{
  "index": "1976-B-2",
  "type": "ALG",
  "tag": [
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "B-2. Suppose that \\( G \\) is a group generated by elements \\( A \\) and \\( B \\), that is, every element of \\( G \\) can be written as a finite \"word\" \\( A^{n_{1}} B^{n_{2}} A^{n_{1}} \\cdots B^{n_{k}} \\), where \\( n_{1}, \\ldots, n_{k} \\) are any integers, and \\( A^{0}=B^{n}=1 \\) as usual. Also, suppose that \\( A^{4}=B^{7}=A B A^{-1} B=1, A^{2} \\neq 1 \\), and \\( B \\neq 1 \\).\n(a) How many elements of \\( G \\) are of the form \\( C^{2} \\) with \\( C \\) in \\( G \\) ?\n(b) Write each such square as a word in \\( A \\) and \\( B \\).",
  "solution": "B-2.\nThe answers are (a) 8 ; (b) \\( 1, A^{2}, B, B^{2}, B^{3}, B^{4}, B^{5}, B^{6} \\). Since \\( B=\\left(B^{4}\\right)^{2}, B^{3}=\\left(B^{5}\\right)^{2}, B^{5}=\\left(B^{6}\\right)^{2} \\), the elements in the answer to (b) are all squares in \\( G \\). They are distinct since \\( B \\) has order 7 and \\( A \\) has order 4. To show that there are no other squares, we first note that \\( A B A^{-1} B=1 \\) implies \\( A B=B^{-1} A \\). Then\n\\[\nA B^{2}=\\left(B^{-1} A\\right) B=B^{-1}(A B)=B^{-1}\\left(B^{-1} A\\right)=B^{-2} A .\n\\]\n\nSimilarly \\( A B^{n}=B^{-n} A \\) for the other \\( n \\) 's in \\( \\{0,1, \\ldots, 6\\} \\) and so for all integers \\( n \\). With this, one obtains\n\\[\n\\left(B^{\\prime} A^{\\prime}\\right)\\left(B^{h} A^{k}\\right)=B^{u} A^{v} \\text { with } u=i+(-1)^{\\prime} h, v=j+k .\n\\]\n\nThus the set \\( S \\) of elements of the form \\( B^{\\prime} A^{\\prime} \\) is closed under multiplication. \\( S \\) is finite since \\( i \\) and \\( j \\) may be restricted to \\( 0 \\leqq i \\leqq 6 \\) and \\( 0 \\leqq j \\leqq 3 \\). Hence \\( S \\) is a group and so \\( S=G \\). It then follows from (P) that the squares in \\( G \\) are the \\( B^{\\prime \\prime} A^{v} \\) with \\( u=i\\left[1+(-1)^{\\prime}\\right] \\) and \\( v=2 j \\). If \\( j \\) is odd, \\( u=0 \\) and \\( v \\equiv 2(\\bmod 4) \\). If \\( j \\) is even, \\( v \\equiv 0(\\bmod 4) \\). Thus there are no squares other than those listed above.",
  "vars": [
    "n",
    "n_1",
    "n_2",
    "k",
    "h",
    "i",
    "j",
    "u",
    "v",
    "C"
  ],
  "params": [
    "A",
    "B",
    "G",
    "S",
    "P"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "intvar",
        "n_1": "firstint",
        "n_2": "secondint",
        "k": "wordcount",
        "h": "exponenth",
        "i": "exponenti",
        "j": "exponentj",
        "u": "exponentu",
        "v": "exponentv",
        "C": "elementc",
        "A": "generatora",
        "B": "generatorb",
        "G": "groupg",
        "S": "subgroup",
        "P": "propertyp"
      },
      "question": "B-2. Suppose that \\( groupg \\) is a group generated by elements \\( generatora \\) and \\( generatorb \\), that is, every element of \\( groupg \\) can be written as a finite \"word\" \\( generatora^{firstint} generatorb^{secondint} generatora^{firstint} \\cdots generatorb^{intvar_{wordcount}} \\), where \\( firstint, \\ldots , intvar_{wordcount} \\) are any integers, and \\( generatora^{0}=generatorb^{intvar}=1 \\) as usual. Also, suppose that \\( generatora^{4}=generatorb^{7}=generatora generatorb generatora^{-1} generatorb=1, generatora^{2} \\neq 1 \\), and \\( generatorb \\neq 1 \\).\n(a) How many elements of \\( groupg \\) are of the form \\( elementc^{2} \\) with \\( elementc \\) in \\( groupg \\) ?\n(b) Write each such square as a word in \\( generatora \\) and \\( generatorb \\).",
      "solution": "B-2.\nThe answers are (a) 8 ; (b) \\( 1, generatora^{2}, generatorb, generatorb^{2}, generatorb^{3}, generatorb^{4}, generatorb^{5}, generatorb^{6} \\). Since \\( generatorb=\\left(generatorb^{4}\\right)^{2},\\; generatorb^{3}=\\left(generatorb^{5}\\right)^{2},\\; generatorb^{5}=\\left(generatorb^{6}\\right)^{2} \\), the elements in the answer to (b) are all squares in \\( groupg \\). They are distinct since \\( generatorb \\) has order 7 and \\( generatora \\) has order 4. To show that there are no other squares, we first note that \\( generatora\\,generatorb\\,generatora^{-1}\\,generatorb=1 \\) implies \\( generatora\\,generatorb=generatorb^{-1}\\,generatora \\). Then\n\\[\ngeneratora\\,generatorb^{2}=\\left(generatorb^{-1}\\,generatora\\right)generatorb=generatorb^{-1}(generatora\\,generatorb)=generatorb^{-1}\\left(generatorb^{-1}\\,generatora\\right)=generatorb^{-2}\\,generatora .