui(zxcvbnmq-1)+asdfghjk poiuytre-1}{asdfghjk} \\right\\rfloor \\\\\n&= \\left\\lfloor \\frac{qwertyui zxcvbnmq-1}{asdfghjk} \\right\\rfloor - \\left\\lfloor \\frac{qwertyui(zxcvbnmq-1)-1}{asdfghjk} \\right\\rfloor\n\\end{align*}\nand so\n\\begin{align*}\npolikujm_{zxcvbnmq-lkjhgfds} - polikujm_{zxcvbnmq} &= \\left( \\left\\lfloor \\frac{qwertyui zxcvbnmq-1}{asdfghjk} \\right\\rfloor - \\left\\lfloor \\frac{qwertyui zxcvbnmq}{asdfghjk} \\right\\rfloor \\right)\\\\\n&\\quad\n- \\left( \\left\\lfloor \\frac{qwertyui(zxcvbnmq-1)-1}{asdfghjk} \\right\\rfloor - \\left\\lfloor \\frac{qwertyui(zxcvbnmq-1)}{asdfghjk} \\right\\rfloor \\right).\n\\end{align*}\nThe first parenthesized expression equals 1 if $asdfghjk$ divides $qwertyui zxcvbnmq$, or equivalently $asdfghjk$ divides $zxcvbnmq$, and 0 otherwise.\nSimilarly, the second parenthesized expression equals 1 if $asdfghjk$ divides $zxcvbnmq-1$ and 0 otherwise. This proves the stated identity.\n\nWe now use the given relation $mwsxvzlk nhytrewq^{polikujm_1} \\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk}} = e$ to write\n\\begin{align*}\n mwsxvzlk nhytrewq mwsxvzlk^{-1} nhytrewq^{-1} &= mwsxvzlk nhytrewq( nhytrewq^{polikujm_1} mwsxvzlk nhytrewq^{polikujm_2} \\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk-1}} mwsxvzlk nhytrewq^{polikujm_{asdfghjk}})nhytrewq^{-1} \\\\\n&= mwsxvzlk nhytrewq^{polikujm_1+1} mwsxvzlk nhytrewq^{polikujm_2} \\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk}-1} \\\\\n&= mwsxvzlk nhytrewq^{polikujm_{1-lkjhgfds}} \\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk-lkjhgfds}} \\\\\n&= (mwsxvzlk nhytrewq^{polikujm_{asdfghjk+1-lkjhgfds}} \\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk}}) (mwsxvzlk nhytrewq^{polikujm_1} \\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk-lkjhgfds}}).\n\\end{align*}\nThe two parenthesized expressions multiply in the opposite order to $mwsxvzlk nhytrewq^{polikujm_1} \\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk}} = e$, so they must be\n(two-sided) inverses of each other. We deduce that $mwsxvzlk nhytrewq mwsxvzlk^{-1} nhytrewq^{-1} = e$, meaning that $mwsxvzlk$ and $nhytrewq$ commute.\n\n\\noindent\n\\textbf{Third solution.} (by Sucharit Sarkar)\nLet $bvcxzlkj$ denote the torus $\\mathbb{R}^2/\\mathbb{Z}^2$. The line segments from $(0,0)$ to $(1,0)$ and from $(0,0)$ to $(0,1)$ are closed loops in $bvcxzlkj$, and we denote them by $mwsxvzlk$ and $nhytrewq$ respectively. Now let $asplkjmn$ be the (image of the) point $(\\epsilon,-\\epsilon)$ in $bvcxzlkj$ for some $0<\\epsilon\\ll 1$. The punctured torus $bvcxzlkj \\setminus \\{asplkjmn\\}$ deformation retracts onto the union of the loops $mwsxvzlk$ and $nhytrewq$, and so $\\pi_1(bvcxzlkj\\setminus\\{asplkjmn\\})$, the fundamental group of $bvcxzlkj\\setminus\\{asplkjmn\\}$ based at $(0,0)$, is the free group on two generators, $\\langle mwsxvzlk,nhytrewq\\rangle$.\n\nLet $\\gamma$ and $\\tilde{\\gamma}$ denote the following loops based at $(0,0)$ in $bvcxzlkj$: $\\gamma$ is the image of the line segment from $(0,0)$ to $(asdfghjk,qwertyui)$ under the projection $\\mathbb{R}^2 \\to bvcxzlkj$, and $\\tilde{\\gamma}$ is the image of the lattice walk from $(0,0)$ to $(asdfghjk,qwertyui)$, staying just below the line $y=qwertyui x/asdfghjk$, that was described in the first solution. There is a straight-line homotopy with fixed endpoints between the two paths in $\\mathbb{R}^2$ from $(0,0)$ to $(asdfghjk,qwertyui)$, the line segment and the lattice walk, and this