{
  "index": "1994-B-6",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "For any integer $n$, set\n\\[\nn_a = 101a - 100\\cdot 2^a.\n\\]\nShow that for $0 \\leq a,b,c,d \\leq 99$, $n_a + n_b \\equiv\nn_c + n_d \\pmod{10100}$ implies $\\{a,b\\} = \\{c,d\\}$.\n\n\\end{itemize}\n\\end{document}",
  "solution": "Lemma. \\( 2^{a} \\equiv 1(\\bmod 101) \\) if and only if \\( a \\) is divisible by 100.\nProof. We need to show that the order \\( m \\) of the image of 2 in the group \\( (\\mathbb{Z} / 101 \\mathbb{Z})^{*} \\) is 100 . Since the group has order \\( 100, m \\) divides 100 . If \\( m \\) were a proper divisor of 100 , then \\( m \\) would divide either \\( 100 / 2=50 \\) or \\( 100 / 5=20 \\), so either \\( 2^{50} \\) or \\( 2^{20} \\) would be 1 modulo 101. But we compute\n\\[\n\\begin{array}{l}\n2^{10}=1024 \\equiv 14 \\quad(\\bmod 101) \\\\\n2^{20} \\equiv 14^{2} \\equiv-6 \\quad(\\bmod 101) \\\\\n2^{40} \\equiv(-6)^{2} \\equiv 36 \\quad(\\bmod 101) \\\\\n2^{50} \\equiv 36 \\cdot 14 \\equiv-1 \\quad(\\bmod 101)\n\\end{array}\n\\]\n\nCorollary 1. If \\( a \\) and \\( b \\) are nonnegative integers such that \\( 2^{a} \\equiv 2^{b}(\\bmod 101) \\), then \\( a \\equiv b(\\bmod 100) \\).\n\nProof. Without loss of generality, assume \\( a \\geq b \\). If 101 divides \\( 2^{a}-2^{b}=2^{b}\\left(2^{a-b}-1\\right) \\), then \\( 2^{a-b} \\equiv 1(\\bmod 101) \\), so \\( a-b \\equiv 0(\\bmod 100) \\) by the lemma.\n\nSolution. By the Chinese Remainder Theorem, \\( n_{a}+n_{b} \\equiv n_{c}+n_{d}(\\bmod 10100) \\) is equivalent to\n\\[\na+b \\equiv c+d \\quad(\\bmod 100)\n\\]\nand\n\\[\n2^{a}+2^{b} \\equiv 2^{c}+2^{d} \\quad(\\bmod 101)\n\\]\n\nBy Fermat's Little Theorem, (1) implies \\( 2^{a+b} \\equiv 2^{c+d}(\\bmod 101) \\), or equivalently\n\\[\n2^{a} 2^{b} \\equiv 2^{c} 2^{d} \\quad(\\bmod 101)\n\\]\n\nSolve for \\( 2^{b} \\) in (2) and substitute into (3) to obtain\n\\[\n2^{a}\\left(2^{c}+2^{d}-2^{a}\\right) \\equiv 2^{c} 2^{d} \\quad(\\bmod 101)\n\\]\nor equivalently\n\\[\n0 \\equiv\\left(2^{a}-2^{c}\\right)\\left(2^{a}-2^{d}\\right) \\quad(\\bmod 101)\n\\]\n(That such a factorization exists could have been guessed from the desired conclusion that \\( a=c \\) or \\( a=d \\).) Hence \\( 2^{a} \\equiv 2^{c}(\\bmod 101) \\) or \\( 2^{a} \\equiv 2^{d}(\\bmod 101) \\). By the corollary above, \\( a \\) is congruent to \\( c \\) or \\( d \\) modulo 100 . Then by ( 1 ), \\( b \\) is congruent to the other. But \\( 0 \\leq a, b, c, d \\leq 99 \\), so these congruences are equalities.",
  "vars": [
    "n",
    "n_a",
    "n_b",
    "n_c",
    "n_d",
    "a",
    "b",
    "c",
    "d",
    "m"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "intvarn",
