{
  "index": "1948-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "3. If \\( n \\) is a positive integer, prove that\n\\[\n[\\sqrt{n}+\\sqrt{n+1}]=[\\sqrt{4 n+2}]\n\\]\nwhere \\( [x] \\) denotes as usual the greatest integer not exceeding \\( x . \\quad \\)",
  "solution": "Solution. Since \\( \\sqrt{x} \\) has negative second derivative for \\( x>0 \\), its graph is concave downward and\n\\[\n\\frac{\\sqrt{x}+\\sqrt{x+1}}{2}<\\sqrt{x+\\frac{1}{2}} \\text { for all } x \\geq 0\n\\]\n\nThus \\( \\sqrt{x}+\\sqrt{x+1}<\\sqrt{4 x+2} \\) for all \\( x \\geq 0 \\), and hence \\( [\\sqrt{x}+\\sqrt{x+1}] \\) \\( \\leq[\\sqrt{4 x+2}] \\).\n\nSuppose that for some positive integer \\( n,[\\sqrt{n}+\\sqrt{n+1}] \\neq[\\sqrt{4 n+2}] \\).\nLet \\( p=[\\sqrt{4 n+2}] \\). Then\n\\[\n\\sqrt{n}+\\sqrt{n+1}<p \\leq \\sqrt{4 n+2}\n\\]\n\nSquaring, we get\n\\[\n2 n+1+2 \\sqrt{n(n+1)}<p^{2} \\leq 4 n+2\n\\]\n\nTherefore \\( 2 \\sqrt{n(n+1)}<p^{2}-2 n-1 \\leq 2 n+1 \\). Squaring again gives\n\\[\n4 n(n+1)<\\left(p^{2}-2 n-1\\right)^{2} \\leq 4 n^{2}+4 n+1\n\\]\n\nSince the outer numbers are consecutive integers, and since the middle number in the inequality is also an integer, we have \\( \\left(p^{2}-2 n-1\\right)^{2}= \\) \\( (2 n+1)^{2} \\) and therefore \\( p^{2}=4 n+2 \\). But this last equation is impossible, for the square of an integer cannot be congruent to 2 modulo 4 .\n\nThis contradiction proves that \\( [\\sqrt{n}+\\sqrt{n+1}]=[\\sqrt{4 n+2}] \\) for every positive integer \\( n \\).",
  "vars": [
    "n",
    "x",
    "p"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "integer",
        "x": "variable",
        "p": "floorval"
      },
      "question": "3. If \\( integer \\) is a positive integer, prove that\n\\[\n[\\sqrt{integer}+\\sqrt{integer+1}]=[\\sqrt{4 integer+2}]\n\\]\nwhere \\( [\\,variable\\,] \\) denotes as usual the greatest integer not exceeding \\( variable . \\quad \\)",
      "solution": "Solution. Since \\( \\sqrt{variable} \\) has negative second derivative for \\( variable>0 \\), its graph is concave downward and\n\\[\n\\frac{\\sqrt{variable}+\\sqrt{variable+1}}{2}<\\sqrt{variable+\\frac{1}{2}} \\text { for all } variable \\geq 0\n\\]\n\nThus \\( \\sqrt{variable}+\\sqrt{variable+1}<\\sqrt{4 variable+2} \\) for all \\( variable \\geq 0 \\), and hence \\( [\\sqrt{variable}+\\sqrt{variable+1}] \\leq[\\sqrt{4 variable+2}] \\).\n\nSuppose that for some positive integer \\( integer,[\\sqrt{integer}+\\sqrt{integer+1}] \\neq[\\sqrt{4 integer+2}] \\).\nLet \\( floorval=[\\sqrt{4 integer+2}] \\). Then\n\\[\n\\sqrt{integer}+\\sqrt{integer+1}<floorval \\leq \\sqrt{4 integer+2}\n\\]\n\nSquaring, we get\n\\[\n2\\,integer+1+2 \\sqrt{integer(integer+1)}<floorval^{2} \\leq 4\\,integer+2\n\\]\n\nTherefore \\( 2 \\sqrt{integer(integer+1)}<floorval^{2}-2\\,integer-1 \\leq 2\\,integer+1 \\). Squaring again gives\n\\[\n4\\,integer(integer+1)<\\left(floorval^{2}-2\\,integer-1\\right)^{2} \\leq 4\\,integer^{2}+4\\,integer+1\n\\]\n\nSince the outer numbers are consecutive integers, and since the middle number in the inequality is also an integer, we have \\( \\left(floorval^{2}-2\\,integer-1\\right)^{2}= (2\\,integer+1)^{2} \\) and therefore \\( floorval^{2}=4\\,integer+2 \\). But this last equation is impossible, for the square of an integer cannot be congruent to 2 modulo 4 .\n\nThis contradiction proves that \\( [\\sqrt{integer}+\\sqrt{integer+1}]=[\\sqrt{4 integer+2}] \\) for every positive integer \\( integer \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "limestone",