homotopy does not pass through any point of the form $(a+\\epsilon,b-\\epsilon)$ for $a,b\\in\\mathbb{Z}$ by the construction of the lattice walk. It follows that $\\gamma$ and $\\tilde{\\gamma}$ are homotopic loops in $bvcxzlkj \\setminus \\{asplkjmn\\}$. Since the class of $\\tilde{\\gamma}$ in $\\pi_1(bvcxzlkj\\setminus\\{asplkjmn\\})$ is evidently $mwsxvzlk nhytrewq^{polikujm_1}mwsxvzlk nhytrewq^{polikujm_2}\\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk}}$, it follows that the class of $\\gamma$ in $\\pi_1(bvcxzlkj\\setminus\\{asplkjmn\\})$ is the same.\n\nNow since $\\gcd(qwertyui,asdfghjk)=1$, there is an element $\\phi \\in GL_2(\\mathbb{Z})$ sending $(asdfghjk,qwertyui)$ to $(1,0)$, which then sends the line segment from $(0,0)$ to $(asdfghjk,qwertyui)$ to the segment from $(0,0)$ to $(1,0)$. Then $\\phi$ induces a homeomorphism of $bvcxzlkj$ sending $\\gamma$ to $mwsxvzlk$, which in turn induces an isomorphism $\\phi_* :\\thinspace \\pi_1(bvcxzlkj\\setminus \\{asplkjmn\\}) \\to \\pi_1(bvcxzlkj\\setminus \\{\\phi^{-1}(asplkjmn)\\})$. Both fundamental groups are equal to $\\langle mwsxvzlk,nhytrewq\\rangle$, and we conclude that $\\phi_*$ sends $mwsxvzlk nhytrewq^{polikujm_1}mwsxvzlk nhytrewq^{polikujm_2}\\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk}}$ to $mwsxvzlk$. It follows that $\\phi_*$ induces an isomorphism\n\\[\n\\langle mwsxvzlk,nhytrewq\\,|\\,mwsxvzlk nhytrewq^{polikujm_1}mwsxvzlk nhytrewq^{polikujm_2}\\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk}} \\rangle \\to \\langle mwsxvzlk,nhytrewq \\,|\\, mwsxvzlk\\rangle \\cong \\langle nhytrewq\\rangle \\cong \\mathbb{Z}.\n\\]\n\nSince $\\mathbb{Z}$ is abelian, $mwsxvzlk$ and $nhytrewq$ must commute in $\\langle mwsxvzlk,nhytrewq\\,|\\,mwsxvzlk nhytrewq^{polikujm_1}mwsxvzlk nhytrewq^{polikujm_2}\\cdots mwsxvzlk nhytrewq^{polikujm_{asdfghjk}}\\rangle$, whence they must also commute in $jklmnbvf$.}",
      "confidence": 0.08
    },
    "kernel_variant": {
      "question": "Let m and n be positive integers with gcd(m,n)=1 and define\n\na_k = \\left\\lfloor \\frac{mk}{n} \\right\\rfloor-\\left\\lfloor \\frac{m(k-1)}{n} \\right\\rfloor\\qquad (k=1,2,\\dots ,n).\n\nLet g,h be elements of a group G such that the word\n\ngh^{a_1}\\,gh^{a_2}\\,\\dots\\,gh^{a_n}=e\\qquad (e \\text{ the identity of }G)\n\nholds in G.  Prove that g and h commute, i.e. gh = hg.",
      "solution": "We give a completely algebraic proof that avoids the incorrect manipulation pointed out in the review.  The idea is to show that the subgroup generated by g and h is in fact cyclic, whence it is automatically abelian.\n\n------------------------------------------------------------\n1.  Passage to a two–generator, one–relator group\n------------------------------------------------------------\nPut F := ⟨g,h⟩, the free group of rank 2 on the letters g and h, and\n\n      R := gh^{a_1}gh^{a_2}\\cdots gh^{a_n} ∈ F.\n\nLet\n\n      Γ := F/⟨⟨R⟩⟩\n\nbe the one–relator quotient.  Any pair (g,h) in an arbitrary group G\nthat satisfies the displayed relation is the image of the symbols g,h\nunder a homomorphism F → G that kills R; equivalently, it factors\nthrough the canonical map F → Γ.  Consequently it suffices to prove\nthat the two canonical generators commute in Γ.  (If they commute in\nΓ, their images commute in every quotient of Γ, in particular in our\noriginal group G.)\n\n------------------------------------------------------------\n2.  