{
  "index": "1977-A-5",
  "type": "NT",
  "tag": [
    "NT",
    "COMB"
  ],
  "difficulty": "",
  "question": "Problem A-5\nProve that\n\\[\n\\binom{p a}{p b} \\equiv\\binom{a}{b}(\\bmod p)\n\\]\nfor all integers \\( p, a \\), and \\( b \\) with \\( p \\) a prime, \\( p>0 \\), and \\( a>b>0 \\).\nNotation: \\( \\binom{m}{n} \\) denotes the binomial coefficient \\( \\frac{m!}{n!(m-n)!} \\).",
  "solution": "A-5.\nIt is well known that \\( \\binom{p}{i} \\equiv 0 \\bmod p \\) for \\( i=1,2, \\cdots, p-1 \\) or equivalently that in \\( Z_{p}[x] \\) one has \\( (1+x)^{p}=1+x^{p} \\), where \\( Z_{p} \\) is the field of the integers modulo \\( p \\). Thus in \\( Z_{p}[x] \\),\n\\[\n\\sum_{k=0}^{p a}\\binom{p a}{k} x^{k}=(1+x)^{p a}=\\left[(1+x)^{p}\\right]^{a}=\\left[1+x^{p}\\right]^{a}=\\sum_{j=0}^{a}\\binom{a}{j} x^{j p} .\n\\]\n\nSince coefficients of like powers must be congruent modulo \\( p \\) in the equality\n\\[\n\\sum_{k=0}^{p a}\\binom{p a}{k} x^{k}=\\sum_{j=0}^{c}\\binom{a}{j} x^{j p}\n\\]\nin \\( Z_{p}[x] \\), one sees that\n\\[\n\\binom{p a}{p b} \\equiv\\binom{a}{b}(\\bmod p)\n\\]\nfor \\( b=0,1, \\ldots, a \\).",
  "vars": [
    "i",
    "k",
    "j",
    "m",
    "n",
    "x"
  ],
  "params": [
    "p",
    "a",
    "b",
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "p": "primevar",
        "a": "biggerval",
        "b": "smallint",
        "c": "charparam",
        "k": "counterk",
        "j": "indexj",
        "m": "integerem",
        "n": "integeren",
        "x": "variablex"
      },
      "question": "Problem A-5\nProve that\n\\[\n\\binom{primevar biggerval}{primevar smallint} \\equiv\\binom{biggerval}{smallint}(\\bmod primevar)\n\\]\nfor all integers \\( primevar, biggerval \\), and \\( smallint \\) with \\( primevar \\) a prime, \\( primevar>0 \\), and \\( biggerval>smallint>0 \\).\nNotation: \\( \\binom{integerem}{integeren} \\) denotes the binomial coefficient \\( \\frac{integerem!}{integeren!(integerem-integeren)!} \\).",
      "solution": "A-5.\nIt is well known that \\( \\binom{primevar}{i} \\equiv 0 \\\\bmod primevar \\) for \\( i=1,2, \\cdots, primevar-1 \\) or equivalently that in \\( Z_{primevar}[variablex] \\) one has \\( (1+variablex)^{primevar}=1+variablex^{primevar} \\), where \\( Z_{primevar} \\) is the field of the integers modulo \\( primevar \\). Thus in \\( Z_{primevar}[variablex] \\),\n\\[\n\\sum_{counterk=0}^{primevar biggerval}\\binom{primevar biggerval}{counterk} variablex^{counterk}=(1+variablex)^{primevar biggerval}=\\left[(1+variablex)^{primevar}\\right]^{biggerval}=\\left[1+variablex^{primevar}\\right]^{biggerval}=\\sum_{indexj=0}^{biggerval}\\binom{biggerval}{indexj} variablex^{indexj primevar} .\n\\]\n\nSince coefficients of like powers must be congruent modulo \\( primevar \\) in the equality\n\\[\n\\sum_{counterk=0}^{primevar biggerval}\\binom{primevar biggerval}{counterk} variablex^{counterk}=\\sum_{indexj=0}^{charparam}\\binom{biggerval}{indexj} variablex^{indexj primevar}\n\\]\nin \\( Z_{primevar}[variablex] \\), one sees that\n\\[\n\\binom{primevar biggerval}{primevar smallint} \\equiv\\binom{biggerval}{smallint}(\\bmod primevar)\n\\]\nfor \\( smallint=0,1, \\ldots, biggerval \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "k": "thimbleweed",
        "j": "butterglory",
        "m": "cloudanchor",
        "n": "pebbletrail",
        "x": "quillsparrow",
        "p": "lanternbloom",
        "a": "driftwooden",
        "b": "starlitridge",
        "c": "moonlilac"
      },
      "question": "Problem A-5\nProve that\n\\[\n\\binom{lanternbloom driftwooden}{lanternbloom starlitridge} \\equiv\\binom{driftwooden}{starlitridge}(\\bmod lanternbloom)\n\\]\nfor all integers \\( lanternbloom, driftwooden \\), and \\( starlitridge \\) with \\( lanternbloom \\) a prime, \\( lanternbloom>0 \\), and \\( driftwooden>starlitridge>0 \\).\nNotation: \\( \\binom{cloudanchor}{pebbletrail} \\) denotes the binomial coefficient \\( \\frac{cloudanchor!}{pebbletrail!(cloudanchor-pebbletrail)!} \\).",