\n\\]\nSimilarly \\( generatora\\,generatorb^{intvar}=generatorb^{-intvar}\\,generatora \\) for the other \\( intvar \\)'s in \\( \\{0,1, \\ldots , 6\\} \\) and so for all integers \\( intvar \\). With this, one obtains\n\\[\n\\left(generatorb^{exponenti}\\,generatora^{exponentj}\\right)\\left(generatorb^{exponenth}\\,generatora^{wordcount}\\right)=generatorb^{exponentu}\\,generatora^{exponentv} \\text { with } exponentu=exponenti+(-1)^{exponentj}\\,exponenth,\\quad exponentv=exponentj+wordcount .\n\\]\nThus the set \\( subgroup \\) of elements of the form \\( generatorb^{exponenti}\\,generatora^{exponentj} \\) is closed under multiplication. \\( subgroup \\) is finite since \\( exponenti \\) and \\( exponentj \\) may be restricted to \\( 0 \\leqq exponenti \\leqq 6 \\) and \\( 0 \\leqq exponentj \\leqq 3 \\). Hence \\( subgroup \\) is a group and so \\( subgroup=groupg \\). It then follows from (propertyp) that the squares in \\( groupg \\) are the \\( generatorb^{\\prime \\prime}\\,generatora^{exponentv} \\) with \\( exponentu=exponenti\\left[1+(-1)^{exponentj}\\right] \\) and \\( exponentv=2\\,exponentj \\). If \\( exponentj \\) is odd, \\( exponentu=0 \\) and \\( exponentv \\equiv 2(\\bmod 4) \\). If \\( exponentj \\) is even, \\( exponentv \\equiv 0(\\bmod 4) \\). Thus there are no squares other than those listed above."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "lighthouse",
        "n_1": "sandcastle",
        "n_2": "raincloud",
        "k": "hummingbird",
        "h": "bridgework",
        "i": "daydream",
        "j": "paintbrush",
        "u": "starlight",
        "v": "moonstone",
        "C": "buttercup",
        "A": "sunflower",
        "B": "watermelon",
        "G": "cheeseboard",
        "S": "dragonfly",
        "P": "parchment"
      },
      "question": "Suppose that \\( cheeseboard \\) is a group generated by elements \\( sunflower \\) and \\( watermelon \\), that is, every element of \\( cheeseboard \\) can be written as a finite \"word\" \\( sunflower^{sandcastle} watermelon^{raincloud} sunflower^{sandcastle} \\cdots watermelon^{hummingbird} \\), where \\( sandcastle, \\ldots, hummingbird \\) are any integers, and \\( sunflower^{0}=watermelon^{lighthouse}=1 \\) as usual. Also, suppose that \\( sunflower^{4}=watermelon^{7}=sunflower watermelon sunflower^{-1} watermelon=1, sunflower^{2} \\neq 1 \\), and \\( watermelon \\neq 1 \\).\n(a) How many elements of \\( cheeseboard \\) are of the form \\( buttercup^{2} \\) with \\( buttercup \\) in \\( cheeseboard \\) ?\n(b) Write each such square as a word in \\( sunflower \\) and \\( watermelon \\).",
      "solution": "The answers are (a) 8 ; (b) \\( 1, sunflower^{2}, watermelon, watermelon^{2}, watermelon^{3}, watermelon^{4}, watermelon^{5}, watermelon^{6} \\). Since \\( watermelon=\\left(watermelon^{4}\\right)^{2}, watermelon^{3}=\\left(watermelon^{5}\\right)^{2}, watermelon^{5}=\\left(watermelon^{6}\\right)^{2} \\), the elements in the answer to (b) are all squares in \\( cheeseboard \\). They are distinct since \\( watermelon \\) has order 7 and \\( sunflower \\) has order 4. To show that there are no other squares, we first note that \\( sunflower watermelon sunflower^{-1} watermelon=1 \\) implies \\( sunflower watermelon=watermelon^{-1} sunflower \\). Then\n\\[\nsunflower watermelon^{2}=\\left(watermelon^{-1} sunflower\\right) watermelon=watermelon^{-1}(sunflower watermelon)=watermelon^{-1}\\left(watermelon^{-1} sunflower\\right)=watermelon^{-2} sunflower .\n\\]\nSimilarly \\( sunflower watermelon^{lighthouse}=watermelon^{-lighthouse} sunflower \\) for the other \\( lighthouse \\)'s in \\( \\{0,1, \\ldots, 6\\} \\) and so for all integers \\( lighthouse \\). With this, one obtains\n\\[\n\\left(watermelon^{daydream} sunflower^{paintbrush}\\right)\\left(watermelon^{bridgework} sunflower^{hummingbird}\\right)=watermelon^{starlight} sunflower^{moonstone} \\text { with } starlight=daydream+(-1)^{paintbrush} bridgework, moonstone=paintbrush+hummingbird .