        "x": "sailcloth",
        "p": "drumstick"
      },
      "question": "3. If \\( limestone \\) is a positive integer, prove that\n\\[\n[\\sqrt{limestone}+\\sqrt{limestone+1}]=[\\sqrt{4 limestone+2}]\n\\]\nwhere \\( [sailcloth] \\) denotes as usual the greatest integer not exceeding \\( sailcloth . \\quad \\)",
      "solution": "Solution. Since \\( \\sqrt{sailcloth} \\) has negative second derivative for \\( sailcloth>0 \\), its graph is concave downward and\n\\[\n\\frac{\\sqrt{sailcloth}+\\sqrt{sailcloth+1}}{2}<\\sqrt{sailcloth+\\frac{1}{2}} \\text { for all } sailcloth \\geq 0\n\\]\n\nThus \\( \\sqrt{sailcloth}+\\sqrt{sailcloth+1}<\\sqrt{4 sailcloth+2} \\) for all \\( sailcloth \\geq 0 \\), and hence \\( [\\sqrt{sailcloth}+\\sqrt{sailcloth+1}] \\) \\( \\leq[\\sqrt{4 sailcloth+2}] \\).\n\nSuppose that for some positive integer \\( limestone,[\\sqrt{limestone}+\\sqrt{limestone+1}] \\neq[\\sqrt{4 limestone+2}] \\).\nLet \\( drumstick=[\\sqrt{4 limestone+2}] \\). Then\n\\[\n\\sqrt{limestone}+\\sqrt{limestone+1}<drumstick \\leq \\sqrt{4 limestone+2}\n\\]\n\nSquaring, we get\n\\[\n2 limestone+1+2 \\sqrt{limestone(limestone+1)}<drumstick^{2} \\leq 4 limestone+2\n\\]\n\nTherefore \\( 2 \\sqrt{limestone(limestone+1)}<drumstick^{2}-2 limestone-1 \\leq 2 limestone+1 \\). Squaring again gives\n\\[\n4 limestone(limestone+1)<\\left(drumstick^{2}-2 limestone-1\\right)^{2} \\leq 4 limestone^{2}+4 limestone+1\n\\]\n\nSince the outer numbers are consecutive integers, and since the middle number in the inequality is also an integer, we have \\( \\left(drumstick^{2}-2 limestone-1\\right)^{2}= \\) \\( (2 limestone+1)^{2} \\) and therefore \\( drumstick^{2}=4 limestone+2 \\). But this last equation is impossible, for the square of an integer cannot be congruent to 2 modulo 4 .\n\nThis contradiction proves that \\( [\\sqrt{limestone}+\\sqrt{limestone+1}]=[\\sqrt{4 limestone+2}] \\) for every positive integer \\( limestone \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "negativeint",
        "x": "fixedvalue",
        "p": "irrational"
      },
      "question": "3. If \\( negativeint \\) is a positive integer, prove that\n\\[\n[\\sqrt{negativeint}+\\sqrt{negativeint+1}]=[\\sqrt{4 negativeint+2}]\n\\]\nwhere \\( [fixedvalue] \\) denotes as usual the greatest integer not exceeding \\( fixedvalue . \\quad \\)",