      "solution": "A-5.\nIt is well known that \\( \\binom{lanternbloom}{i} \\equiv 0 \\bmod lanternbloom \\) for \\( i=1,2, \\cdots, lanternbloom-1 \\) or equivalently that in \\( Z_{lanternbloom}[quillsparrow] \\) one has \\( (1+quillsparrow)^{lanternbloom}=1+quillsparrow^{lanternbloom} \\), where \\( Z_{lanternbloom} \\) is the field of the integers modulo \\( lanternbloom \\). Thus in \\( Z_{lanternbloom}[quillsparrow] \\),\n\\[\n\\sum_{thimbleweed=0}^{lanternbloom driftwooden}\\binom{lanternbloom driftwooden}{thimbleweed} quillsparrow^{thimbleweed}=(1+quillsparrow)^{lanternbloom driftwooden}=\\left[(1+quillsparrow)^{lanternbloom}\\right]^{driftwooden}=\\left[1+quillsparrow^{lanternbloom}\\right]^{driftwooden}=\\sum_{butterglory=0}^{driftwooden}\\binom{driftwooden}{butterglory} quillsparrow^{butterglory lanternbloom} .\n\\]\n\nSince coefficients of like powers must be congruent modulo \\( lanternbloom \\) in the equality\n\\[\n\\sum_{thimbleweed=0}^{lanternbloom driftwooden}\\binom{lanternbloom driftwooden}{thimbleweed} quillsparrow^{thimbleweed}=\\sum_{butterglory=0}^{moonlilac}\\binom{driftwooden}{butterglory} quillsparrow^{butterglory lanternbloom}\n\\]\nin \\( Z_{lanternbloom}[quillsparrow] \\), one sees that\n\\[\n\\binom{lanternbloom driftwooden}{lanternbloom starlitridge} \\equiv\\binom{driftwooden}{starlitridge}(\\bmod lanternbloom)\n\\]\nfor \\( starlitridge=0,1, \\ldots, driftwooden \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "i": "unindexed",
        "k": "singlevalue",
        "j": "fixedpoint",
        "m": "constant",
        "n": "immutable",
        "x": "steadystate",
        "p": "composite",
        "a": "terminal",
        "b": "external",
        "c": "undefined"
      },
      "question": "Problem A-5\nProve that\n\\[\n\\binom{composite terminal}{composite external} \\equiv\\binom{terminal}{external}(\\bmod composite)\n\\]\nfor all integers \\( composite, terminal \\), and \\( external \\) with \\( composite \\) a prime, \\( composite>0 \\), and \\( terminal>external>0 \\).\nNotation: \\( \\binom{constant}{immutable} \\) denotes the binomial coefficient \\( \\frac{constant!}{immutable!(constant-immutable)!} \\).",
      "solution": "A-5.\nIt is well known that \\( \\binom{composite}{unindexed} \\equiv 0 \\bmod composite \\) for \\( unindexed=1,2, \\cdots, composite-1 \\) or equivalently that in \\( Z_{composite}[steadystate] \\) one has \\( (1+steadystate)^{composite}=1+steadystate^{composite} \\), where \\( Z_{composite} \\) is the field of the integers modulo \\( composite \\). Thus in \\( Z_{composite}[steadystate] \\),\n\\[\n\\sum_{singlevalue=0}^{composite\\ terminal}\\binom{composite\\ terminal}{singlevalue} steadystate^{singlevalue}=(1+steadystate)^{composite\\ terminal}=\\left[(1+steadystate)^{composite}\\right]^{terminal}=\\left[1+steadystate^{composite}\\right]^{terminal}=\\sum_{fixedpoint=0}^{terminal}\\binom{terminal}{fixedpoint} steadystate^{fixedpoint\\ composite} .\n\\]\n\nSince coefficients of like powers must be congruent modulo \\( composite \\) in the equality\n\\[\n\\sum_{singlevalue=0}^{composite\\ terminal}\\binom{composite\\ terminal}{singlevalue} steadystate^{singlevalue}=\\sum_{fixedpoint=0}^{undefined}\\binom{terminal}{fixedpoint} steadystate^{fixedpoint\\ composite}\n\\]\nin \\( Z_{composite}[steadystate] \\), one sees that\n\\[\n\\binom{composite\\ terminal}{composite\\ external} \\equiv\\binom{terminal}{external}(\\bmod composite)\n\\]\nfor \\( external=0,1, \\ldots, terminal \\)."