\n\\]\nThus the set \\( dragonfly \\) of elements of the form \\( watermelon^{daydream} sunflower^{paintbrush} \\) is closed under multiplication. \\( dragonfly \\) is finite since \\( daydream \\) and \\( paintbrush \\) may be restricted to \\( 0 \\leqq daydream \\leqq 6 \\) and \\( 0 \\leqq paintbrush \\leqq 3 \\). Hence \\( dragonfly \\) is a group and so \\( dragonfly=cheeseboard \\). It then follows from (parchment) that the squares in \\( cheeseboard \\) are the \\( watermelon^{starlight} sunflower^{moonstone} \\) with \\( starlight=daydream\\left[1+(-1)^{paintbrush}\\right] \\) and \\( moonstone=2 paintbrush \\). If \\( paintbrush \\) is odd, \\( starlight=0 \\) and \\( moonstone \\equiv 2(\\bmod 4) \\). If \\( paintbrush \\) is even, \\( moonstone \\equiv 0(\\bmod 4) \\). Thus there are no squares other than those listed above."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "irrational",
        "n_1": "fractionalfirst",
        "n_2": "fractionalsecond",
        "k": "boundless",
        "h": "deepness",
        "i": "realistic",
        "j": "solidness",
        "u": "knownness",
        "v": "scalarvalue",
        "C": "variableelem",
        "A": "negating",
        "B": "affirming",
        "G": "chaosset",
        "S": "endlessset",
        "P": "misreason"
      },
      "question": "B-2. Suppose that \\( chaosset \\) is a group generated by elements \\( negating \\) and \\( affirming \\), that is, every element of \\( chaosset \\) can be written as a finite \"word\" \\( negating^{fractionalfirst} affirming^{fractionalsecond} negating^{fractionalfirst} \\cdots affirming^{irrational_{boundless}} \\), where \\( fractionalfirst, \\ldots, irrational_{boundless} \\) are any integers, and \\( negating^{0}=affirming^{irrational}=1 \\) as usual. Also, suppose that \\( negating^{4}=affirming^{7}=negating affirming negating^{-1} affirming=1, negating^{2} \\neq 1 \\), and \\( affirming \\neq 1 \\).\n(a) How many elements of \\( chaosset \\) are of the form \\( variableelem^{2} \\) with \\( variableelem \\) in \\( chaosset \\) ?\n(b) Write each such square as a word in \\( negating \\) and \\( affirming \\).",
      "solution": "B-2.\nThe answers are (a) 8 ; (b) \\( 1, negating^{2}, affirming, affirming^{2}, affirming^{3}, affirming^{4}, affirming^{5}, affirming^{6} \\). Since \\( affirming=\\left(affirming^{4}\\right)^{2}, affirming^{3}=\\left(affirming^{5}\\right)^{2}, affirming^{5}=\\left(affirming^{6}\\right)^{2} \\), the elements in the answer to (b) are all squares in \\( chaosset \\). They are distinct since \\( affirming \\) has order 7 and \\( negating \\) has order 4. To show that there are no other squares, we first note that \\( negating affirming negating^{-1} affirming=1 \\) implies \\( negating affirming=affirming^{-1} negating \\). Then\n\\[\nnegating affirming^{2}=\\left(affirming^{-1} negating\\right) affirming=affirming^{-1}(negating affirming)=affirming^{-1}\\left(affirming^{-1} negating\\right)=affirming^{-2} negating .\n\\]\n\nSimilarly \\( negating affirming^{irrational}=affirming^{-\\irrational} negating \\) for the other \\( irrational \\)'s in \\( \\{0,1, \\ldots, 6\\} \\) and so for all integers \\( irrational \\). With this, one obtains\n\\[\n\\left(affirming^{\\prime} negating^{\\prime}\\right)\\left(affirming^{deepness} negating^{boundless}\\right)=affirming^{knownness} negating^{scalarvalue} \\text { with } knownness=realistic+(-1)^{\\prime} deepness, scalarvalue=solidness+boundless .\n\\]\n\nThus the set \\( endlessset \\) of elements of the form \\( affirming^{\\prime} negating^{\\prime} \\) is closed under multiplication. \\( endlessset \\) is finite since \\( realistic \\) and \\( solidness \\) may be restricted to \\( 0 \\leqq realistic \\leqq 6 \\) and \\( 0 \\leqq solidness \\leqq 3 \\). Hence \\( endlessset \\) is a group and so \\( endlessset=chaosset \\). It then follows from (misreason) that the squares in \\( chaosset \\) are the \\( affirming^{\\prime \\prime} negating^{scalarvalue} \\) with \\( knownness=realistic\\left[1+(-1)^{\\prime}\\right] \\) and \\( scalarvalue=2 solidness \\). If \\( solidness \\) is odd, \\( knownness=0 \\) and \\( scalarvalue \\equiv 2(\\bmod 4) \\). If \\( solidness \\) is even, \\( scalarvalue \\equiv 0(\\bmod 4) \\). Thus there are no squares other than those listed above."