The relator in the abelianisation\n------------------------------------------------------------\nThe abelianisation of F is the free abelian group Z² with basis\n[g] , [h].  In that abelianisation the element R maps to\n\n      [R] = n[g] + m[h]\\;∈ Z²,\n\nbecause g occurs exactly n times and the total exponent of h equals\nm (this is an immediate telescoping calculation).  Since gcd(m,n)=1,\n[R] is a primitive element of Z².\n\n------------------------------------------------------------\n3.  Nielsen–Schreier–Tietze re-parametrisation\n------------------------------------------------------------\nA primitive element of Z² can be completed to a basis.  Hence there is\nsome A ∈ GL₂(Z) that sends (n,m) to (1,0).  It is a classical fact\n(from Nielsen theory) that any matrix in GL₂(Z) lifts to an\nautomorphism Φ of the free group F.  Concretely, Φ acts on the free\nbasis {g,h} by a sequence of elementary Nielsen transformations:\n\n  • g ↦ g^{±1}, h ↦ h  (inversion),\n  • g ↦ gh^{±1}, h ↦ h (right multiplication),\n  • g ↦ h, h ↦ g      (swap).\n\nTherefore we may choose such a Φ with the property\n\n      Φ(R) = g.\n\n------------------------------------------------------------\n4.  The resulting presentation\n------------------------------------------------------------\nApplying Φ to the defining presentation of Γ we obtain an isomorphic\npresentation\n\n      Γ ≅ ⟨ g, h \\mid g ⟩.\n\nBut the quotient of the free group F by the normal closure of g is\nclearly an infinite cyclic group generated by the image of h.  In this\npresentation g is trivial and h survives, so Γ is isomorphic to Z and\nhence abelian.  In particular the images of g and h commute in Γ; as\nexplained in Section 1, this forces gh = hg in the original group G.\n\n------------------------------------------------------------\n5.  A pleasant by-product\n------------------------------------------------------------\nBecause Γ ≅ Z, there is an element k in Γ with k ↦ h under the above\nidentification.  Tracing back through Φ one finds integers r,s such\nthat\n\n      g = k^{m},   h = k^{-n}.\n\nThus, once commutativity has been established, g and h are opposite,\nco-prime powers of a single group element – exactly the additional\ninformation mentioned in the original statement.",
      "_meta": {
        "core_steps": [
          "Induct on the sum m+n, base case n=1 (⇒ gh=e ⇒ g,h inverses ⇒ commute).",
          "If m>n, replace (g,h,m,n) by (gh , h , m−n , n) so the same relation holds with a smaller sum.",
          "If m<n, note a_k∈{0,1}, switch to (h⁻¹ , g⁻¹ , n , m) to obtain an analogous relation with smaller sum.",
          "Apply the induction hypothesis to the reduced pair to get commutation there.",
          "Lift commutation back to the original elements, concluding gh=hg."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Rounding convention used in the definition of a_k; replacing ⌊·⌋ by ⌈·⌉ (or any shift giving identical successive differences) leaves the proof unchanged.",
            "original": "floor function ⌊ mk/n ⌋"
          },
          "slot2": {
            "description": "Indexing offset for the sequence {a_k} and the product; any contiguous block of n indices (e.g., k=0…n−1) works identically.",
            "original": "k runs from 1 to n"
          },
          "slot3": {
            "description": "Amount subtracted in the Euclidean reduction step when m>n; using m mod n instead of m−n still decreases m+n and keeps the induction valid.",
            "original": "m−n"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}