    },
    "garbled_string": {
      "map": {
        "i": "bvncmars",
        "k": "tzlrqwish",
        "j": "soccusdn",
        "m": "xjkwprem",
        "n": "uvyfteop",
        "x": "rlkdpaqv",
        "p": "qzxwvtnp",
        "a": "hjgrksla",
        "b": "vkdplqtw",
        "c": "fmszwxne"
      },
      "question": "Problem A-5\nProve that\n\\[\n\\binom{qzxwvtnp hjgrksla}{qzxwvtnp vkdplqtw} \\equiv\\binom{hjgrksla}{vkdplqtw}(\\bmod qzxwvtnp)\n\\]\nfor all integers \\( qzxwvtnp, hjgrksla \\), and \\( vkdplqtw \\) with \\( qzxwvtnp \\) a prime, \\( qzxwvtnp>0 \\), and \\( hjgrksla>vkdplqtw>0 \\).\nNotation: \\( \\binom{xjkwprem}{uvyfteop} \\) denotes the binomial coefficient \\( \\frac{xjkwprem!}{uvyfteop!(xjkwprem-uvyfteop)!} \\).",
      "solution": "A-5.\nIt is well known that \\( \\binom{qzxwvtnp}{bvncmars} \\equiv 0 \\bmod qzxwvtnp \\) for \\( bvncmars=1,2, \\cdots, qzxwvtnp-1 \\) or equivalently that in \\( Z_{qzxwvtnp}[rlkdpaqv] \\) one has \\( (1+rlkdpaqv)^{qzxwvtnp}=1+rlkdpaqv^{qzxwvtnp} \\), where \\( Z_{qzxwvtnp} \\) is the field of the integers modulo \\( qzxwvtnp \\). Thus in \\( Z_{qzxwvtnp}[rlkdpaqv] \\),\n\\[\n\\sum_{tzlrqwish=0}^{qzxwvtnp hjgrksla}\\binom{qzxwvtnp hjgrksla}{tzlrqwish} rlkdpaqv^{tzlrqwish}=(1+rlkdpaqv)^{qzxwvtnp hjgrksla}=\\left[(1+rlkdpaqv)^{qzxwvtnp}\\right]^{hjgrksla}=\\left[1+rlkdpaqv^{qzxwvtnp}\\right]^{hjgrksla}=\\sum_{soccusdn=0}^{hjgrksla}\\binom{hjgrksla}{soccusdn} rlkdpaqv^{soccusdn qzxwvtnp} .\n\\]\n\nSince coefficients of like powers must be congruent modulo \\( qzxwvtnp \\) in the equality\n\\[\n\\sum_{tzlrqwish=0}^{qzxwvtnp hjgrksla}\\binom{qzxwvtnp hjgrksla}{tzlrqwish} rlkdpaqv^{tzlrqwish}=\\sum_{soccusdn=0}^{fmszwxne}\\binom{hjgrksla}{soccusdn} rlkdpaqv^{soccusdn qzxwvtnp}\n\\]\nin \\( Z_{qzxwvtnp}[rlkdpaqv] \\), one sees that\n\\[\n\\binom{qzxwvtnp hjgrksla}{qzxwvtnp vkdplqtw} \\equiv\\binom{hjgrksla}{vkdplqtw}(\\bmod qzxwvtnp)\n\\]\nfor \\( vkdplqtw=0,1, \\ldots, hjgrksla \\)."
    },
    "kernel_variant": {
      "question": "(Ljunggren–Jacobsthal super–congruence — sharp exponent, block–multinomial extension)\n\nFix a prime number $p\\ge 5$ and integers $r,s$ with $r\\ge s\\ge 0$.\n\nA.  (Jacobsthal–Ljunggren modulo $p^{3}$)  \nShow that  \n\\[\n\\binom{p\\,r}{p\\,s}\\equiv\\binom{r}{s}\\pmod{p^{3}}. \\tag{1}\n\\]\n\nB.  (How large can the modulus be?)\n\n1.  Prove that  \n\\[\nv_{p}\\!\\Bigl(\\binom{2p}{p}-2\\Bigr)=\n3+v_{p}\\!\\bigl(B_{p-3}\\bigr), \\tag{2}\n\\]\nwhere $B_{k}$ denotes the $k$–th Bernoulli number and $v_{p}$ is\nthe $p$–adic valuation.\n\n2.  Put $\\lambda:=v_{p}(B_{p-3})$.\n\n   (a)  Show that  \n   \\[\n   v_{p}\\!\\Bigl(\\binom{2p}{p}-2\\Bigr)=3+\\lambda .\n   \\]\n\n   (b)  Deduce that for the pair $(r,s)=(2,1)$ the largest exponent\n   $E=E(p)$ for which  \n   \\[\n   \\binom{p\\,r}{p\\,s}\\equiv\\binom{r}{s}\\pmod{p^{E}}\n   \\]\n   holds is precisely $E=3+\\lambda$.  \n   Conclude in particular\n\n   •  If $\\lambda=0$ (equivalently $p\\nmid B_{p-3}$, the generic\n   case) then the modulus $p^{3}$ in (1) is optimal.\n\n   •  If $p$ is a Wolstenholme prime (so that $\\lambda\\ge 1$) the\n   modulus improves to $p^{4}$ or higher according as\n   $\\lambda=1,2,\\dots$, and the optimal exponent is $3+\\lambda$.\n   In particular, for the two currently known Wolstenholme primes\n   one has $\\lambda=1$ and the best modulus is $p^{4}$.\n\nC.  ($d$–fold block–multinomial version)  \nLet a fixed integer $d\\ge 2$ be given.  \nFor vectors\n\\[\n\\mathbf r=(r_{1},\\dots ,r_{d}),\\quad\n\\mathbf s=(s_{1},\\dots ,s_{d}),\\qquad\n0\\le s_{i}\\le r_{i},\\;\\;\n\\sum_{i=1}^{d}r_{i}=r,\\;\n\\sum_{i=1}^{d}s_{i}=s,\n\\]\nwrite $\\alpha_{i}:=r_{i}-s_{i}$ and set\n\\[\nM_{p}(\\mathbf r,\\mathbf s):=\\frac{(p\\,r)!}{\\displaystyle\\prod_{i=1}^{d}(p\\,s_{i})!\\,(p\\,\\alpha_{i})!},\n\\qquad\nM_{1}(\\mathbf r,\\mathbf s):=\\frac{r!}{\\displaystyle\\prod_{i=1}^{d}s_{i}!\\,\\alpha_{i}!}.\n\\]\nProve the super–congruence  \n\\[\nM_{p}(\\mathbf r,\\mathbf s)\\equiv M_{1}(\\mathbf r,\\mathbf s)\\pmod{p^{3}}. \\tag{3}\n\\]\n\n--------------------------------------------------------------------",