    },
    "garbled_string": {
      "map": {
        "n": "yptkmdsvo",
        "n_1": "ldfugzoaie",
        "n_2": "rmpqbncku",
        "k": "hqwznsote",
        "h": "vcaypldex",
        "i": "eglhmqrat",
        "j": "fbzxowrni",
        "u": "tskipvaer",
        "v": "nikjdubza",
        "C": "wqtrlyeps",
        "A": "plmdqrexc",
        "B": "sjkhtvopa",
        "G": "zxiwybrnm",
        "S": "qudajblfe",
        "P": "otniwcrsg"
      },
      "question": "B-2. Suppose that \\( zxiwybrnm \\) is a group generated by elements \\( plmdqrexc \\) and \\( sjkhtvopa \\), that is, every element of \\( zxiwybrnm \\) can be written as a finite \"word\" \\( plmdqrexc^{ldfugzoaie} sjkhtvopa^{rmpqbncku} plmdqrexc^{ldfugzoaie} \\cdots sjkhtvopa^{yptkmdsvo_{hqwznsote}} \\), where \\( ldfugzoaie, \\ldots, yptkmdsvo_{hqwznsote} \\) are any integers, and \\( plmdqrexc^{0}=sjkhtvopa^{yptkmdsvo}=1 \\) as usual. Also, suppose that \\( plmdqrexc^{4}=sjkhtvopa^{7}=plmdqrexc sjkhtvopa plmdqrexc^{-1} sjkhtvopa=1, plmdqrexc^{2} \\neq 1 \\), and \\( sjkhtvopa \\neq 1 \\).\n(a) How many elements of \\( zxiwybrnm \\) are of the form \\( wqtrlyeps^{2} \\) with \\( wqtrlyeps \\) in \\( zxiwybrnm \\) ?\n(b) Write each such square as a word in \\( plmdqrexc \\) and \\( sjkhtvopa \\).",
      "solution": "B-2.\nThe answers are (a) 8 ; (b) \\( 1, plmdqrexc^{2}, sjkhtvopa, sjkhtvopa^{2}, sjkhtvopa^{3}, sjkhtvopa^{4}, sjkhtvopa^{5}, sjkhtvopa^{6} \\). Since \\( sjkhtvopa=\\left(sjkhtvopa^{4}\\right)^{2}, sjkhtvopa^{3}=\\left(sjkhtvopa^{5}\\right)^{2}, sjkhtvopa^{5}=\\left(sjkhtvopa^{6}\\right)^{2} \\), the elements in the answer to (b) are all squares in \\( zxiwybrnm \\). They are distinct since \\( sjkhtvopa \\) has order 7 and \\( plmdqrexc \\) has order 4. To show that there are no other squares, we first note that \\( plmdqrexc sjkhtvopa plmdqrexc^{-1} sjkhtvopa=1 \\) implies \\( plmdqrexc sjkhtvopa=sjkhtvopa^{-1} plmdqrexc \\). Then\n\\[\nplmdqrexc sjkhtvopa^{2}=\\left(sjkhtvopa^{-1} plmdqrexc\\right) sjkhtvopa=sjkhtvopa^{-1}(plmdqrexc sjkhtvopa)=sjkhtvopa^{-1}\\left(sjkhtvopa^{-1} plmdqrexc\\right)=sjkhtvopa^{-2} plmdqrexc .\n\\]\n\nSimilarly \\( plmdqrexc sjkhtvopa^{yptkmdsvo}=sjkhtvopa^{-yptkmdsvo} plmdqrexc \\) for the other \\( yptkmdsvo \\) 's in \\( \\{0,1, \\ldots, 6\\} \\) and so for all integers \\( yptkmdsvo \\). With this, one obtains\n\\[\n\\left(sjkhtvopa^{\\prime} plmdqrexc^{\\prime}\\right)\\left(sjkhtvopa^{vcaypldex} plmdqrexc^{hqwznsote}\\right)=sjkhtvopa^{tskipvaer} plmdqrexc^{nikjdubza} \\text { with } tskipvaer=eglhmqrat+(-1)^{\\prime} vcaypldex, nikjdubza=fbzxowrni+hqwznsote .\n\\]\n\nThus the set \\( qudajblfe \\) of elements of the form \\( sjkhtvopa^{\\prime} plmdqrexc^{\\prime} \\) is closed under multiplication. \\( qudajblfe \\) is finite since \\( eglhmqrat \\) and \\( fbzxowrni \\) may be restricted to \\( 0 \\leqq eglhmqrat \\leqq 6 \\) and \\( 0 \\leqq fbzxowrni \\leqq 3 \\). Hence \\( qudajblfe \\) is a group and so \\( qudajblfe=zxiwybrnm \\). It then follows from (otniwcrsg) that the squares in \\( zxiwybrnm \\) are the \\( sjkhtvopa^{\\prime \\prime} plmdqrexc^{nikjdubza} \\) with \\( tskipvaer=eglhmqrat\\left[1+(-1)^{\\prime}\\right] \\) and \\( nikjdubza=2 fbzxowrni \\). If \\( fbzxowrni \\) is odd, \\( tskipvaer=0 \\) and \\( nikjdubza \\equiv 2(\\bmod 4) \\). If \\( fbzxowrni \\) is even, \\( nikjdubza \\equiv 0(\\bmod 4) \\). Thus there are no squares other than those listed above."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nG=\\langle\\,A,B\\mid A^{30}=1,\\;B^{31}=1,\\;A B A^{-1}=B^{11},\\;\nA^{15}\\neq1,\\;B\\neq1\\rangle .\n\\]\n\n(Because \\(11\\) is a primitive root modulo \\(31\\), conjugation by \\(A\\) induces on\n\\(\\langle B\\rangle\\cong\\mathbf Z_{31}\\) the automorphism\n\\(\\varphi(x)=x^{11}\\); hence \\(G\\cong\\mathbf Z_{31}\\rtimes_{\\varphi}\\mathbf Z_{30}\\).)\n\na)  Prove that every element of \\(G\\) can be written uniquely in the normal form  \n\\[\nA^{\\,i}B^{\\,j}\\qquad(0\\le i<30,\\;0\\le j<31).\n\\]\n\nb)  Let \n\\[\n\\mathrm{Sq}(G)=\\{\\,g^{2}\\mid g\\in G\\,\\}.\n\\]\nDetermine \\(|\\mathrm{Sq}(G)|\\).\n\nc)  Let \n\\[\n\\mathrm{Cu}(G)=\\{\\,g^{3}\\mid g\\in G\\,\\}.\n\\]\nDetermine \\(|\\mathrm{Cu}(G)|\\).\n\nd)  Describe explicitly, as words in \\(A\\) and \\(B\\), all elements that lie in\n\\(\\mathrm{Sq}(G)\\cap\\mathrm{Cu}(G)\\).",