        "n_a": "itemnaaa",
        "n_b": "itemnabb",
        "n_c": "itemnacc",
        "n_d": "itemnadd",
        "a": "indexone",
        "b": "indextwo",
        "c": "indexthr",
        "d": "indexfou",
        "m": "orderlen"
      },
      "question": "For any integer $intvarn$, set\n\\[\nitemnaaa = 101indexone - 100\\cdot 2^{indexone}.\n\\]\nShow that for $0 \\leq indexone,indextwo,indexthr,indexfou \\leq 99$, $itemnaaa + itemnabb \\equiv\nitemnacc + itemnadd \\pmod{10100}$ implies $\\{indexone,indextwo\\} = \\{indexthr,indexfou\\}$.\n\n\\end{itemize}\n\\end{document}",
      "solution": "Lemma. \\( 2^{indexone} \\equiv 1(\\bmod 101) \\) if and only if \\( indexone \\) is divisible by 100.\nProof. We need to show that the order \\( orderlen \\) of the image of 2 in the group \\( (\\mathbb{Z} / 101 \\mathbb{Z})^{*} \\) is 100 . Since the group has order 100, orderlen divides 100. If orderlen were a proper divisor of 100, then orderlen would divide either $100 / 2=50$ or $100 / 5=20$, so either $2^{50}$ or $2^{20}$ would be 1 modulo 101. But we compute\n\\[\n\\begin{array}{l}\n2^{10}=1024 \\equiv 14 \\quad(\\bmod 101) \\\\\n2^{20} \\equiv 14^{2} \\equiv-6 \\quad(\\bmod 101) \\\\\n2^{40} \\equiv(-6)^{2} \\equiv 36 \\quad(\\bmod 101) \\\\\n2^{50} \\equiv 36 \\cdot 14 \\equiv-1 \\quad(\\bmod 101)\n\\end{array}\n\\]\n\nCorollary 1. If \\( indexone \\) and \\( indextwo \\) are nonnegative integers such that \\( 2^{indexone} \\equiv 2^{indextwo}(\\bmod 101) \\), then \\( indexone \\equiv indextwo(\\bmod 100) \\).\n\nProof. Without loss of generality, assume \\( indexone \\geq indextwo \\). If 101 divides \\( 2^{indexone}-2^{indextwo}=2^{indextwo}\\left(2^{indexone-indextwo}-1\\right) \\), then \\( 2^{indexone-indextwo} \\equiv 1(\\bmod 101) \\), so \\( indexone-indextwo \\equiv 0(\\bmod 100) \\) by the lemma.\n\nSolution. By the Chinese Remainder Theorem, \\( itemnaaa+itemnabb \\equiv itemnacc+itemnadd(\\bmod 10100) \\) is equivalent to\n\\[\nindexone+indextwo \\equiv indexthr+indexfou \\quad(\\bmod 100)\n\\]\nand\n\\[\n2^{indexone}+2^{indextwo} \\equiv 2^{indexthr}+2^{indexfou} \\quad(\\bmod 101)\n\\]\n\nBy Fermat's Little Theorem, (1) implies \\( 2^{indexone+indextwo} \\equiv 2^{indexthr+indexfou}(\\bmod 101) \\), or equivalently\n\\[\n2^{indexone} 2^{indextwo} \\equiv 2^{indexthr} 2^{indexfou} \\quad(\\bmod 101)\n\\]\n\nSolve for \\( 2^{indextwo} \\) in (2) and substitute into (3) to obtain\n\\[\n2^{indexone}\\left(2^{indexthr}+2^{indexfou}-2^{indexone}\\right) \\equiv 2^{indexthr} 2^{indexfou} \\quad(\\bmod 101)\n\\]\nor equivalently\n\\[\n0 \\equiv\\left(2^{indexone}-2^{indexthr}\\right)\\left(2^{indexone}-2^{indexfou}\\right) \\quad(\\bmod 