      "solution": "Throughout $p\\ge 5$ is fixed, $v_{p}$ denotes the $p$–adic valuation, and  \n\\[\nH_{n,m}:=\\sum_{k=1}^{n}\\frac1{k^{m}},\\qquad H_{n}:=H_{n,1}.\n\\]\nBernoulli numbers are defined by  \n\\[\n\\frac{t}{e^{t}-1}=\\sum_{k=0}^{\\infty}B_{k}\\frac{t^{k}}{k!}.\n\\]\nWe shall make repeated use of  \n\n(F1)  (Wolstenholme) $H_{p-1}\\equiv 0\\pmod{p^{2}}$,\n\n(F2)  $H_{p-1,2}\\equiv 0\\pmod{p}$,\n\n(F3)  $\\displaystyle\\sum_{k=1}^{p-1}k^{m}\\equiv 0\\pmod{p}$ whenever\n$p-1\\nmid m$,\n\nand of the classical congruences due to Glaisher  \n\n(G1)  $H_{p-1}\\equiv-\\dfrac{p^{2}}{3}B_{p-3}\\pmod{p^{3}}$,\n\n(G2)  $H_{p-1,2}\\equiv\\dfrac{2p}{3}B_{p-3}\\pmod{p^{2}}$.\n\n--------------------------------------------------------------------\nA.  Proof of (1)\n\nDefine  \n\\[\nF(r,s):=\\frac{\\binom{p\\,r}{p\\,s}}{\\binom{r}{s}},\\qquad r\\ge s.\n\\]\n\nStep 1.  Block decomposition.  \nWriting the factorials in blocks of length $p$ one obtains the exact\nequality  \n\\[\n\\frac{\\binom{p\\,r}{p\\,s}}{\\binom{r}{s}}\n    =\\prod_{j=1}^{s}\\prod_{t=1}^{p-1}\n      \\frac{p(r-s)+p(j-1)+t}{p(j-1)+t}. \\tag{4}\n\\]\n\nPut  \n\\[\nx_{j,t}:=\\frac{p(r-s)}{p(j-1)+t},\\qquad |x_{j,t}|_{p}=p^{-1}.\n\\]\nBecause $|x_{j,t}|_{p}<1$, the $p$–adic logarithm converges, giving\n\\[\n\\log F(r,s)=\n\\sum_{m\\ge 1}\\frac{(-1)^{m+1}}{m}\\,\n            p^{m}(r-s)^{m}\n            \\sum_{j=1}^{s}\\sum_{t=1}^{p-1}(p(j-1)+t)^{-m}. \\tag{5}\n\\]\n\nStep 2.  Valuation of the inner sums.  \nSet  \n\\[\nS_{m}(s):=\\sum_{u=0}^{s-1}\\sum_{t=1}^{p-1}(p\\,u+t)^{-m}.\n\\]\n\nLemma 1.  One has $v_{p}\\!\\bigl(S_{1}(s)\\bigr)\\ge 2$ and\n$v_{p}\\!\\bigl(S_{2}(s)\\bigr)\\ge 1$.\n\nProof.  \nExpand $(p\\,u+t)^{-1}$ in a geometric series and use (F1)–(F2); the\ndetails are identical to the original solution. ∎\n\nStep 3.  Finishing the valuation.  \nFrom (5) and Lemma 1 one finds that every term in the series for\n$\\log F(r,s)$ has $p$–adic valuation at least $3$, hence\n\\[\nv_{p}\\bigl(\\log F(r,s)\\bigr)\\ge 3,\\qquad\n\\log F(r,s)\\equiv 0\\pmod{p^{3}}.\n\\]\nBecause $p^{3}\\mid\\log F(r,s)$ we may truncate the $p$–adic\nexponential and conclude $F(r,s)\\equiv 1\\pmod{p^{3}}$, i.e. (1). ∎\n\n\n\n--------------------------------------------------------------------\nB.  Sharpness of the exponent\n\nB.1  Exact valuation of $\\displaystyle\\binom{2p}{p}-2$\n\nSet  \n\\[\nG:=\\frac12\\binom{2p}{p}\n     =\\binom{2p-1}{\\,p-1\\,}\n     =\\prod_{k=1}^{p-1}\\Bigl(1+\\frac{p}{k}\\Bigr). \\tag{6}\n\\]\nTaking the $p$–adic logarithm and expanding as in (5) gives  \n\\[\n\\log G=\np\\,H_{p-1}-\\frac{p^{2}}{2}H_{p-1,2}\n+\\frac{p^{3}}{3}H_{p-1,3}\n-\\bigl(\\text{terms divisible by }p^{4}\\bigr). \\tag{7}\n\\]\nInsert (G1)–(G2) and use $p\\mid H_{p-1,3}$ (from (F3)):\n\\[\n\\log G\\equiv\n-\\frac{2}{3}\\,p^{3}B_{p-3}\\pmod{p^{4}}. \\tag{8}\n\\]\nBecause $p^{4}\\mid(\\log G)^{2}$, the $p$–adic exponential truncates\nafter the linear term:\n\\[\nG=\\exp(\\log G)\\equiv\n1-\\frac{2}{3}\\,p^{3}B_{p-3}\\pmod{p^{4}}. \\tag{9}\n\\]\nMultiplying by $2$ yields\n\\[\n\\binom{2p}{p}\\equiv\n2-\\frac{4}{3}\\,p^{3}B_{p-3}\\pmod{p^{4}}. \\tag{10}\n\\]\nSince $-\\dfrac43$ is a $p$–adic unit,\n\\[\nv_{p}\\!\\Bigl(\\binom{2p}{p}-2\\Bigr)=3+v_{p}\\!