      "solution": "Throughout, exponents of \\(A\\) are taken modulo \\(30\\) and those of \\(B\\) modulo \\(31\\).\n\nWrite  \n\\[\n\\varphi(x)=x^{11}\\in\\operatorname{Aut}\\!\\bigl(\\mathbf Z_{31}\\bigr) ,\n\\qquad \n\\varphi^{30}=\\operatorname{id}\n\\]\nbecause \\(11^{30}\\equiv1\\pmod{31}\\).\n\n--------------------------------------------------\na)  Normal form and its uniqueness\n--------------------------------------------------\nFrom the defining relation \\(A B A^{-1}=B^{11}\\) we get for every \\(i\\ge0\\)\n\\[\nA^{i} B A^{-i}=B^{11^{\\,i}}.\n\\]\nHence\n\\[\nB\\,A^{\\,i}=A^{\\,i}B^{\\,11^{\\,i}}\\quad\\Longrightarrow\\quad\nB^{\\,j}A^{\\,i}=A^{\\,i}B^{\\,j\\,11^{\\,i}},\\qquad\n0\\le i<30,\\;0\\le j<31. \\tag{1}\n\\]\n\nFormula (1) allows any word to be rewritten with all \\(A\\)-powers on the\nleft and all \\(B\\)-powers on the right, i.e. in the form \\(A^{\\,i}B^{\\,j}\\).\n\nUniqueness.\nAssume \\(A^{\\,i}B^{\\,j}=A^{\\,i'}B^{\\,j'}\\).\nThen\n\\[\nA^{\\,i-i'}=B^{\\,j'-j}.\n\\]\nThe left-hand side lies in \n\\(\\langle A\\rangle\\cong\\mathbf Z_{30}\\) and the right-hand side in\n\\(\\langle B\\rangle\\cong\\mathbf Z_{31}\\).\nBecause \\(\\gcd(30,31)=1\\), their intersection is trivial, so\n\\(i\\equiv i' \\pmod{30}\\) and \\(j\\equiv j'\\pmod{31}\\).  The normal form is thus unique.\n\n--------------------------------------------------\nb)  Counting the squares\n--------------------------------------------------\nLet \\(g=A^{\\,i}B^{\\,j}\\).  Using (1) once,\n\\[\ng^{2}=A^{\\,i}B^{\\,j}\\,A^{\\,i}B^{\\,j}=A^{\\,2i}\\,B^{\\,j\\bigl(1+11^{\\,i}\\bigr)}. \\tag{2}\n\\]\nPut  \n\\[\ns(i)=2i\\pmod{30},\\qquad c(i)=1+11^{\\,i}\\pmod{31}. \\tag{3}\n\\]\nFormula (2) shows\n\\[\n\\mathrm{Sq}(G)=\n\\bigl\\{\\,A^{\\,s(i)}B^{\\,c(i)j}\\;\\bigm|\\;i\\in\\mathbf Z_{30},\\,j\\in\\mathbf Z_{31}\\bigr\\}. \\tag{4}\n\\]\n\nStep 1.  Zeros of \\(c(i)\\).\n\\[\nc(i)=0\\iff 11^{\\,i}\\equiv-1\\pmod{31}.\n\\]\nBecause \\(11\\) has order \\(30\\) modulo \\(31\\), this happens exactly for\n\\(i\\equiv15\\).\n\nStep 2.  Image sizes.\nIf \\(i=15\\) then \\(c(i)=0\\) and (2) gives \\(g^{2}=A^{30}=1\\) for every \\(j\\).\nIf \\(i\\neq15\\) then \\(c(i)\\neq0\\); multiplication by \\(c(i)\\) acts\nbijectively on \\(\\mathbf Z_{31}\\), so varying \\(j\\) produces all\n\\(31\\) exponents of \\(B\\).\n\nStep 3.  Identifying equal \\(A\\)-exponents.\nBecause \\(\\gcd(2,30)=2\\), the map \\(i\\mapsto s(i)=2i\\) hits the \\(15\\) even\nresidues of \\(\\mathbf Z_{30}\\); each even residue \\(s\\) has two preimages \\(i\\)\nand \\(i+15\\).\nFor \\(s=0\\) the two preimages are \\(i=0\\) and \\(i=15\\).\nThe case \\(i=15\\) contributes only the identity, which is already produced\nwhen \\(i=0\\) and \\(j=0\\).\nEvery other even residue \\(s\\) arises from two indices \\(i\\) whose\n\\(c(i)\\) are both non–zero, hence lead to the same set\n\\(\\{\\,A^{\\,s}B^{\\,k}\\mid k\\in\\mathbf Z_{31}\\}\\).\n\nTherefore  \n\\[\n|\\mathrm{Sq}(G)|=15\\times31=465. \\tag{5}\n\\]\n\nExplicitly,\n\\[\n\\mathrm{Sq}(G)=\n\\bigl\\{\\,A^{\\,2r}B^{\\,k}\\;\\bigm|\\;0\\le r<15,\\;0\\le k<31\\bigr\\}. \\tag{6}\n\\]\n\n--------------------------------------------------\nc)  Counting the cubes\n--------------------------------------------------\nWith the same notation,\n\\[\ng^{3}=A^{\\,i}B^{\\,j}\\,A^{\\,i}B^{\\,j}\\,A^{\\,i}B^{\\,j}\n       =A^{\\,3i}\\,B^{\\,j\\bigl(1+11^{\\,i}+11^{\\,2i}\\bigr)}. \\tag{7}\n\\]\nPut  \n\\[\nt(i)=3i\\pmod{30},\\qquad d(i)=1+11^{\\,i}+11^{\\,2i}\\pmod{31}. \\tag{8}\n\\]\n\nStep 1.  Zeros of \\(d(i)\\).\nSet \\(x=11^{\\,i}\\).\nThen \\(d(i)=0\\iff 1+x+x^{2}=0\\iff x^{3}=1\\) and \\(x\\neq1\\).\nIn the cyclic group \\((\\mathbf Z_{31})^{\\times}\\) of order \\(30\\)\nthere are exactly two elements of order \\(3\\),\nnamely \\(11^{10}\\) and \\(11^{20}\\); hence\n\\[\nd(i)=0\\iff i\\equiv10,20\\pmod{30}. \\tag{9}\n\\]\n\nStep 2.  