      "solution": "Solution. Since \\( \\sqrt{fixedvalue} \\) has negative second derivative for \\( fixedvalue>0 \\), its graph is concave downward and\n\\[\n\\frac{\\sqrt{fixedvalue}+\\sqrt{fixedvalue+1}}{2}<\\sqrt{fixedvalue+\\frac{1}{2}} \\text { for all } \\fixedvalue \\geq 0\n\\]\n\nThus \\( \\sqrt{fixedvalue}+\\sqrt{fixedvalue+1}<\\sqrt{4 fixedvalue+2} \\) for all \\( fixedvalue \\geq 0 \\), and hence \\( [\\sqrt{fixedvalue}+\\sqrt{fixedvalue+1}] \\) \\( \\leq[\\sqrt{4 fixedvalue+2}] \\).\n\nSuppose that for some positive integer \\( negativeint,[\\sqrt{negativeint}+\\sqrt{negativeint+1}] \\neq[\\sqrt{4 negativeint+2}] \\).\nLet \\( irrational=[\\sqrt{4 negativeint+2}] \\). Then\n\\[\n\\sqrt{negativeint}+\\sqrt{negativeint+1}<irrational \\leq \\sqrt{4 negativeint+2}\n\\]\n\nSquaring, we get\n\\[\n2 negativeint+1+2 \\sqrt{negativeint(negativeint+1)}<irrational^{2} \\leq 4 negativeint+2\n\\]\n\nTherefore \\( 2 \\sqrt{negativeint(negativeint+1)}<irrational^{2}-2 negativeint-1 \\leq 2 negativeint+1 \\). Squaring again gives\n\\[\n4 negativeint(negativeint+1)<\\left(irrational^{2}-2 negativeint-1\\right)^{2} \\leq 4 negativeint^{2}+4 negativeint+1\n\\]\n\nSince the outer numbers are consecutive integers, and since the middle number in the inequality is also an integer, we have \\( \\left(irrational^{2}-2 negativeint-1\\right)^{2}= \\) \\( (2 negativeint+1)^{2} \\) and therefore \\( irrational^{2}=4 negativeint+2 \\). But this last equation is impossible, for the square of an integer cannot be congruent to 2 modulo 4 .\n\nThis contradiction proves that \\( [\\sqrt{negativeint}+\\sqrt{negativeint+1}]=[\\sqrt{4 negativeint+2}] \\) for every positive integer \\( negativeint \\)."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "x": "hjgrksla",
        "p": "vdqkrnfo"
      },
      "question": "3. If \\( qzxwvtnp \\) is a positive integer, prove that\n\\[\n[\\sqrt{qzxwvtnp}+\\sqrt{qzxwvtnp+1}]=[\\sqrt{4 qzxwvtnp+2}]\n\\]\nwhere \\( [hjgrksla] \\) denotes as usual the greatest integer not exceeding \\( hjgrksla . \\quad \\)",
      "solution": "Solution. Since \\( \\sqrt{hjgrksla} \\) has negative second derivative for \\( hjgrksla>0 \\), its graph is concave downward and\n\\[\n\\frac{\\sqrt{hjgrksla}+\\sqrt{hjgrksla+1}}{2}<\\sqrt{hjgrksla+\\frac{1}{2}} \\text { for all } hjgrksla \\geq 0\n\\]\n\nThus \\( \\sqrt{hjgrksla}+\\sqrt{hjgrksla+1}<\\sqrt{4 hjgrksla+2} \\) for all \\( hjgrksla \\geq 0 \\), and hence \\( [\\sqrt{hjgrksla}+\\sqrt{hjgrksla+1}] \\) \\( \\leq[\\sqrt{4 hjgrksla+2}] \\).\n\nSuppose that for some positive integer \\( qzxwvtnp,[\\sqrt{qzxwvtnp}+\\sqrt{qzxwvtnp+1}] \\neq[\\sqrt{4 qzxwvtnp+2}] \\).\nLet \\( vdqkrnfo=[\\sqrt{4 qzxwvtnp+2}] \\). Then\n\\[\n\\sqrt{qzxwvtnp}+\\sqrt{qzxwvtnp+1}<vdqkrnfo \\leq \\sqrt{4 qzxwvtnp+2}\n\\]\n\nSquaring, we get\n\\[\n2 qzxwvtnp+1+2 \\sqrt{qzxwvtnp(qzxwvtnp+1)}<vdqkrnfo^{2} \\leq 4 qzxwvtnp+2\n\\]\n\nTherefore \\( 2 \\sqrt{qzxwvtnp(qzxwvtnp+1)}<vdqkrnfo^{2}-2 qzxwvtnp-1 \\leq 2 qzxwvtnp+1 \\). Squaring again gives\n\\[\n4 qzxwvtnp(qzxwvtnp+1)<\\left(vdqkrnfo^{2}-2 qzxwvtnp-1\\right)^{2} \\leq 4 qzxwvtnp^{2}+4 qzxwvtnp+1\n\\]\n\nSince the outer numbers are consecutive integers, and since the middle number in the inequality is also an integer, we have \\( \\left(vdqkrnfo^{2}-2 qzxwvtnp-1\\right)^{2}= (2 qzxwvtnp+1)^{2} \\) and therefore \\( vdqkrnfo^{2}=4 qzxwvtnp+2 \\). But this last equation is impossible, for the square of an integer cannot be congruent to 2 modulo 4 .\n\nThis contradiction proves that \\( [\\sqrt{qzxwvtnp}+\\sqrt{qzxwvtnp+1}]=[\\sqrt{4 qzxwvtnp+2}] \\) for every positive integer \\( qzxwvtnp \\)."