\\bigl(B_{p-3}\\bigr),\n\\]\nestablishing (2). ∎\n\n\n\nB.2  Consequences for the optimal exponent  \n\nLet $\\lambda:=v_{p}(B_{p-3})$; then by (2)\n\\[\nv_{p}\\!\\Bigl(\\binom{2p}{p}-2\\Bigr)=3+\\lambda . \\tag{11}\n\\]\n\nFor the pair $(r,s)=(2,1)$ we have  \n\\[\n\\binom{p\\,r}{p\\,s}-\\binom{r}{s}=\\binom{2p}{p}-2,\n\\]\nso (11) gives\n\\[\nv_{p}\\!\\Bigl(\\binom{p\\,r}{p\\,s}-\\binom{r}{s}\\Bigr)=3+\\lambda .\n\\]\nTherefore\n\n•  $\\displaystyle\\binom{p\\,r}{p\\,s}\\equiv\\binom{r}{s}\\pmod{p^{3+\\lambda}}$,  \n\n•  $\\displaystyle\\binom{p\\,r}{p\\,s}\\not\\equiv\\binom{r}{s}\\pmod{p^{4+\\lambda}}$.\n\nHence the largest exponent for which the congruence (1) can hold in\ngeneral is $E=3+\\lambda$.\n\nSpecial cases.\n\n(i)  Generic primes ($\\lambda=0$).  \nThen $E=3$, so the modulus $p^{3}$ in (1) is best possible.\n\n(ii)  Wolstenholme primes ($\\lambda\\ge 1$).  \nThe modulus improves to $p^{4}$ or higher; its exact value is\n$p^{3+\\lambda}$.  \nFor the two known Wolstenholme primes $16843$ and $2124679$ one has\n$\\lambda=1$, whence the optimal exponent is $4$. ∎\n\n\n\n--------------------------------------------------------------------\nC.  Proof of the block–multinomial congruence (3)\n\nStep 1.  Iterated–binomial decomposition.  \nIntroduce\n\\[\nk_{2i-1}:=s_{i},\\qquad\nk_{2i}:=\\alpha_{i}=r_{i}-s_{i}\\quad(1\\le i\\le d),\\qquad\nm:=2d,\n\\]\nand partial sums $L_{j}:=\\sum_{i=1}^{j-1}k_{i}$ ($L_{1}=0$).  The\nclassical identity\n\\[\n\\binom{n}{k_{1},\\dots ,k_{m}}=\\prod_{j=1}^{m-1}\\binom{n-L_{j}}{k_{j}}\n\\tag{12}\n\\]\ngives\n\\[\nM_{1}(\\mathbf r,\\mathbf s)\n      =\\prod_{j=1}^{m-1}\\binom{r-L_{j}}{k_{j}},\\qquad\nM_{p}(\\mathbf r,\\mathbf s)\n      =\\prod_{j=1}^{m-1}\\binom{p\\,r-p\\,L_{j}}{p\\,k_{j}}. \\tag{13}\n\\]\n\nStep 2.  Application of Part A.  \nFor every $j$ one has $r-L_{j}\\ge k_{j}\\ge 0$, hence (1) yields  \n\\[\n\\binom{p(r-L_{j})}{p\\,k_{j}}\\equiv\n\\binom{r-L_{j}}{k_{j}}\\pmod{p^{3}}.\n\\]\nDividing (13) by (12) we get\n\\[\n\\frac{M_{p}(\\mathbf r,\\mathbf s)}{M_{1}(\\mathbf r,\\mathbf s)}\n      =\\prod_{j=1}^{m-1}\\bigl(1+p^{3}\\theta_{j}\\bigr)\n      \\equiv 1\\pmod{p^{3}}, \\tag{14}\n\\]\nwith $\\theta_{j}\\in\\mathbb Z_{p}$.  Multiplying by\n$M_{1}(\\mathbf r,\\mathbf s)$ proves (3). ∎\n\n\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.631593",
        "was_fixed": false,
        "difficulty_analysis": "1.  The original problem required only a first-level Freshman-Dream argument modulo p.  \n    The enhanced variant asks for a congruence two orders deeper (mod p³) and is known classically as Ljunggren’s theorem; its proof forces the solver to manipulate p-adic logarithms/ exponentials or, equivalently, to perform delicate denominator–sums that do not appear at all in the original exercise.\n\n2.  Optimality (part B) obliges the contestant not only to prove a congruence but also to understand its exact strength, constructing explicit counter-examples beyond p³.  This uses detailed expansions of binomial quotients and harmonic sums mod high powers of p.\n\n3.  Part C couples the higher-order binomial congruence with a multidimensional generalisation, intertwining multinomial coefficients and the earlier p-adic analysis.  Handling several interacting indices simultaneously and showing that all error terms cancel demands an even subtler bookkeeping of congruences.\n\n4.  Techniques needed now include  \n   • $p$-adic valuations and carries (Kummer theory),  \n   • $p$-adic logarithm/exponential series,  \n   • harmonic-sum congruences (Wolstenholme type), and  \n   • factorisations of multinomial quotients,  \nall of which are far beyond the elementary polynomial manipulation sufficient for the original problem.  \nThus the new kernel variant is substantially, demonstrably harder."
