{
  "index": "1986-B-4",
  "type": "NT",
  "tag": [
    "NT",
    "ANA"
  ],
  "difficulty": "",
  "question": "For a positive real number $r$, let $G(r)$ be the minimum value of $|r\n- \\sqrt{m^2+2n^2}|$ for all integers $m$ and $n$. Prove or disprove\nthe assertion that $\\lim_{r\\to \\infty}G(r)$ exists and equals 0.",
  "solution": "Solution (Doug Jungreis). First,\n\\[\n0 \\leq\\left|r-\\sqrt{m^{2}+2 n^{2}}\\right|=\\frac{\\left|r^{2}-m^{2}-2 n^{2}\\right|}{r+\\sqrt{m^{2}+2 n^{2}}} \\leq \\frac{\\left|r^{2}-m^{2}-2 n^{2}\\right|}{r},\n\\]\nso it will suffice to bound the latter expression. Select the largest integer \\( m \\geq 0 \\) such that \\( r^{2}-m^{2} \\geq 0 \\). Then \\( m^{2} \\leq r^{2}<(m+1)^{2} \\), so \\( m \\leq r \\) and \\( r^{2}-m^{2}<2 m+1 \\). Next select the largest integer \\( n \\geq 0 \\) such that \\( r^{2}-m^{2}-2 n^{2} \\geq 0 \\). Then \\( 2 n^{2} \\leq r^{2}-m^{2}< \\) \\( 2(n+1)^{2} \\). This implies \\( n \\leq \\sqrt{\\left(r^{2}-m^{2}\\right) / 2} \\) and\n\\[\n\\begin{aligned}\n\\left|r^{2}-m^{2}-2 n^{2}\\right| & =r^{2}-m^{2}-2 n^{2} \\\\\n& <2(2 n+1) \\\\\n& \\leq 2+4 \\sqrt{\\left(r^{2}-m^{2}\\right) / 2} \\\\\n& \\leq 2+\\sqrt{2 m+1} \\\\\n& \\leq 2+\\sqrt{2 r+1}\n\\end{aligned}\n\\]\n\nHence \\( G(r) \\leq(2+\\sqrt{2 r+1}) / r \\) and \\( \\lim _{r \\rightarrow \\infty} G(r)=0 \\).",
  "vars": [
    "r",
    "m",
    "n"
  ],
  "params": [
    "G"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "r": "realvar",
        "m": "integerfirst",
        "n": "integersecond",
        "G": "gapfunc"
      },
      "question": "For a positive real number \\(realvar\\), let \\(gapfunc(realvar)\\) be the minimum value of \\(\\left|realvar - \\sqrt{integerfirst^{2}+2 integersecond^{2}}\\right|\\) for all integers \\(integerfirst\\) and \\(integersecond\\). Prove or disprove the assertion that \\(\\lim_{realvar\\to \\infty}gapfunc(realvar)\\) exists and equals 0.",
      "solution": "Solution (Doug Jungreis). First,\n\\[\n0 \\leq\\left|realvar-\\sqrt{integerfirst^{2}+2 integersecond^{2}}\\right|=\\frac{\\left|realvar^{2}-integerfirst^{2}-2 integersecond^{2}\\right|}{realvar+\\sqrt{integerfirst^{2}+2 integersecond^{2}}} \\leq \\frac{\\left|realvar^{2}-integerfirst^{2}-2 integersecond^{2}\\right|}{realvar},\n\\]\nso it will suffice to bound the latter expression. Select the largest integer \\( integerfirst \\geq 0 \\) such that \\( realvar^{2}-integerfirst^{2} \\geq 0 \\). Then \\( integerfirst^{2} \\leq realvar^{2}<(integerfirst+1)^{2} \\), so \\( integerfirst \\leq realvar \\) and \\( realvar^{2}-integerfirst^{2}<2 integerfirst+1 \\). Next select the largest integer \\( integersecond \\geq 0 \\) such that \\( realvar^{2}-integerfirst^{2}-2 integersecond^{2} \\geq 0 \\). Then \\( 2 integersecond^{2} \\leq realvar^{2}-integerfirst^{2}< 2(integersecond+1)^{2} \\). This implies \\( integersecond \\leq \\sqrt{\\left(realvar^{2}-integerfirst^{2}\\right) / 2} \\) and\n\\[\n\\begin{aligned}\n\\left|realvar^{2}-integerfirst^{2}-2 integersecond^{2}\\right| & =realvar^{2}-integerfirst^{2}-2 integersecond^{2} \\\\\n& <2(2 integersecond+1) \\\\\n& \\leq 2+4 \\sqrt{\\left(realvar^{2}-integerfirst^{2}\\right) / 2} \\\\\n& \\leq 2+\\sqrt{2 integerfirst+1} \\\\\n& \\leq 2+\\sqrt{2 realvar+1}\n\\end{aligned}\n\\]\n\nHence \\( gapfunc(realvar) \\leq(2+\\sqrt{2 realvar+1}) / realvar \\) and \\( \\lim _{realvar \\rightarrow \\infty} gapfunc(realvar)=0 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "r": "meadowland",