    },
    "kernel_variant": {
      "question": "Let n be a positive integer and set  \n\n  S(n)=√n+√(n+1)+√(n+2),   T(n)=√(9n+9)=3√(n+1).\n\n1.  Prove the sharpened two–sided estimate  \n\n   0 < T(n)-S(n)< 1/(2√n) < 1  for every n ≥ 1.  \n\n2.  Denote by ε(n) the characteristic function of the set of shifted perfect squares,  \n\n  ε(n)=1 iff n = t²–1 (t∈ℕ, t≥1), ε(n)=0 otherwise.  \n\n   Show that  \n\n  ⌊√n+√(n+1)+√(n+2)⌋ = ⌊√(9n+9)⌋ − ε(n).\n\nIn other words  \n\n  ⌊S(n)⌋ = ⌊T(n)⌋ exactly when n+1 is not a perfect square,  \n  ⌊S(n)⌋ = ⌊T(n)⌋−1 when n+1 happens to be a perfect square.",
      "solution": "Throughout put  \n\n  Δ_k := √(k+1)−√k = 1/(√k+√(k+1)), k≥0.\n\nStep 1.  A precise upper bound for T(n)−S(n).  \nBecause Δ_k is strictly decreasing (√x is concave),\n\n  T(n)−S(n)=3√(n+1)−(√n+√(n+1)+√(n+2))\n       = (Δ_n) − (Δ_{\\,n+1})                                (1)\n\nand hence  \n\n  T(n)−S(n)= (√(n+2)−√n) / [(√n+√(n+1))(√(n+1)+√(n+2))].        (2)\n\nWrite A=√n, B=√(n+1), C=√(n+2).  Then  \n\n  C−A = 2/(A+C) ≤ 1/A  because A+C ≥ 2A,\n\nwhile  \n\n  (A+B)(B+C) ≥ (A+B)(B+A) > (2A)(2A)=4A² = 4n.\n\nSubstituting in (2) gives the uniform bound  \n\n  0 < T(n)−S(n) < 1/(4n^{3/2}) < 1/(2√n) < 1, n≥1.            (3)\n\n(The weak form stated in the problem, 1/(2√n), still follows, but the\nsharper 1/(4n^{3/2}) will be useful below.)\n\nStep 2.  A universal lower bound for the fractional part of T(n).  \nPut  \n\n  m := n+1, s := √m, k := ⌊3s⌋, δ := {3s}=3s−k∈(0,1).\n\nIf m is a perfect square then δ=0.  Assume henceforth that m is not a square.  \nBecause (3s)²–k² = 9m−k² is a positive integer, we have\n\n  9m−k² = (3s−k)(3s+k) = δ(3s+k) ≥ 1.                          (4)\n\nMoreover k < 3s < k+1 ⇒ 3s+k < 6s, and therefore  \n\n  δ = (9m−k²)/(3s+k) ≥ 1/(6s) = 1/(6√m) = 1/(6√{n+1}).         (5)\n\nThus  \n\n  {T(n)} = δ ≥ 1/(6√{n+1}),  if n+1 is not a perfect square.   (6)\n\nStep 3.  Comparing T(n)−S(n) with {T(n)}.  \nFor n ≥ 4 we combine (3) and (6):\n\n  T(n)−S(n) < 1/(4n^{3/2})  \n                        < 1/(12√n) ≤ 1/(6√{n+1}) ≤ {T(n)}.       (7)\n\nHence  \n\n  S(n) = T(n) − (T(n)−S(n)) > ⌊T(n)⌋,  n ≥ 4, n+1 non-square, (8)\n\nso ⌊S(n)⌋ = ⌊T(n)⌋.\n\nStep 4.  The three small integers n = 1,2,3.  \nDirect calculation shows\n\n n=1: S≈4.146, T≈4.242, {T}=0.242>0.096=T−S;   \n n=2: S≈5.146, T≈5.196, {T}=0.196>0.050=T−S;   \n n=3: n+1=4 is a square ⇒ {T}=0 (exceptional case).\n\nThus the conclusion of Step 3 remains true for n=1,2, whereas n=3\nbelongs to the “perfect-square” family treated next.\n\nStep 5.  The exceptional integers.  \nEquation (6) shows δ=0 exactly when n+1 is a perfect square.  In that\ncase T(n)=3√(n+1) is an integer, while (3) still gives S(n)<T(n)<S(n)+1,\nso necessarily  \n\n  ⌊S(n)⌋ = ⌊T(n)⌋ − 1  if n+1 is a perfect square.            (9)\n\nStep 6.  Conclusion.  \nCombining (8) and (9) we finally obtain for every positive integer n\n\n  ⌊√n+√(n+1)+√(n+2)⌋ =\n   ⌊√(9n+9)⌋     if n+1 is not a square,\n   ⌊√(9n+9)⌋–1  if n+1 is a square,\n\nequivalently  \n\n  ⌊S(n)⌋ = ⌊T(n)⌋ − ε(n). ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.420384",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimensional interaction:  three square–root terms must be\nhandled simultaneously, not just two as in the original problem.\n\n• Sharper analytic bounds:  the proof requires a delicate two–sided\nestimate 0 < T(n)–S(n) < 1 obtained through monotonicity and concavity\narguments, whereas the original identity only needed one‐sided\nconcavity.\n\n• Diophantine classification:  detecting when T(n) is integral entails\nsolving 9(n+1)=m², an extra quadratic–Diophantine step absent from the\nkernel problem.\n\n• Piecewise answer:  the floor equality now depends on the arithmetic\nnature of n (whether n+1 is a perfect square), introducing an\n“exceptional set’’ and making the final statement conditional.\n\nThese additional layers demand a longer chain of reasoning, blending\ncalculus‐style inequalities with elementary number theory, and give a\nsubstantially more intricate problem than both the original and the\ncurrent kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let n be a positive integer and set  \n\n  S(n)=√n+√(n+1)+√(n+2),   T(n)=√(9n+9)=3√(n+1).\n\n1.  Prove the sharpened two–sided estimate  \n\n   0 < T(n)-S(n)< 1/(2√n) < 1  for every n ≥ 1.  \n\n2.  Denote by ε(n) the characteristic function of the set of shifted perfect squares,  \n\n  ε(n)=1 iff n = t²–1 (t∈ℕ, t≥1), ε(n)=0 otherwise.  \n\n   Show that  \n\n  ⌊√n+√(n+1)+√(n+2)⌋ = ⌊√(9n+9)⌋ − ε(n).\n\nIn other words  \n\n  ⌊S(n)⌋ = ⌊T(n)⌋ exactly when n+1 is not a perfect square,  \n  ⌊S(n)⌋ = ⌊T(n)⌋−1 when n+1 happens to be a perfect square.",
      "solution": "Throughout put  \n\n  Δ_k := √(k+1)−√k = 1/(√k+√(k+1)), k≥0.\n\nStep 1.  A precise upper bound for T(n)−S(n).  \nBecause Δ_k is strictly decreasing (√x is concave),\n\n  T(n)−S(n)=3√(n+1)−(√n+√(n+1)+√(n+2))\n       = (Δ_n) − (Δ_{\\,n+1})                                (1)\n\nand hence  \n\n  T(n)−S(n)= (√(n+2)−√n) / [(√n+√(n+1))(√(n+1)+√(n+2))].        (2)\n\nWrite A=√n, B=√(n+1), C=√(n+2).  Then  \n\n  C−A = 2/(A+C) ≤ 1/A  because A+C ≥ 2A,\n\nwhile  \n\n  (A+B)(B+C) ≥ (A+B)(B+A) > (2A)(2A)=4A² = 4n.\n\nSubstituting in (2) gives the uniform bound  \n\n  0 < T(n)−S(n) < 1/(4n^{3/2}) < 1/(2√n) < 1, n≥1.            (3)\n\n(The weak form stated in the problem, 1/(2√n), still follows, but the\nsharper 1/(4n^{3/2}) will be useful below.)\n\nStep 2.  A universal lower bound for the fractional part of T(n).  \nPut  \n\n  m := n+1, s := √m, k := ⌊3s⌋, δ := {3s}=3s−k∈(0,1).\n\nIf m is a perfect square then δ=0.  Assume henceforth that m is not a square.  \nBecause (3s)²–k² = 9m−k² is a positive integer, we have\n\n  9m−k² = (3s−k)(3s+k) = δ(3s+k) ≥ 1.                          (4)\n\nMoreover k < 3s < k+1 ⇒ 3s+k < 6s, and therefore  \n\n  δ = (9m−k²)/(3s+k) ≥ 1/(6s) = 1/(6√m) = 1/(6√{n+1}).         (5)\n\nThus  \n\n  {T(n)} = δ ≥ 1/(6√{n+1}),  if n+1 is not a perfect square.   (6)\n\nStep 3.  Comparing T(n)−S(n) with {T(n)}.  \nFor n ≥ 4 we combine (3) and (6):\n\n  T(n)−S(n) < 1/(4n^{3/2})  \n                        < 1/(12√n) ≤ 1/(6√{n+1}) ≤ {T(n)}.       (7)\n\nHence  \n\n  S(n) = T(n) − (T(n)−S(n)) > ⌊T(n)⌋,  n ≥ 4, n+1 non-square, (8)\n\nso ⌊S(n)⌋ = ⌊T(n)⌋.\n\nStep 4.  The three small integers n = 1,2,3.  \nDirect calculation shows\n\n n=1: S≈4.146, T≈4.242, {T}=0.242>0.096=T−S;   \n n=2: S≈5.146, T≈5.196, {T}=0.196>0.050=T−S;   \n n=3: n+1=4 is a square ⇒ {T}=0 (exceptional case).\n\nThus the conclusion of Step 3 remains true for n=1,2, whereas n=3\nbelongs to the “perfect-square” family treated next.\n\nStep 5.  The exceptional integers.  \nEquation (6) shows δ=0 exactly when n+1 is a perfect square.  In that\ncase T(n)=3√(n+1) is an integer, while (3) still gives S(n)<T(n)<S(n)+1,\nso necessarily  \n\n  ⌊S(n)⌋ = ⌊T(n)⌋ − 1  if n+1 is a perfect square.            (9)\n\nStep 6.  Conclusion.  \nCombining (8) and (9) we finally obtain for every positive integer n\n\n  ⌊√n+√(n+1)+√(n+2)⌋ =\n   ⌊√(9n+9)⌋     if n+1 is not a square,\n   ⌊√(9n+9)⌋–1  if n+1 is a square,\n\nequivalently  \n\n  ⌊S(n)⌋ = ⌊T(n)⌋ − ε(n). ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.365532",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimensional interaction:  three square–root terms must be\nhandled simultaneously, not just two as in the original problem.\n\n• Sharper analytic bounds:  the proof requires a delicate two–sided\nestimate 0 < T(n)–S(n) < 1 obtained through monotonicity and concavity\narguments, whereas the original identity only needed one‐sided\nconcavity.\n\n• Diophantine classification:  detecting when T(n) is integral entails\nsolving 9(n+1)=m², an extra quadratic–Diophantine step absent from the\nkernel problem.\n\n• Piecewise answer:  the floor equality now depends on the arithmetic\nnature of n (whether n+1 is a perfect square), introducing an\n“exceptional set’’ and making the final statement conditional.\n\nThese additional layers demand a longer chain of reasoning, blending\ncalculus‐style inequalities with elementary number theory, and give a\nsubstantially more intricate problem than both the original and the\ncurrent kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}