Non–zero \\(d(i)\\).\nFor the remaining \\(28\\) indices \\(d(i)\\neq0\\); varying \\(j\\) runs through all \\(31\\) exponents of \\(B\\).\n\nStep 3.  Identifying equal \\(A\\)-exponents.\nThe map \\(i\\mapsto t(i)=3i\\) has kernel \\(\\{0,10,20\\}\\) and image the ten\nresidues \\(0,3,6,\\dots,27\\).  Each residue occurs for three different\n\\(i\\)’s, and at least one of them satisfies \\(d(i)\\neq0\\).\nThus every coset \\(\\{\\,A^{\\,t}B^{\\,k}\\mid k\\}\\) with\n\\(t\\equiv0,3,6,\\dots,27\\) appears exactly once.\n\nConsequently  \n\\[\n|\\mathrm{Cu}(G)|=10\\times31=310. \\tag{10}\n\\]\n\nExplicitly,\n\\[\n\\mathrm{Cu}(G)=\n\\bigl\\{\\,A^{\\,3r}B^{\\,k}\\;\\bigm|\\;0\\le r<10,\\;0\\le k<31\\bigr\\}. \\tag{11}\n\\]\n\n--------------------------------------------------\nd)  Squares that are also cubes\n--------------------------------------------------\nAn element \\(A^{\\,\\ell}B^{\\,k}\\) lies in \\(\\mathrm{Sq}(G)\\) iff \\(\\ell\\) is even,\nand in \\(\\mathrm{Cu}(G)\\) iff \\(\\ell\\equiv0\\pmod{3}\\).\nTherefore\n\\[\n\\ell\\equiv0\\pmod{2},\\quad\\ell\\equiv0\\pmod{3}\n\\;\\Longrightarrow\\;\n\\ell\\equiv0\\pmod{6}.\n\\]\nWithin \\(0\\le\\ell<30\\) this gives\n\\[\n\\ell\\in\\{0,6,12,18,24\\}. \\tag{12}\n\\]\nFor each such \\(\\ell\\) the descriptions (6) and (11) both provide every\nexponent of \\(B\\), so for every \\(k\\) there exist elements whose square and whose cube equal \\(A^{\\,\\ell}B^{\\,k}\\).\nHence\n\\[\n\\mathrm{Sq}(G)\\cap\\mathrm{Cu}(G)=\n\\bigl\\{\\,A^{\\,\\ell}B^{\\,k}\\;\\bigm|\\;\n\\ell\\in\\{0,6,12,18,24\\},\\;0\\le k<31\\bigr\\}, \\tag{13}\n\\]\nand\n\\[\n|\\mathrm{Sq}(G)\\cap\\mathrm{Cu}(G)|=5\\times31=155. \\tag{14}\n\\]\n\n--------------------------------------------------\nSummary\n--------------------------------------------------\n1.  Unique normal form \\(A^{\\,i}B^{\\,j}\\) with \\((i,j)\\in[0,29]\\times[0,30]\\).  \n2.  Number of squares:  \\(|\\mathrm{Sq}(G)|=465\\).  \n3.  Number of cubes:    \\(|\\mathrm{Cu}(G)|=310\\).  \n4.  Simultaneous squares and cubes: \\(155\\) elements, explicitly listed above.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.626008",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-order relations.  The exponents 30 and 31 and the primitive-root\n   conjugation A B A^{-1}=B^{11} force use of cyclic-group automorphisms\n   of order 30, far beyond the inversion relation of the original problems.\n\n2. Multiple interacting concepts.  Solving parts (b)–(d) requires\n   simultaneous control of three ingredients:\n   • semidirect-product normal forms,\n   • arithmetic in (ℤ/31ℤ)^×, including primitive roots and roots of unity,\n   • counting images of non-trivial endomorphisms of finite abelian groups.\n\n3. Deeper theoretical tools.  One must exploit properties of cyclic groups,\n   orders of automorphisms, geometric-series identities in group rings, and\n   Chinese-remainder–type counting (e.g. lcm considerations to intersect the\n   square and cube sets).\n\n4. Substantially longer solution path.  Where the original problem needed\n   only a single conjugation identity, this variant demands four main steps:\n   establishing normal form, analysing squares, analysing cubes, and then\n   intersecting the two large sets, each step involving non-trivial number-theoretic\n   calculations.\n\n5. Much larger search space.  The group now has 930 elements (vs. 66 or 132\n   in the originals), and the answer sets (465, 310, 155 elements) cannot be\n   listed by inspection; a structural approach is indispensable.\n\nThese additions create a kernel variant that is markedly more intricate and\nconceptually demanding than both the original and the current kernel versions."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nG=\\langle\\,A,B\\mid A^{30}=1,\\;B^{31}=1,\\;A B A^{-1}=B^{11},\\;\nA^{15}\\neq1,\\;B\\neq1\\rangle .\n\\]\n\n(Because \\(11\\) is a primitive root modulo \\(31\\), conjugation by \\(A\\) induces on\n\\(\\langle B\\rangle\\cong\\mathbf Z_{31}\\) the automorphism\n\\(\\varphi(x)=x^{11}\\); hence \\(G\\cong\\mathbf Z_{31}\\rtimes_{\\varphi}\\mathbf Z_{30}\\).)\n\na)  Prove that every element of \\(G\\) can be written uniquely in the normal form  \n\\[\nA^{\\,i}B^{\\,j}\\qquad(0\\le i<30,\\;0\\le j<31).\n\\]\n\nb)  Let \n\\[\n\\mathrm{Sq}(G)=\\{\\,g^{2}\\mid g\\in G\\,\\}.\n\\]\nDetermine \\(|\\mathrm{Sq}(G)|\\).\n\nc)  Let \n\\[\n\\mathrm{Cu}(G)=\\{\\,g^{3}\\mid g\\in G\\,\\}.\n\\]\nDetermine \\(|\\mathrm{Cu}(G)|\\).\n\nd)  Describe explicitly, as words in \\(A\\) and \\(B\\), all elements that lie in\n\\(\\mathrm{Sq}(G)\\cap\\mathrm{Cu}(G)\\).",