      }
    },
    "original_kernel_variant": {
      "question": "(Ljunggren–Jacobsthal super–congruence — sharp exponent, block–multinomial extension)\n\nFix a prime number $p\\ge 5$ and integers $r,s$ with $r\\ge s\\ge 0$.\n\nA.  (Jacobsthal–Ljunggren modulo $p^{3}$)  \nShow that  \n\\[\n\\binom{p\\,r}{p\\,s}\\equiv\\binom{r}{s}\\pmod{p^{3}}. \\tag{1}\n\\]\n\nB.  (How large can the modulus be?)\n\n1.  Prove that  \n\\[\nv_{p}\\!\\Bigl(\\binom{2p}{p}-2\\Bigr)=\n3+v_{p}\\!\\bigl(B_{p-3}\\bigr), \\tag{2}\n\\]\nwhere $B_{k}$ denotes the $k$–th Bernoulli number and $v_{p}$ is\nthe $p$–adic valuation.\n\n2.  Put $\\lambda:=v_{p}(B_{p-3})$.\n\n   (a)  Show that  \n   \\[\n   v_{p}\\!\\Bigl(\\binom{2p}{p}-2\\Bigr)=3+\\lambda .\n   \\]\n\n   (b)  Deduce that for the pair $(r,s)=(2,1)$ the largest exponent\n   $E=E(p)$ for which  \n   \\[\n   \\binom{p\\,r}{p\\,s}\\equiv\\binom{r}{s}\\pmod{p^{E}}\n   \\]\n   holds is precisely $E=3+\\lambda$.  \n   Conclude in particular\n\n   •  If $\\lambda=0$ (equivalently $p\\nmid B_{p-3}$, the generic\n   case) then the modulus $p^{3}$ in (1) is optimal.\n\n   •  If $p$ is a Wolstenholme prime (so that $\\lambda\\ge 1$) the\n   modulus improves to $p^{4}$ or higher according as\n   $\\lambda=1,2,\\dots$, and the optimal exponent is $3+\\lambda$.\n   In particular, for the two currently known Wolstenholme primes\n   one has $\\lambda=1$ and the best modulus is $p^{4}$.\n\nC.  ($d$–fold block–multinomial version)  \nLet a fixed integer $d\\ge 2$ be given.  \nFor vectors\n\\[\n\\mathbf r=(r_{1},\\dots ,r_{d}),\\quad\n\\mathbf s=(s_{1},\\dots ,s_{d}),\\qquad\n0\\le s_{i}\\le r_{i},\\;\\;\n\\sum_{i=1}^{d}r_{i}=r,\\;\n\\sum_{i=1}^{d}s_{i}=s,\n\\]\nwrite $\\alpha_{i}:=r_{i}-s_{i}$ and set\n\\[\nM_{p}(\\mathbf r,\\mathbf s):=\\frac{(p\\,r)!}{\\displaystyle\\prod_{i=1}^{d}(p\\,s_{i})!\\,(p\\,\\alpha_{i})!},\n\\qquad\nM_{1}(\\mathbf r,\\mathbf s):=\\frac{r!}{\\displaystyle\\prod_{i=1}^{d}s_{i}!\\,\\alpha_{i}!}.\n\\]\nProve the super–congruence  \n\\[\nM_{p}(\\mathbf r,\\mathbf s)\\equiv M_{1}(\\mathbf r,\\mathbf s)\\pmod{p^{3}}. \\tag{3}\n\\]\n\n--------------------------------------------------------------------",
      "solution": "Throughout $p\\ge 5$ is fixed, $v_{p}$ denotes the $p$–adic valuation, and  \n\\[\nH_{n,m}:=\\sum_{k=1}^{n}\\frac1{k^{m}},\\qquad H_{n}:=H_{n,1}.\n\\]\nBernoulli numbers are defined by  \n\\[\n\\frac{t}{e^{t}-1}=\\sum_{k=0}^{\\infty}B_{k}\\frac{t^{k}}{k!}.\n\\]\nWe shall make repeated use of  \n\n(F1)  (Wolstenholme) $H_{p-1}\\equiv 0\\pmod{p^{2}}$,\n\n(F2)  $H_{p-1,2}\\equiv 0\\pmod{p}$,\n\n(F3)  $\\displaystyle\\sum_{k=1}^{p-1}k^{m}\\equiv 0\\pmod{p}$ whenever\n$p-1\\nmid m$,\n\nand of the classical congruences due to Glaisher  \n\n(G1)  $H_{p-1}\\equiv-\\dfrac{p^{2}}{3}B_{p-3}\\pmod{p^{3}}$,\n\n(G2)  $H_{p-1,2}\\equiv\\dfrac{2p}{3}B_{p-3}\\pmod{p^{2}}$.\n\n--------------------------------------------------------------------\nA.  Proof of (1)\n\nDefine  \n\\[\nF(r,s):=\\frac{\\binom{p\\,r}{p\\,s}}{\\binom{r}{s}},\\qquad r\\ge s.\n\\]\n\nStep 1.  Block decomposition.  \nWriting the factorials in blocks of length $p$ one obtains the exact\nequality  \n\\[\n\\frac{\\binom{p\\,r}{p\\,s}}{\\binom{r}{s}}\n    =\\prod_{j=1}^{s}\\prod_{t=1}^{p-1}\n      \\frac{p(r-s)+p(j-1)+t}{p(j-1)+t}. \\tag{4}\n\\]\n\nPut  \n\\[\nx_{j,t}:=\\frac{p(r-s)}{p(j-1)+t},\\qquad |x_{j,t}|_{p}=p^{-1}.\n\\]\nBecause $|x_{j,t}|_{p}<1$, the $p$–adic logarithm converges, giving\n\\[\n\\log F(r,s)=\n\\sum_{m\\ge 1}\\frac{(-1)^{m+1}}{m}\\,\n            p^{m}(r-s)^{m}\n            \\sum_{j=1}^{s}\\sum_{t=1}^{p-1}(p(j-1)+t)^{-m}. \\tag{5}\n\\]\n\nStep 2.  Valuation of the inner sums.  \nSet  \n\\[\nS_{m}(s):=\\sum_{u=0}^{s-1}\\sum_{t=1}^{p-1}(p\\,u+t)^{-m}.\n\\]\n\nLemma 1.  One has $v_{p}\\!\\bigl(S_{1}(s)\\bigr)\\ge 2$ and\n$v_{p}\\!\\bigl(S_{2}(s)\\bigr)\\ge 1$.\n\nProof.  \nExpand $(p\\,u+t)^{-1}$ in a geometric series and use (F1)–(F2); the\ndetails are identical to the original solution. ∎\n\nStep 3.  Finishing the valuation.  \nFrom (5) and Lemma 1 one finds that every term in the series for\n$\\log F(r,s)$ has $p$–adic valuation at least $3$, hence\n\\[\nv_{p}\\bigl(\\log F(r,s)\\bigr)\\ge 3,\\qquad\n\\log F(r,s)\\equiv 0\\pmod{p^{3}}.\n\\]\nBecause $p^{3}\\mid\\log F(r,s)$ we may truncate the $p$–adic\nexponential and conclude $F(r,s)\\equiv 1\\pmod{p^{3}}$, i.e. (1). ∎\n\n\n\n--------------------------------------------------------------------\nB.  Sharpness of the exponent\n\nB.1  Exact valuation of $\\displaystyle\\binom{2p}{p}-2$\n\nSet  \n\\[\nG:=\\frac12\\binom{2p}{p}\n     =\\binom{2p-1}{\\,p-1\\,}\n     =\\prod_{k=1}^{p-1}\\Bigl(1+\\frac{p}{k}\\Bigr). \\tag{6}\n\\]\nTaking the $p$–adic logarithm and expanding as in (5) gives  \n\\[\n\\log G=\np\\,H_{p-1}-\\frac{p^{2}}{2}H_{p-1,2}\n+\\frac{p^{3}}{3}H_{p-1,3}\n-\\bigl(\\text{terms divisible by }p^{4}\\bigr). \\tag{7}\n\\]\nInsert (G1)–(G2) and use $p\\mid H_{p-1,3}$ (from (F3)):\n\\[\n\\log G\\equiv\n-\\frac{2}{3}\\,p^{3}B_{p-3}\\pmod{p^{4}}. \\tag{8}\n\\]\nBecause $p^{4}\\mid(\\log G)^{2}$, the $p$–adic exponential truncates\nafter the linear term:\n\\[\nG=\\exp(\\log G)\\equiv\n1-\\frac{2}{3}\\,p^{3}B_{p-3}\\pmod{p^{4}}. \\tag{9}\n\\]\nMultiplying by $2$ yields\n\\[\n\\binom{2p}{p}\\equiv\n2-\\frac{4}{3}\\,p^{3}B_{p-3}\\pmod{p^{4}}. \\tag{10}\n\\]\nSince $-\\dfrac43$ is a $p$–adic unit,\n\\[\nv_{p}\\!\\Bigl(\\binom{2p}{p}-2\\Bigr)=3+v_{p}\\!\\bigl(B_{p-3}\\bigr),\n\\]\nestablishing (2). ∎\n\n\n\nB.2  Consequences for the optimal exponent  \n\nLet $\\lambda:=v_{p}(B_{p-3})$; then by (2)\n\\[\nv_{p}\\!\\Bigl(\\binom{2p}{p}-2\\Bigr)=3+\\lambda . \\tag{11}\n\\]\n\nFor the pair $(r,s)=(2,1)$ we have  \n\\[\n\\binom{p\\,r}{p\\,s}-\\binom{r}{s}=\\binom{2p}{p}-2,\n\\]\nso (11) gives\n\\[\nv_{p}\\!\\Bigl(\\binom{p\\,r}{p\\,s}-\\binom{r}{s}\\Bigr)=3+\\lambda .\n\\]\nTherefore\n\n•  $\\displaystyle\\binom{p\\,r}{p\\,s}\\equiv\\binom{r}{s}\\pmod{p^{3+\\lambda}}$,  \n\n•  $\\displaystyle\\binom{p\\,r}{p\\,s}\\not\\equiv\\binom{r}{s}\\pmod{p^{4+\\lambda}}$.\n\nHence the largest exponent for which the congruence (1) can hold in\ngeneral is $E=3+\\lambda$.\n\nSpecial cases.\n\n(i)  Generic primes ($\\lambda=0$).  \nThen $E=3$, so the modulus $p^{3}$ in (1) is best possible.\n\n(ii)  Wolstenholme primes ($\\lambda\\ge 1$).  \nThe modulus improves to $p^{4}$ or higher; its exact value is\n$p^{3+\\lambda}$.  \nFor the two known Wolstenholme primes $16843$ and $2124679$ one has\n$\\lambda=1$, whence the optimal exponent is $4$. ∎\n\n\n\n--------------------------------------------------------------------\nC.  Proof of the block–multinomial congruence (3)\n\nStep 1.  Iterated–binomial decomposition.  \nIntroduce\n\\[\nk_{2i-1}:=s_{i},\\qquad\nk_{2i}:=\\alpha_{i}=r_{i}-s_{i}\\quad(1\\le i\\le d),\\qquad\nm:=2d,\n\\]\nand partial sums $L_{j}:=\\sum_{i=1}^{j-1}k_{i}$ ($L_{1}=0$).  The\nclassical identity\n\\[\n\\binom{n}{k_{1},\\dots ,k_{m}}=\\prod_{j=1}^{m-1}\\binom{n-L_{j}}{k_{j}}\n\\tag{12}\n\\]\ngives\n\\[\nM_{1}(\\mathbf r,\\mathbf s)\n      =\\prod_{j=1}^{m-1}\\binom{r-L_{j}}{k_{j}},\\qquad\nM_{p}(\\mathbf r,\\mathbf s)\n      =\\prod_{j=1}^{m-1}\\binom{p\\,r-p\\,L_{j}}{p\\,k_{j}}. \\tag{13}\n\\]\n\nStep 2.  Application of Part A.  \nFor every $j$ one has $r-L_{j}\\ge k_{j}\\ge 0$, hence (1) yields  \n\\[\n\\binom{p(r-L_{j})}{p\\,k_{j}}\\equiv\n\\binom{r-L_{j}}{k_{j}}\\pmod{p^{3}}.\n\\]\nDividing (13) by (12) we get\n\\[\n\\frac{M_{p}(\\mathbf r,\\mathbf s)}{M_{1}(\\mathbf r,\\mathbf s)}\n      =\\prod_{j=1}^{m-1}\\bigl(1+p^{3}\\theta_{j}\\bigr)\n      \\equiv 1\\pmod{p^{3}}, \\tag{14}\n\\]\nwith $\\theta_{j}\\in\\mathbb Z_{p}$.  Multiplying by\n$M_{1}(\\mathbf r,\\mathbf s)$ proves (3). ∎\n\n\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.503578",
        "was_fixed": false,
        "difficulty_analysis": "1.  The original problem required only a first-level Freshman-Dream argument modulo p.  \n    The enhanced variant asks for a congruence two orders deeper (mod p³) and is known classically as Ljunggren’s theorem; its proof forces the solver to manipulate p-adic logarithms/ exponentials or, equivalently, to perform delicate denominator–sums that do not appear at all in the original exercise.\n\n2.  Optimality (part B) obliges the contestant not only to prove a congruence but also to understand its exact strength, constructing explicit counter-examples beyond p³.  This uses detailed expansions of binomial quotients and harmonic sums mod high powers of p.\n\n3.  Part C couples the higher-order binomial congruence with a multidimensional generalisation, intertwining multinomial coefficients and the earlier p-adic analysis.  Handling several interacting indices simultaneously and showing that all error terms cancel demands an even subtler bookkeeping of congruences.\n\n4.  Techniques needed now include  \n   • $p$-adic valuations and carries (Kummer theory),  \n   • $p$-adic logarithm/exponential series,  \n   • harmonic-sum congruences (Wolstenholme type), and  \n   • factorisations of multinomial quotients,  \nall of which are far beyond the elementary polynomial manipulation sufficient for the original problem.  \nThus the new kernel variant is substantially, demonstrably harder."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}