        "m": "turnpike",
        "n": "paperback",
        "G": "labyrinth"
      },
      "question": "For a positive real number $meadowland$, let $labyrinth(meadowland)$ be the minimum value of $|meadowland\n- \\sqrt{turnpike^2+2paperback^2}|$ for all integers $turnpike$ and $paperback$. Prove or disprove\nthe assertion that $\\lim_{meadowland\\to \\infty}labyrinth(meadowland)$ exists and equals 0.",
      "solution": "Solution (Doug Jungreis). First,\n\\[\n0 \\leq\\left|meadowland-\\sqrt{turnpike^{2}+2 paperback^{2}}\\right|=\\frac{\\left|meadowland^{2}-turnpike^{2}-2 paperback^{2}\\right|}{meadowland+\\sqrt{turnpike^{2}+2 paperback^{2}}} \\leq \\frac{\\left|meadowland^{2}-turnpike^{2}-2 paperback^{2}\\right|}{meadowland},\n\\]\nso it will suffice to bound the latter expression. Select the largest integer \\( turnpike \\geq 0 \\) such that \\( meadowland^{2}-turnpike^{2} \\geq 0 \\). Then \\( turnpike^{2} \\leq meadowland^{2}<(turnpike+1)^{2} \\), so \\( turnpike \\leq meadowland \\) and \\( meadowland^{2}-turnpike^{2}<2 turnpike+1 \\). Next select the largest integer \\( paperback \\geq 0 \\) such that \\( meadowland^{2}-turnpike^{2}-2 paperback^{2} \\geq 0 \\). Then \\( 2 paperback^{2} \\leq meadowland^{2}-turnpike^{2}< 2(paperback+1)^{2} \\). This implies \\( paperback \\leq \\sqrt{\\left(meadowland^{2}-turnpike^{2}\\right) / 2} \\) and\n\\[\n\\begin{aligned}\n\\left|meadowland^{2}-turnpike^{2}-2 paperback^{2}\\right| & =meadowland^{2}-turnpike^{2}-2 paperback^{2} \\\\\n& <2(2 paperback+1) \\\\\n& \\leq 2+4 \\sqrt{\\left(meadowland^{2}-turnpike^{2}\\right) / 2} \\\\\n& \\leq 2+\\sqrt{2 turnpike+1} \\\\\n& \\leq 2+\\sqrt{2 meadowland+1}\n\\end{aligned}\n\\]\n\nHence \\( labyrinth(meadowland) \\leq(2+\\sqrt{2 meadowland+1}) / meadowland \\) and \\( \\lim _{meadowland \\rightarrow \\infty} labyrinth(meadowland)=0 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "r": "staticvalue",
        "m": "irrationalvalue",
        "n": "fractionalvalue",
        "G": "maximalfunc"
      },
      "question": "For a positive real number $staticvalue$, let $maximalfunc(staticvalue)$ be the minimum value of $|staticvalue - \\sqrt{irrationalvalue^2+2fractionalvalue^2}|$ for all integers $irrationalvalue$ and $fractionalvalue$. Prove or disprove the assertion that $\\lim_{staticvalue\\to \\infty}maximalfunc(staticvalue)$ exists and equals 0.",