101)\n\\]\n(That such a factorization exists could have been guessed from the desired conclusion that \\( indexone=indexthr \\) or \\( indexone=indexfou \\).) Hence \\( 2^{indexone} \\equiv 2^{indexthr}(\\bmod 101) \\) or \\( 2^{indexone} \\equiv 2^{indexfou}(\\bmod 101) \\). By the corollary above, \\( indexone \\) is congruent to \\( indexthr \\) or \\( indexfou \\) modulo 100. Then by (1), \\( indextwo \\) is congruent to the other. But \\( 0 \\leq indexone, indextwo, indexthr, indexfou \\leq 99 \\), so these congruences are equalities."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "chandelier",
        "n_a": "marigolds",
        "n_b": "bricklayer",
        "n_c": "candlewick",
        "n_d": "lighthouse",
        "a": "saffronia",
        "b": "thundervo",
        "c": "peppermill",
        "d": "snowdrift",
        "m": "patchwork"
      },
      "question": "For any integer $chandelier$, set\n\\[\nmarigolds = 101saffronia - 100\\cdot 2^{saffronia}.\n\\]\nShow that for $0 \\leq saffronia,thundervo,peppermill,snowdrift \\leq 99$, $marigolds + bricklayer \\equiv\ncandlewick + lighthouse \\pmod{10100}$ implies $\\{saffronia,thundervo\\} = \\{peppermill,snowdrift\\}.\n\n\\end{itemize}\n\\end{document}",
      "solution": "Lemma. \\( 2^{saffronia} \\equiv 1(\\bmod 101) \\) if and only if \\( saffronia \\) is divisible by 100.\nProof. We need to show that the order \\( patchwork \\) of the image of 2 in the group \\( (\\mathbb{Z} / 101 \\mathbb{Z})^{*} \\) is 100 . Since the group has order 100, patchwork divides 100. If patchwork were a proper divisor of 100, then patchwork would divide either $100/2=50$ or $100/5=20$, so either $2^{50}$ or $2^{20}$ would be 1 modulo 101. But we compute\n\\[\n\\begin{array}{l}\n2^{10}=1024 \\equiv 14 \\quad(\\bmod 101) \\\\\n2^{20} \\equiv 14^{2} \\equiv-6 \\quad(\\bmod 101) \\\\\n2^{40} \\equiv(-6)^{2} \\equiv 36 \\quad(\\bmod 101) \\\\\n2^{50} \\equiv 36 \\cdot 14 \\equiv-1 \\quad(\\bmod 101)\n\\end{array}\n\\]\n\nCorollary 1. If \\( saffronia \\) and \\( thundervo \\) are nonnegative integers such that \\( 2^{saffronia} \\equiv 2^{thundervo}(\\bmod 101) \\), then \\( saffronia \\equiv thundervo(\\bmod 100) \\).\n\nProof. Without loss of generality, assume \\( saffronia \\geq thundervo \\). If 101 divides $2^{saffronia}-2^{thundervo}=2^{thundervo}(2^{saffronia-thundervo}-1)$, then \\(2^{saffronia-thundervo} \\equiv 1(\\bmod 101)\\), so $saffronia-thundervo \\equiv 0(\\bmod 100)$ by the lemma.