      "solution": "Throughout, exponents of \\(A\\) are taken modulo \\(30\\) and those of \\(B\\) modulo \\(31\\).\n\nWrite  \n\\[\n\\varphi(x)=x^{11}\\in\\operatorname{Aut}\\!\\bigl(\\mathbf Z_{31}\\bigr) ,\n\\qquad \n\\varphi^{30}=\\operatorname{id}\n\\]\nbecause \\(11^{30}\\equiv1\\pmod{31}\\).\n\n--------------------------------------------------\na)  Normal form and its uniqueness\n--------------------------------------------------\nFrom the defining relation \\(A B A^{-1}=B^{11}\\) we get for every \\(i\\ge0\\)\n\\[\nA^{i} B A^{-i}=B^{11^{\\,i}}.\n\\]\nHence\n\\[\nB\\,A^{\\,i}=A^{\\,i}B^{\\,11^{\\,i}}\\quad\\Longrightarrow\\quad\nB^{\\,j}A^{\\,i}=A^{\\,i}B^{\\,j\\,11^{\\,i}},\\qquad\n0\\le i<30,\\;0\\le j<31. \\tag{1}\n\\]\n\nFormula (1) allows any word to be rewritten with all \\(A\\)-powers on the\nleft and all \\(B\\)-powers on the right, i.e. in the form \\(A^{\\,i}B^{\\,j}\\).\n\nUniqueness.\nAssume \\(A^{\\,i}B^{\\,j}=A^{\\,i'}B^{\\,j'}\\).\nThen\n\\[\nA^{\\,i-i'}=B^{\\,j'-j}.\n\\]\nThe left-hand side lies in \n\\(\\langle A\\rangle\\cong\\mathbf Z_{30}\\) and the right-hand side in\n\\(\\langle B\\rangle\\cong\\mathbf Z_{31}\\).\nBecause \\(\\gcd(30,31)=1\\), their intersection is trivial, so\n\\(i\\equiv i' \\pmod{30}\\) and \\(j\\equiv j'\\pmod{31}\\).  The normal form is thus unique.\n\n--------------------------------------------------\nb)  Counting the squares\n--------------------------------------------------\nLet \\(g=A^{\\,i}B^{\\,j}\\).  Using (1) once,\n\\[\ng^{2}=A^{\\,i}B^{\\,j}\\,A^{\\,i}B^{\\,j}=A^{\\,2i}\\,B^{\\,j\\bigl(1+11^{\\,i}\\bigr)}. \\tag{2}\n\\]\nPut  \n\\[\ns(i)=2i\\pmod{30},\\qquad c(i)=1+11^{\\,i}\\pmod{31}. \\tag{3}\n\\]\nFormula (2) shows\n\\[\n\\mathrm{Sq}(G)=\n\\bigl\\{\\,A^{\\,s(i)}B^{\\,c(i)j}\\;\\bigm|\\;i\\in\\mathbf Z_{30},\\,j\\in\\mathbf Z_{31}\\bigr\\}. \\tag{4}\n\\]\n\nStep 1.  Zeros of \\(c(i)\\).\n\\[\nc(i)=0\\iff 11^{\\,i}\\equiv-1\\pmod{31}.\n\\]\nBecause \\(11\\) has order \\(30\\) modulo \\(31\\), this happens exactly for\n\\(i\\equiv15\\).\n\nStep 2.  Image sizes.\nIf \\(i=15\\) then \\(c(i)=0\\) and (2) gives \\(g^{2}=A^{30}=1\\) for every \\(j\\).\nIf \\(i\\neq15\\) then \\(c(i)\\neq0\\); multiplication by \\(c(i)\\) acts\nbijectively on \\(\\mathbf Z_{31}\\), so varying \\(j\\) produces all\n\\(31\\) exponents of \\(B\\).\n\nStep 3.  Identifying equal \\(A\\)-exponents.\nBecause \\(\\gcd(2,30)=2\\), the map \\(i\\mapsto s(i)=2i\\) hits the \\(15\\) even\nresidues of \\(\\mathbf Z_{30}\\); each even residue \\(s\\) has two preimages \\(i\\)\nand \\(i+15\\).\nFor \\(s=0\\) the two preimages are \\(i=0\\) and \\(i=15\\).\nThe case \\(i=15\\) contributes only the identity, which is already produced\nwhen \\(i=0\\) and \\(j=0\\).\nEvery other even residue \\(s\\) arises from two indices \\(i\\) whose\n\\(c(i)\\) are both non–zero, hence lead to the same set\n\\(\\{\\,A^{\\,s}B^{\\,k}\\mid k\\in\\mathbf Z_{31}\\}\\).\n\nTherefore  \n\\[\n|\\mathrm{Sq}(G)|=15\\times31=465. \\tag{5}\n\\]\n\nExplicitly,\n\\[\n\\mathrm{Sq}(G)=\n\\bigl\\{\\,A^{\\,2r}B^{\\,k}\\;\\bigm|\\;0\\le r<15,\\;0\\le k<31\\bigr\\}. \\tag{6}\n\\]\n\n--------------------------------------------------\nc)  Counting the cubes\n--------------------------------------------------\nWith the same notation,\n\\[\ng^{3}=A^{\\,i}B^{\\,j}\\,A^{\\,i}B^{\\,j}\\,A^{\\,i}B^{\\,j}\n       =A^{\\,3i}\\,B^{\\,j\\bigl(1+11^{\\,i}+11^{\\,2i}\\bigr)}. \\tag{7}\n\\]\nPut  \n\\[\nt(i)=3i\\pmod{30},\\qquad d(i)=1+11^{\\,i}+11^{\\,2i}\\pmod{31}. \\tag{8}\n\\]\n\nStep 1.  