      "solution": "Solution (Doug Jungreis). First,\n\\[\n0 \\leq\\left|staticvalue-\\sqrt{irrationalvalue^{2}+2 fractionalvalue^{2}}\\right|=\\frac{\\left|staticvalue^{2}-irrationalvalue^{2}-2 fractionalvalue^{2}\\right|}{staticvalue+\\sqrt{irrationalvalue^{2}+2 fractionalvalue^{2}}} \\leq \\frac{\\left|staticvalue^{2}-irrationalvalue^{2}-2 fractionalvalue^{2}\\right|}{staticvalue},\n\\]\nso it will suffice to bound the latter expression. Select the largest integer \\( irrationalvalue \\geq 0 \\) such that \\( staticvalue^{2}-irrationalvalue^{2} \\geq 0 \\). Then \\( irrationalvalue^{2} \\leq staticvalue^{2}<(irrationalvalue+1)^{2} \\), so \\( irrationalvalue \\leq staticvalue \\) and \\( staticvalue^{2}-irrationalvalue^{2}<2 irrationalvalue+1 \\). Next select the largest integer \\( fractionalvalue \\geq 0 \\) such that \\( staticvalue^{2}-irrationalvalue^{2}-2 fractionalvalue^{2} \\geq 0 \\). Then \\( 2 fractionalvalue^{2} \\leq staticvalue^{2}-irrationalvalue^{2}< \\) \\( 2(fractionalvalue+1)^{2} \\). This implies \\( fractionalvalue \\leq \\sqrt{\\left(staticvalue^{2}-irrationalvalue^{2}\\right) / 2} \\) and\n\\[\n\\begin{aligned}\n\\left|staticvalue^{2}-irrationalvalue^{2}-2 fractionalvalue^{2}\\right| & =staticvalue^{2}-irrationalvalue^{2}-2 fractionalvalue^{2} \\\\\n& <2(2 fractionalvalue+1) \\\\\n& \\leq 2+4 \\sqrt{\\left(staticvalue^{2}-irrationalvalue^{2}\\right) / 2} \\\\\n& \\leq 2+\\sqrt{2 irrationalvalue+1} \\\\\n& \\leq 2+\\sqrt{2 staticvalue+1}\n\\end{aligned}\n\\]\nHence \\( maximalfunc(staticvalue) \\leq(2+\\sqrt{2 staticvalue+1}) / staticvalue \\) and \\( \\lim _{staticvalue \\rightarrow \\infty} maximalfunc(staticvalue)=0 \\)."
    },
    "garbled_string": {
      "map": {
        "r": "qzxwvtnp",
        "m": "hjgrksla",
        "n": "bplqvsmc",
        "G": "kdfghjql"
      },
      "question": "For a positive real number $qzxwvtnp$, let $kdfghjql(qzxwvtnp)$ be the minimum value of $|qzxwvtnp\n- \\sqrt{hjgrksla^2+2bplqvsmc^2}|$ for all integers $hjgrksla$ and $bplqvsmc$. Prove or disprove\nthe assertion that $\\lim_{qzxwvtnp\\to \\infty}kdfghjql(qzxwvtnp)$ exists and equals 0.",
      "solution": "Solution (Doug Jungreis). First,\n\\[\n0 \\leq\\left|qzxwvtnp-\\sqrt{hjgrksla^{2}+2 bplqvsmc^{2}}\\right|=\\frac{\\left|qzxwvtnp^{2}-hjgrksla^{2}-2 bplqvsmc^{2}\\right|}{qzxwvtnp+\\sqrt{hjgrksla^{2}+2 bplqvsmc^{2}}} \\leq \\frac{\\left|qzxwvtnp^{2}-hjgrksla^{2}-2 bplqvsmc^{2}\\right|}{qzxwvtnp},\n\\]\nso it will suffice to bound the latter expression. Select the largest integer \\( hjgrksla \\geq 0 \\) such that \\( qzxwvtnp^{2}-hjgrksla^{2} \\geq 0 \\). Then \\( hjgrksla^{2} \\leq qzxwvtnp^{2}<(hjgrksla+1)^{2} \\), so \\( hjgrksla \\leq qzxwvtnp \\) and \\( qzxwvtnp^{2}-hjgrksla^{2}<2 hjgrksla+1 \\). Next select the largest integer \\( bplqvsmc \\geq 0 \\) such that \\( qzxwvtnp^{2}-hjgrksla^{2}-2 bplqvsmc^{2} \\geq 0 \\). Then \\( 2 bplqvsmc^{2} \\leq qzxwvtnp^{2}-hjgrksla^{2}< 2(bplqvsmc+1)^{2} \\). This implies \\( bplqvsmc \\leq \\sqrt{\\left(qzxwvtnp^{2}-hjgrksla^{2}\\right) / 2} \\) and\n\\[\n\\begin{aligned}\n\\left|qzxwvtnp^{2}-hjgrksla^{2}-2 bplqvsmc^{2}\\right| & =qzxwvtnp^{2}-hjgrksla^{2}-2 bplqvsmc^{2} \\\\\n& <2(2 bplqvsmc+1) \\\\\n& \\leq 2+4 \\sqrt{\\left(qzxwvtnp^{2}-hjgrksla^{2}\\right) / 2} \\\\\n& \\leq 2+\\sqrt{2 hjgrksla+1} \\\\\n& \\leq 2+\\sqrt{2 qzxwvtnp+1}\n\\end{aligned}\n\\]\n\nHence \\( kdfghjql(qzxwvtnp) \\leq(2+\\sqrt{2 qzxwvtnp+1}) / qzxwvtnp \\) and \\( \\lim _{qzxwvtnp \\rightarrow \\infty} kdfghjql(qzxwvtnp)=0 \\)."