\n\nSolution. By the Chinese Remainder Theorem, \\( marigolds+bricklayer \\equiv candlewick+lighthouse(\\bmod 10100) \\) is equivalent to\n\\[\nsaffronia+thundervo \\equiv peppermill+snowdrift \\quad(\\bmod 100)\n\\]\nand\n\\[\n2^{saffronia}+2^{thundervo} \\equiv 2^{peppermill}+2^{snowdrift} \\quad(\\bmod 101)\n\\]\n\nBy Fermat's Little Theorem, (1) implies \\( 2^{saffronia+thundervo} \\equiv 2^{peppermill+snowdrift}(\\bmod 101) \\), or equivalently\n\\[\n2^{saffronia} 2^{thundervo} \\equiv 2^{peppermill} 2^{snowdrift} \\quad(\\bmod 101)\n\\]\n\nSolve for $2^{thundervo}$ in (2) and substitute into (3) to obtain\n\\[\n2^{saffronia}\\left(2^{peppermill}+2^{snowdrift}-2^{saffronia}\\right) \\equiv 2^{peppermill} 2^{snowdrift} \\quad(\\bmod 101)\n\\]\nor equivalently\n\\[\n0 \\equiv\\left(2^{saffronia}-2^{peppermill}\\right)\\left(2^{saffronia}-2^{snowdrift}\\right) \\quad(\\bmod 101)\n\\]\nHence \\( 2^{saffronia} \\equiv 2^{peppermill}(\\bmod 101) \\) or \\( 2^{saffronia} \\equiv 2^{snowdrift}(\\bmod 101) \\). By the corollary above, $saffronia$ is congruent to $peppermill$ or $snowdrift$ modulo 100. Then by (1), $thundervo$ is congruent to the other. But $0 \\leq saffronia, thundervo, peppermill, snowdrift \\leq 99$, so these congruences are equalities."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "noninteger",
        "n_a": "irrationalvaluea",
        "n_b": "irrationalvalueb",
        "n_c": "irrationalvaluec",
        "n_d": "irrationalvalued",
        "a": "nonletter",
        "b": "antialpha",
        "c": "counterbeta",
        "d": "oppositegamma",
        "m": "disorder"
      },
      "question": "<<<\nFor any integer $noninteger$, set\n\\[\nirrationalvaluea = 101 nonletter - 100\\cdot 2^{nonletter}.\n\\]\nShow that for $0 \\leq nonletter, antialpha, counterbeta, oppositegamma \\leq 99$, $irrationalvaluea + irrationalvalueb \\equiv\nirrationalvaluec + irrationalvalued \\pmod{10100}$ implies $\\{nonletter, antialpha\\} = \\{counterbeta, oppositegamma\\}$.\n\n\\end{itemize}\n\\end{document}\n>>>",
      "solution": "<<<\nLemma. \\( 2^{nonletter} \\equiv 1(\\bmod 101) \\) if and only if \\( nonletter \\) is divisible by 100.\nProof. We need to show that the order \\( disorder \\) of the image of 2 in the group \\( (\\mathbb{Z} / 101 \\mathbb{Z})^{*} \\) is 100. Since the group has order 100, disorder divides 100. If disorder were a proper divisor of 100, then disorder would divide either 100/2 = 50 or 100/5 = 20, so either \\( 2^{50} \\) or \\( 2^{20} \\) would be 1 modulo 101. But we compute\n\\[\n\\begin{array}{l}\n2^{10}=1024 \\equiv 14 \\quad(\\bmod 101) \\\\\n2^{20} \\equiv 14^{2} \\equiv -6 \\quad(\\bmod 101) \\\\\n2^{40} \\equiv (-6)^{2} \\equiv 36 \\quad(\\bmod 101) \\\\\n2^{50} \\equiv 36 \\cdot 14 \\equiv -1 \\quad(\\bmod 101)\n\\end{array}\n\\]\n\nCorollary 1. If \\( nonletter \\) and \\( antialpha \\) are nonnegative integers such that \\( 2^{nonletter} \\equiv 2^{antialpha}(\\bmod 101) \\), then \\( nonletter \\equiv antialpha(\\bmod 100) \\).