Zeros of \\(d(i)\\).\nSet \\(x=11^{\\,i}\\).\nThen \\(d(i)=0\\iff 1+x+x^{2}=0\\iff x^{3}=1\\) and \\(x\\neq1\\).\nIn the cyclic group \\((\\mathbf Z_{31})^{\\times}\\) of order \\(30\\)\nthere are exactly two elements of order \\(3\\),\nnamely \\(11^{10}\\) and \\(11^{20}\\); hence\n\\[\nd(i)=0\\iff i\\equiv10,20\\pmod{30}. \\tag{9}\n\\]\n\nStep 2.  Non–zero \\(d(i)\\).\nFor the remaining \\(28\\) indices \\(d(i)\\neq0\\); varying \\(j\\) runs through all \\(31\\) exponents of \\(B\\).\n\nStep 3.  Identifying equal \\(A\\)-exponents.\nThe map \\(i\\mapsto t(i)=3i\\) has kernel \\(\\{0,10,20\\}\\) and image the ten\nresidues \\(0,3,6,\\dots,27\\).  Each residue occurs for three different\n\\(i\\)’s, and at least one of them satisfies \\(d(i)\\neq0\\).\nThus every coset \\(\\{\\,A^{\\,t}B^{\\,k}\\mid k\\}\\) with\n\\(t\\equiv0,3,6,\\dots,27\\) appears exactly once.\n\nConsequently  \n\\[\n|\\mathrm{Cu}(G)|=10\\times31=310. \\tag{10}\n\\]\n\nExplicitly,\n\\[\n\\mathrm{Cu}(G)=\n\\bigl\\{\\,A^{\\,3r}B^{\\,k}\\;\\bigm|\\;0\\le r<10,\\;0\\le k<31\\bigr\\}. \\tag{11}\n\\]\n\n--------------------------------------------------\nd)  Squares that are also cubes\n--------------------------------------------------\nAn element \\(A^{\\,\\ell}B^{\\,k}\\) lies in \\(\\mathrm{Sq}(G)\\) iff \\(\\ell\\) is even,\nand in \\(\\mathrm{Cu}(G)\\) iff \\(\\ell\\equiv0\\pmod{3}\\).\nTherefore\n\\[\n\\ell\\equiv0\\pmod{2},\\quad\\ell\\equiv0\\pmod{3}\n\\;\\Longrightarrow\\;\n\\ell\\equiv0\\pmod{6}.\n\\]\nWithin \\(0\\le\\ell<30\\) this gives\n\\[\n\\ell\\in\\{0,6,12,18,24\\}. \\tag{12}\n\\]\nFor each such \\(\\ell\\) the descriptions (6) and (11) both provide every\nexponent of \\(B\\), so for every \\(k\\) there exist elements whose square and whose cube equal \\(A^{\\,\\ell}B^{\\,k}\\).\nHence\n\\[\n\\mathrm{Sq}(G)\\cap\\mathrm{Cu}(G)=\n\\bigl\\{\\,A^{\\,\\ell}B^{\\,k}\\;\\bigm|\\;\n\\ell\\in\\{0,6,12,18,24\\},\\;0\\le k<31\\bigr\\}, \\tag{13}\n\\]\nand\n\\[\n|\\mathrm{Sq}(G)\\cap\\mathrm{Cu}(G)|=5\\times31=155. \\tag{14}\n\\]\n\n--------------------------------------------------\nSummary\n--------------------------------------------------\n1.  Unique normal form \\(A^{\\,i}B^{\\,j}\\) with \\((i,j)\\in[0,29]\\times[0,30]\\).  \n2.  Number of squares:  \\(|\\mathrm{Sq}(G)|=465\\).  \n3.  Number of cubes:    \\(|\\mathrm{Cu}(G)|=310\\).  \n4.  Simultaneous squares and cubes: \\(155\\) elements, explicitly listed above.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.499905",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-order relations.  The exponents 30 and 31 and the primitive-root\n   conjugation A B A^{-1}=B^{11} force use of cyclic-group automorphisms\n   of order 30, far beyond the inversion relation of the original problems.\n\n2. Multiple interacting concepts.  Solving parts (b)–(d) requires\n   simultaneous control of three ingredients:\n   • semidirect-product normal forms,\n   • arithmetic in (ℤ/31ℤ)^×, including primitive roots and roots of unity,\n   • counting images of non-trivial endomorphisms of finite abelian groups.\n\n3. Deeper theoretical tools.  One must exploit properties of cyclic groups,\n   orders of automorphisms, geometric-series identities in group rings, and\n   Chinese-remainder–type counting (e.g. lcm considerations to intersect the\n   square and cube sets).\n\n4. Substantially longer solution path.  Where the original problem needed\n   only a single conjugation identity, this variant demands four main steps:\n   establishing normal form, analysing squares, analysing cubes, and then\n   intersecting the two large sets, each step involving non-trivial number-theoretic\n   calculations.\n\n5. Much larger search space.  The group now has 930 elements (vs. 66 or 132\n   in the originals), and the answer sets (465, 310, 155 elements) cannot be\n   listed by inspection; a structural approach is indispensable.\n\nThese additions create a kernel variant that is markedly more intricate and\nconceptually demanding than both the original and the current kernel versions."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}