    },
    "kernel_variant": {
      "question": "Fix an integer $d\\ge 3$ and positive integers  \n\\[\n0<c_{1}<c_{2}<\\dots <c_{d}.\n\\]\n(They need not be coprime.)  \nFor every real number $r>0$ define  \n\\[\n\\boxed{%\nF_{d}(r)=\\min_{(m_{1},\\dots ,m_{d})\\in\\mathbb Z^{\\,d}}\n      \\Bigl|\\,r-\\sqrt{c_{1}m_{1}^{2}+c_{2}m_{2}^{2}+\\dots +c_{d}m_{d}^{2}}\\,\\Bigr|\n}.\n\\]\n\n(a)  Show that the limit  \n\\[\n\\lim_{r\\to\\infty}F_{d}(r)\n\\]\nexists and equals $0$.\n\n(b)  Prove that there is a constant $C=C(d,\\mathbf c)>0$ (depending only on $d$ and the fixed coefficients $c_{1},\\dots ,c_{d}$) such that for all $r\\ge 1$\n\\[\nF_{d}(r)\\le C\\,r^{-1+2^{\\,1-d}}.\n\\]\n\nConsequently every sufficiently large positive real number can be approximated arbitrarily well by a square‐root of the quadratic form  \n\\[\nQ_{d}(m_{1},\\dots ,m_{d})=\\sum_{i=1}^{d}c_{i}m_{i}^{2},\n\\]\nand the approximation error decays strictly faster than $r^{-1/2}$, approaching the optimal exponent $-1$ as $d\\to\\infty$.",
      "solution": "Throughout we assume $r\\ge 1$.  Constants $A_{k}$, $C$ depend only on $d$ and the fixed weights $c_{1},\\dots ,c_{d}$.\n\nStep 0 – A one–dimensional estimate  \nFor $t\\ge 0$ put $n=\\lfloor\\sqrt t\\rfloor$.  Then consecutive squares differ by $(n+1)^{2}-n^{2}=2n+1$, hence\n\\[\n\\lvert\\,t-n^{2}\\,\\rvert=t-n^{2}\\le 2n+1\\le 2\\sqrt t+1. \\tag{1}\n\\]\n\nWe shall invoke (1) once for every coefficient $c_{k}$.\n\n\\bigskip\nStep 1 – Constructing the integers $m_{1},\\dots ,m_{d}$  \n\nLet\n\\[\n\\delta_{0}:=r^{2}.\n\\]\n\nFor $k=1,2,\\dots ,d$ set  \n\\[\nm_{k}:=\\Bigl\\lfloor\\sqrt{\\delta_{k-1}/c_{k}}\\Bigr\\rfloor\n\\quad\\bigl(\\text{so }c_{k}m_{k}^{2}\\le\\delta_{k-1}<c_{k}(m_{k}+1)^{2}\\bigr),\n\\]\nand define\n\\[\n\\delta_{k}:=\\delta_{k-1}-c_{k}m_{k}^{2}\\ge 0.\n\\]\n\nApplying (1) with $t=\\delta_{k-1}/c_{k}$ gives\n\\[\n\\delta_{k}\n =\\delta_{k-1}-c_{k}m_{k}^{2}\n \\le c_{k}\\bigl(2\\sqrt{\\tfrac{\\delta_{k-1}}{c_{k}}}+1\\bigr)\n =2\\sqrt{c_{k}\\,\\delta_{k-1}}+c_{k}. \\tag{2}\n\\]\n\n\\bigskip\nStep 2 – A uniform bound for $\\delta_{k}$  \n\nPut\n\\[\n\\beta_{k}:=2^{\\,1-k}\\qquad(0\\le k\\le d).\n\\]\nObserve $\\beta_{0}=2$, and $\\beta_{k}=\\tfrac12\\,\\beta_{k-1}$.\n\nWe claim that there exist positive constants $A_{k}$ such that\n\\[\n\\boxed{\\;\\;\\delta_{k}\\le A_{k}\\,r^{\\beta_{k}}\\quad(0\\le k\\le d).\\;} \\tag{3}\n\\]\n\nProof by induction on $k$.\n\nBase case $k=0$.  Choosing $A_{0}:=1$ gives $\\delta_{0}=r^{2}=A_{0}\\,r^{\\beta_{0}}$.\n\nInductive step.  \nAssume $\\delta_{k-1}\\le A_{k-1}\\,r^{\\beta_{k-1}}$.  \nFrom (2) we get\n\\[\n\\begin{aligned}\n\\delta_{k}\n &\\le 2\\sqrt{c_{k}A_{k-1}}\\,r^{\\beta_{k-1}/2}+c_{k}\\\\\n &\\;= 2\\sqrt{c_{k}A_{k-1}}\\,r^{\\beta_{k}}+c_{k}.\n\\end{aligned}\n\\]\nBecause $r\\ge 1$ and $\\beta_{k}>0$, $r^{\\beta_{k}}\\ge 1$, so\n\\[\n\\delta_{k}\\le\\bigl(2\\sqrt{c_{k}A_{k-1}}+c_{k}\\bigr)\\,r^{\\beta_{k}}\n           =:A_{k}\\,r^{\\beta_{k}}.\n\\]\nThus (3) is proved for all $k$.\n\n\\bigskip\nStep 3 – Bounding $F_{d}(r)$  \n\nLet\n\\[\nQ_{d}(m_{1},\\dots ,m_{d})=\\sum_{i=1}^{d}c_{i}m_{i}^{2},\n\\qquad\nM:=(m_{1},\\dots ,m_{d})\n\\]\nbe the vector constructed above.  Then\n\\[\nQ_{d}(M)=r^{2}-\\delta_{d}.\n\\]\nTherefore\n\\[\nF_{d}(r)\\le \\Bigl|\\,r-\\sqrt{r^{2}-\\delta_{d}}\\,\\Bigr|\n           =\\frac{\\delta_{d}}{r+\\sqrt{r^{2}-\\delta_{d}}}\n           \\le\\frac{\\delta_{d}}{r}. \\tag{4}\n\\]\n\nUsing (3) with $k=d$ and $\\beta_{d}=2^{\\,1-d}$ we find\n\\[\n\\delta_{d}\\le A_{d}\\,r^{\\,2^{\\,1-d}}.\n\\]\nInsert this into (4):\n\\[\nF_{d}(r)\\le A_{d}\\,r^{-1+2^{\\,1-d}}=:C\\,r^{-1+2^{\\,1-d}}. \\tag{5}\n\\]\n\n\\bigskip\nStep 4 – Existence of the limit (part (a))  \n\nBecause $F_{d}(r)\\ge 0$ and (5) yields\n\\[\n0\\le F_{d}(r)\\le C\\,r^{-1+2^{\\,1-d}}\\xrightarrow[r\\to\\infty]{}0,\n\\]\nwe have\n\\[\n\\boxed{\\displaystyle\\lim_{r\\to\\infty}F_{d}(r)=0}.\n\\]\n\nInequality (5) completes the proof of both parts (a) and (b). \\qed",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.696128",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension / more variables – the problem involves an arbitrary number \\(d\\ge 3\\) of integer variables instead of two, vastly enlarging the search space.\n\n2.  Additional requirements – besides showing that the limit is \\(0\\), the solver must derive an explicit decay rate \\(O\\!\\bigl(r^{-1+2^{1-d}}\\bigr)\\), which tightens the quantitative control dramatically.\n\n3.  Sophisticated structures – the proof has to organise an iterative approximation scheme for a general positive–definite quadratic form with pairwise coprime coefficients; it cannot rely on simple “one-axis” tricks.\n\n4.  Deeper theory – one needs elementary geometry-of-numbers ideas (successive refinement, bounding gaps between lattice norms) and a careful inductive analysis to track the error term through \\(d\\) refinement stages.\n\n5.  Multiple interacting concepts – the solution juggles lattice points, quadratic forms, successive minima, and inductive error propagation simultaneously; each refinement step interacts with the previous ones through inequality (2).\n\nAll these layers of complexity render the enhanced variant substantially more challenging than both the original and the previous kernel versions."