\n\nProof. Without loss of generality, assume \\( nonletter \\geq antialpha \\). If 101 divides \\( 2^{nonletter}-2^{antialpha}=2^{antialpha}\\left(2^{nonletter-antialpha}-1\\right) \\), then \\( 2^{nonletter-antialpha} \\equiv 1(\\bmod 101) \\), so \\( nonletter-antialpha \\equiv 0(\\bmod 100) \\) by the lemma.\n\nSolution. By the Chinese Remainder Theorem, \\( irrationalvaluea+irrationalvalueb \\equiv irrationalvaluec+irrationalvalued(\\bmod 10100) \\) is equivalent to\n\\[\nnonletter+antialpha \\equiv counterbeta+oppositegamma \\quad(\\bmod 100)\n\\]\nand\n\\[\n2^{nonletter}+2^{antialpha} \\equiv 2^{counterbeta}+2^{oppositegamma} \\quad(\\bmod 101)\n\\]\n\nBy Fermat's Little Theorem, (1) implies \\( 2^{nonletter+antialpha} \\equiv 2^{counterbeta+oppositegamma}(\\bmod 101) \\), or equivalently\n\\[\n2^{nonletter} 2^{antialpha} \\equiv 2^{counterbeta} 2^{oppositegamma} \\quad(\\bmod 101)\n\\]\n\nSolve for \\( 2^{antialpha} \\) in (2) and substitute into (3) to obtain\n\\[\n2^{nonletter}\\left(2^{counterbeta}+2^{oppositegamma}-2^{nonletter}\\right) \\equiv 2^{counterbeta} 2^{oppositegamma} \\quad(\\bmod 101)\n\\]\nor equivalently\n\\[\n0 \\equiv \\left(2^{nonletter}-2^{counterbeta}\\right)\\left(2^{nonletter}-2^{oppositegamma}\\right) \\quad(\\bmod 101)\n\\]\n(That such a factorization exists could have been guessed from the desired conclusion that \\( nonletter = counterbeta \\) or \\( nonletter = oppositegamma \\).) Hence \\( 2^{nonletter} \\equiv 2^{counterbeta}(\\bmod 101) \\) or \\( 2^{nonletter} \\equiv 2^{oppositegamma}(\\bmod 101) \\). By the corollary above, \\( nonletter \\) is congruent to \\( counterbeta \\) or \\( oppositegamma \\) modulo 100. Then by (1), \\( antialpha \\) is congruent to the other. But \\( 0 \\leq nonletter, antialpha, counterbeta, oppositegamma \\leq 99 \\), so these congruences are equalities.\n>>>"
    },
    "garbled_string": {
      "map": {
        "n": "tzywqplm",
        "n_a": "qzxwvtnp",
        "n_b": "hjgrksla",
        "n_c": "mvplkqsd",
        "n_d": "ftrcjbdh",
        "a": "lkjhgfds",
        "b": "asdfghjk",
        "c": "poiuytre",
        "d": "mnbvcxzq",
        "m": "qazwsxed"
      },
      "question": "For any integer $tzywqplm$, set\n\\[\nqzxwvtnp = 101lkjhgfds - 100\\cdot 2^{lkjhgfds}.\n\\]\nShow that for $0 \\leq lkjhgfds,asdfghjk,poiuytre,mnbvcxzq \\leq 99$, $qzxwvtnp + hjgrksla \\equiv\nmvplkqsd + ftrcjbdh \\pmod{10100}$ implies $\\{lkjhgfds,asdfghjk\\} = \\{poiuytre,mnbvcxzq\\}$.\n\n\\end{itemize}\n\\end{document}",
      "solution": "Lemma. \\( 2^{lkjhgfds} \\equiv 1(\\bmod 101) \\) if and only if \\( lkjhgfds \\) is divisible by 100.\nProof. We need to show that the order \\( qazwsxed \\) of the image of 2 in the group \\( (\\mathbb{Z} / 101 \\mathbb{Z})^{*} \\) is 100 . Since the group has order \\( 100, qazwsxed \\) divides 100 . If \\( qazwsxed \\) were a proper divisor of 100 , then \\( qazwsxed \\) would divide either \\( 100 / 2=50 \\) or \\( 100 / 5=20 \\), so either \\( 2^{50} \\) or \\( 2^{20} \\) would be 1 modulo 101. But we compute\n\\[\n\\begin{array}{l}\n2^{10}=1024 \\equiv 14 \\quad(\\bmod 101) \\\\\n2^{20} \\equiv 14^{2} \\equiv-6 \\quad(\\bmod 101) \\\\\n2^{40} \\equiv(-6)^{2} \\equiv 36 \\quad(\\bmod 101) \\\\\n2^{50} \\equiv 36 \\cdot 14 \\equiv-1 \\quad(\\bmod 101)\n\\end{array}\n\\]\n\nCorollary 1. If \\( lkjhgfds \\) and \\( asdfghjk \\) are nonnegative integers such that \\( 2^{lkjhgfds} \\equiv 2^{asdfghjk}(\\bmod 101) \\), then \\( lkjhgfds \\equiv asdfghjk(\\bmod 100) \\).\n\nProof. Without loss of generality, assume \\( lkjhgfds \\geq asdfghjk \\). If 101 divides \\( 2^{lkjhgfds}-2^{asdfghjk}=2^{asdfghjk}\\left(2^{lkjhgfds-asdfghjk}-1\\right) \\), then \\( 2^{lkjhgfds-asdfghjk} \\equiv 1(\\bmod 101) \\), so \\( lkjhgfds-asdfghjk \\equiv 0(\\bmod 100) \\) by the lemma.\n\nSolution. By the Chinese Remainder Theorem, \\( qzxwvtnp+hjgrksla \\equiv mvplkqsd+ftrcjbdh(\\bmod 10100) \\) is equivalent to\n\\[\nlkjhgfds+asdfghjk \\equiv poiuytre+mnbvcxzq \\quad(\\bmod 100)\n\\]\nand\n\\[\n2^{lkjhgfds}+2^{asdfghjk} \\equiv 2^{poiuytre}+2^{mnbvcxzq} \\quad(\\bmod 101)\n\\]\n\nBy Fermat's Little Theorem, (1) implies \\( 2^{lkjhgfds+asdfghjk} \\equiv 2^{poiuytre+mnbvcxzq}(\\bmod 101) \\), or equivalently\n\\[\n2^{lkjhgfds} 2^{asdfghjk} \\equiv 2^{poiuytre} 2^{mnbvcxzq} \\quad(\\bmod 101)\n\\]\n\nSolve for \\( 2^{asdfghjk} \\) in (2) and substitute into (3) to obtain\n\\[\n2^{lkjhgfds}\\left(2^{poiuytre}+2^{mnbvcxzq}-2^{lkjhgfds}\\right) \\equiv 2^{poiuytre} 2^{mnbvcxzq} \\quad(\\bmod 101)\n\\]\nor equivalently\n\\[\n0 \\equiv\\left(2^{lkjhgfds}-2^{poiuytre}\\right)\\left(2^{lkjhgfds}-2^{mnbvcxzq}\\right) \\quad(\\bmod 101)\n\\]\n(That such a factorization exists could have been guessed from the desired conclusion that \\( lkjhgfds=poiuytre \\) or \\( lkjhgfds=mnbvcxzq \\).) Hence \\( 2^{lkjhgfds} \\equiv 2^{poiuytre}(\\bmod 101) \\) or \\( 2^{lkjhgfds} \\equiv 2^{mnbvcxzq}(\\bmod 101) \\). By the corollary above, \\( lkjhgfds \\) is congruent to \\( poiuytre \\) or \\( mnbvcxzq \\) modulo 100 . Then by ( 1 ), \\( asdfghjk \\) is congruent to the other. But \\( 0 \\leq lkjhgfds, asdfghjk, poiuytre, mnbvcxzq \\leq 99 \\), so these congruences are equalities."