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer $d\\ge 3$ and positive integers  \n\\[\n0<c_{1}<c_{2}<\\dots <c_{d}.\n\\]\n(They need not be coprime.)  \nFor every real number $r>0$ define  \n\\[\n\\boxed{%\nF_{d}(r)=\\min_{(m_{1},\\dots ,m_{d})\\in\\mathbb Z^{\\,d}}\n      \\Bigl|\\,r-\\sqrt{c_{1}m_{1}^{2}+c_{2}m_{2}^{2}+\\dots +c_{d}m_{d}^{2}}\\,\\Bigr|\n}.\n\\]\n\n(a)  Show that the limit  \n\\[\n\\lim_{r\\to\\infty}F_{d}(r)\n\\]\nexists and equals $0$.\n\n(b)  Prove that there is a constant $C=C(d,\\mathbf c)>0$ (depending only on $d$ and the fixed coefficients $c_{1},\\dots ,c_{d}$) such that for all $r\\ge 1$\n\\[\nF_{d}(r)\\le C\\,r^{-1+2^{\\,1-d}}.\n\\]\n\nConsequently every sufficiently large positive real number can be approximated arbitrarily well by a square‐root of the quadratic form  \n\\[\nQ_{d}(m_{1},\\dots ,m_{d})=\\sum_{i=1}^{d}c_{i}m_{i}^{2},\n\\]\nand the approximation error decays strictly faster than $r^{-1/2}$, approaching the optimal exponent $-1$ as $d\\to\\infty$.",
      "solution": "Throughout we assume $r\\ge 1$.  Constants $A_{k}$, $C$ depend only on $d$ and the fixed weights $c_{1},\\dots ,c_{d}$.\n\nStep 0 – A one–dimensional estimate  \nFor $t\\ge 0$ put $n=\\lfloor\\sqrt t\\rfloor$.  Then consecutive squares differ by $(n+1)^{2}-n^{2}=2n+1$, hence\n\\[\n\\lvert\\,t-n^{2}\\,\\rvert=t-n^{2}\\le 2n+1\\le 2\\sqrt t+1. \\tag{1}\n\\]\n\nWe shall invoke (1) once for every coefficient $c_{k}$.\n\n\\bigskip\nStep 1 – Constructing the integers $m_{1},\\dots ,m_{d}$  \n\nLet\n\\[\n\\delta_{0}:=r^{2}.\n\\]\n\nFor $k=1,2,\\dots ,d$ set  \n\\[\nm_{k}:=\\Bigl\\lfloor\\sqrt{\\delta_{k-1}/c_{k}}\\Bigr\\rfloor\n\\quad\\bigl(\\text{so }c_{k}m_{k}^{2}\\le\\delta_{k-1}<c_{k}(m_{k}+1)^{2}\\bigr),\n\\]\nand define\n\\[\n\\delta_{k}:=\\delta_{k-1}-c_{k}m_{k}^{2}\\ge 0.\n\\]\n\nApplying (1) with $t=\\delta_{k-1}/c_{k}$ gives\n\\[\n\\delta_{k}\n =\\delta_{k-1}-c_{k}m_{k}^{2}\n \\le c_{k}\\bigl(2\\sqrt{\\tfrac{\\delta_{k-1}}{c_{k}}}+1\\bigr)\n =2\\sqrt{c_{k}\\,\\delta_{k-1}}+c_{k}. \\tag{2}\n\\]\n\n\\bigskip\nStep 2 – A uniform bound for $\\delta_{k}$  \n\nPut\n\\[\n\\beta_{k}:=2^{\\,1-k}\\qquad(0\\le k\\le d).\n\\]\nObserve $\\beta_{0}=2$, and $\\beta_{k}=\\tfrac12\\,\\beta_{k-1}$.\n\nWe claim that there exist positive constants $A_{k}$ such that\n\\[\n\\boxed{\\;\\;\\delta_{k}\\le A_{k}\\,r^{\\beta_{k}}\\quad(0\\le k\\le d).\\;} \\tag{3}\n\\]\n\nProof by induction on $k$.\n\nBase case $k=0$.  Choosing $A_{0}:=1$ gives $\\delta_{0}=r^{2}=A_{0}\\,r^{\\beta_{0}}$.\n\nInductive step.  \nAssume $\\delta_{k-1}\\le A_{k-1}\\,r^{\\beta_{k-1}}$.  \nFrom (2) we get\n\\[\n\\begin{aligned}\n\\delta_{k}\n &\\le 2\\sqrt{c_{k}A_{k-1}}\\,r^{\\beta_{k-1}/2}+c_{k}\\\\\n &\\;= 2\\sqrt{c_{k}A_{k-1}}\\,r^{\\beta_{k}}+c_{k}.