    },
    "kernel_variant": {
      "question": "Let p be an odd prime, let g be a fixed primitive root modulo p, and set  \n\n  M = p (p − 1).\n\nFor every integer a (not yet reduced mod p − 1) define the residue Nₐ (mod M) by the simultaneous congruences  \n\n  Nₐ ≡ a   (mod p − 1),  \n  Nₐ ≡ gᵃ  (mod p).           (1)\n\nBecause p and p − 1 are coprime, (1) has a unique solution modulo M by the Chinese Remainder Theorem.\n\nProve that for any indices  \n\n  0 ≤ a, b, c, d ≤ p − 2,           (2)\n\nthe congruence  \n\n  Nₐ + N_b ≡ N_c + N_d (mod M)      (3)\n\nforces the multiset equality  \n\n  {a, b} = {c, d}.              (4)\n\nEquivalently, the p − 1 residues N₀, N₁, …, N_{p−2} form a Sidon set in the additive group ℤ/Mℤ: every unordered pair of them has a distinct sum modulo M.",
      "solution": "Step 1.  Consequences of (3) modulo the two CRT–components.  \n• Modulo p − 1.  Using (1) we obtain  \n\n  a + b ≡ c + d  (mod p − 1).      (5)\n\n• Modulo p.  Using (1) again we get  \n\n  gᵃ + gᵇ ≡ gᶜ + gᵈ  (mod p).      (6)\n\nDenote  \n\n  S := gᵃ + gᵇ ≡ gᶜ + gᵈ (mod p),  \n  P := g^{a+b} ≡ g^{c+d} (mod p),    (7)\n\nwhere the second congruence comes from (5) and the fact that g^{p−1} ≡ 1 (mod p).\n\nStep 2.  A polynomial argument inside the field 𝔽_p.  \nConsider the quadratic polynomial  \n\n  f(T) = T² − S T + P ∈ 𝔽_p[T].     (8)\n\nBecause of (7) the pair (gᵃ, gᵇ) satisfies the Vieta relations  \nT₁ + T₂ = S, T₁T₂ = P,  \nhence gᵃ and gᵇ are roots of f.  \nThe same is true for the pair (gᶜ, gᵈ).  But a quadratic polynomial over a field has at most two roots, so the multisets of roots must coincide:\n\n  {gᵃ, gᵇ} = {gᶜ, gᵈ}  in 𝔽_p.    (9)\n\nStep 3.  Translating equality of powers into equality of exponents.  \nSince g is a primitive root, the map  \n\n  ℤ/(p − 1)ℤ  →  𝔽_p^×,  u ↦ gᵘ\n\nis an isomorphism.  Therefore (9) implies\n\n  a ≡ c (mod p − 1) and b ≡ d (mod p − 1)  or  \n  a ≡ d (mod p − 1) and b ≡ c (mod p − 1).  (10)\n\nStep 4.  Concluding with the given index range.  \nBecause all indices lie in the complete system 0, 1, …, p − 2, each congruence in (10) is an ordinary equality.  Hence either (a, b) = (c, d) or (a, b) = (d, c), i.e. (4) holds. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.738861",
        "was_fixed": false,
        "difficulty_analysis": "1. Two independent “kernel” congruences are intertwined.  The solver must keep track of arithmetic in both modulus 1640 and modulus 1332 and understand how to fuse them via the Chinese Remainder Theorem; the original involves only one modulus.\n\n2. Two different primitive roots (6 modulo 41 and 5 modulo 37) and two different orders (40 and 36) appear.  Establishing these orders and working with them demands more number-theoretic computations than the single-root situation of the original problem.\n\n3. The range of the indices is the full lcm(40,36) = 360, so one has to reconcile residue information modulo 40 and modulo 36 and invoke an additional argument with the least common multiple; the original needs only a single residue class consideration.\n\n4. The proof must be carried out twice (once in each modulus) and the conclusions must then be synchronised.  This doubling of structure requires greater organisational clarity and a deeper insight into how simultaneous congruences interact.\n\n5. Overall, solving the enhanced variant compels the contestant to juggle: primitive-root arguments, symmetric-polynomial factorisation, the Chinese Remainder Theorem, and a final lcm-based bounding argument—considerably more machinery than the single-prime, single-order analysis of the original kernel problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let p be an odd prime, let g be a fixed primitive root modulo p, and set  \n\n  M = p (p − 1).\n\nFor every integer a (not yet reduced mod p − 1) define the residue Nₐ (mod M) by the simultaneous congruences  \n\n  Nₐ ≡ a   (mod p − 1),  \n  Nₐ ≡ gᵃ  (mod p).           (1)\n\nBecause p and p − 1 are coprime, (1) has a unique solution modulo M by the Chinese Remainder Theorem.\n\nProve that for any indices  \n\n  0 ≤ a, b, c, d ≤ p − 2,           (2)\n\nthe congruence  \n\n  Nₐ + N_b ≡ N_c + N_d (mod M)      (3)\n\nforces the multiset equality  \n\n  {a, b} = {c, d}.              (4)\n\nEquivalently, the p − 1 residues N₀, N₁, …, N_{p−2} form a Sidon set in the additive group ℤ/Mℤ: every unordered pair of them has a distinct sum modulo M.",
      "solution": "Step 1.  Consequences of (3) modulo the two CRT–components.  \n• Modulo p − 1.  Using (1) we obtain  \n\n  a + b ≡ c + d  (mod p − 1).      (5)\n\n• Modulo p.  Using (1) again we get  \n\n  gᵃ + gᵇ ≡ gᶜ + gᵈ  (mod p).      (6)\n\nDenote  \n\n  S := gᵃ + gᵇ ≡ gᶜ + gᵈ (mod p),  \n  P := g^{a+b} ≡ g^{c+d} (mod p),    (7)\n\nwhere the second congruence comes from (5) and the fact that g^{p−1} ≡ 1 (mod p).\n\nStep 2.  A polynomial argument inside the field 𝔽_p.  \nConsider the quadratic polynomial  \n\n  f(T) = T² − S T + P ∈ 𝔽_p[T].     (8)\n\nBecause of (7) the pair (gᵃ, gᵇ) satisfies the Vieta relations  \nT₁ + T₂ = S, T₁T₂ = P,  \nhence gᵃ and gᵇ are roots of f.  \nThe same is true for the pair (gᶜ, gᵈ).  But a quadratic polynomial over a field has at most two roots, so the multisets of roots must coincide:\n\n  {gᵃ, gᵇ} = {gᶜ, gᵈ}  in 𝔽_p.    (9)\n\nStep 3.  Translating equality of powers into equality of exponents.  \nSince g is a primitive root, the map  \n\n  ℤ/(p − 1)ℤ  →  𝔽_p^×,  u ↦ gᵘ\n\nis an isomorphism.  Therefore (9) implies\n\n  a ≡ c (mod p − 1) and b ≡ d (mod p − 1)  or  \n  a ≡ d (mod p − 1) and b ≡ c (mod p − 1).  (10)\n\nStep 4.  Concluding with the given index range.  \nBecause all indices lie in the complete system 0, 1, …, p − 2, each congruence in (10) is an ordinary equality.  Hence either (a, b) = (c, d) or (a, b) = (d, c), i.e. (4) holds. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.571733",
        "was_fixed": false,
        "difficulty_analysis": "1. Two independent “kernel” congruences are intertwined.  The solver must keep track of arithmetic in both modulus 1640 and modulus 1332 and understand how to fuse them via the Chinese Remainder Theorem; the original involves only one modulus.\n\n2. Two different primitive roots (6 modulo 41 and 5 modulo 37) and two different orders (40 and 36) appear.  Establishing these orders and working with them demands more number-theoretic computations than the single-root situation of the original problem.\n\n3. The range of the indices is the full lcm(40,36) = 360, so one has to reconcile residue information modulo 40 and modulo 36 and invoke an additional argument with the least common multiple; the original needs only a single residue class consideration.\n\n4. The proof must be carried out twice (once in each modulus) and the conclusions must then be synchronised.  This doubling of structure requires greater organisational clarity and a deeper insight into how simultaneous congruences interact.\n\n5. Overall, solving the enhanced variant compels the contestant to juggle: primitive-root arguments, symmetric-polynomial factorisation, the Chinese Remainder Theorem, and a final lcm-based bounding argument—considerably more machinery than the single-prime, single-order analysis of the original kernel problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}