\n\\end{aligned}\n\\]\nBecause $r\\ge 1$ and $\\beta_{k}>0$, $r^{\\beta_{k}}\\ge 1$, so\n\\[\n\\delta_{k}\\le\\bigl(2\\sqrt{c_{k}A_{k-1}}+c_{k}\\bigr)\\,r^{\\beta_{k}}\n           =:A_{k}\\,r^{\\beta_{k}}.\n\\]\nThus (3) is proved for all $k$.\n\n\\bigskip\nStep 3 – Bounding $F_{d}(r)$  \n\nLet\n\\[\nQ_{d}(m_{1},\\dots ,m_{d})=\\sum_{i=1}^{d}c_{i}m_{i}^{2},\n\\qquad\nM:=(m_{1},\\dots ,m_{d})\n\\]\nbe the vector constructed above.  Then\n\\[\nQ_{d}(M)=r^{2}-\\delta_{d}.\n\\]\nTherefore\n\\[\nF_{d}(r)\\le \\Bigl|\\,r-\\sqrt{r^{2}-\\delta_{d}}\\,\\Bigr|\n           =\\frac{\\delta_{d}}{r+\\sqrt{r^{2}-\\delta_{d}}}\n           \\le\\frac{\\delta_{d}}{r}. \\tag{4}\n\\]\n\nUsing (3) with $k=d$ and $\\beta_{d}=2^{\\,1-d}$ we find\n\\[\n\\delta_{d}\\le A_{d}\\,r^{\\,2^{\\,1-d}}.\n\\]\nInsert this into (4):\n\\[\nF_{d}(r)\\le A_{d}\\,r^{-1+2^{\\,1-d}}=:C\\,r^{-1+2^{\\,1-d}}. \\tag{5}\n\\]\n\n\\bigskip\nStep 4 – Existence of the limit (part (a))  \n\nBecause $F_{d}(r)\\ge 0$ and (5) yields\n\\[\n0\\le F_{d}(r)\\le C\\,r^{-1+2^{\\,1-d}}\\xrightarrow[r\\to\\infty]{}0,\n\\]\nwe have\n\\[\n\\boxed{\\displaystyle\\lim_{r\\to\\infty}F_{d}(r)=0}.\n\\]\n\nInequality (5) completes the proof of both parts (a) and (b). \\qed",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.544262",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension / more variables – the problem involves an arbitrary number \\(d\\ge 3\\) of integer variables instead of two, vastly enlarging the search space.\n\n2.  Additional requirements – besides showing that the limit is \\(0\\), the solver must derive an explicit decay rate \\(O\\!\\bigl(r^{-1+2^{1-d}}\\bigr)\\), which tightens the quantitative control dramatically.\n\n3.  Sophisticated structures – the proof has to organise an iterative approximation scheme for a general positive–definite quadratic form with pairwise coprime coefficients; it cannot rely on simple “one-axis” tricks.\n\n4.  Deeper theory – one needs elementary geometry-of-numbers ideas (successive refinement, bounding gaps between lattice norms) and a careful inductive analysis to track the error term through \\(d\\) refinement stages.\n\n5.  Multiple interacting concepts – the solution juggles lattice points, quadratic forms, successive minima, and inductive error propagation simultaneously; each refinement step interacts with the previous ones through inequality (2).\n\nAll these layers of complexity render the enhanced variant substantially more challenging than both the original and the previous kernel versions."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}