{
  "index": "2001-B-2",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "Find all pairs of real numbers $(x,y)$ satisfying the system\nof equations\n\\begin{align*}\n \\frac{1}{x} + \\frac{1}{2y} &= (x^2+3y^2)(3x^2+y^2) \\\\\n   \\frac{1}{x} - \\frac{1}{2y} &= 2(y^4-x^4).\n\\end{align*}",
  "solution": "By adding and subtracting the two given equations, we obtain\nthe equivalent pair of equations\n\\begin{align*}\n2/x &= x^4 + 10x^2y^2 + 5y^4 \\\\\n1/y &= 5x^4 + 10x^2y^2 + y^4.\n\\end{align*}\nMultiplying the former by\n$x$ and the latter by $y$, then adding and subtracting the two\nresulting equations, we obtain another pair of equations equivalent\nto the given ones,\n\\[\n3 = (x+y)^5, \\qquad 1 = (x-y)^5.\n\\]\nIt follows that\n$x = (3^{1/5}+1)/2$ and $y = (3^{1/5}-1)/2$ is the unique solution\nsatisfying the given equations.",
  "vars": [
    "x",
    "y"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "realvarx",
        "y": "realvary"
      },
      "question": "Find all pairs of real numbers $(realvarx,realvary)$ satisfying the system\nof equations\n\\begin{align*}\n \\frac{1}{realvarx} + \\frac{1}{2realvary} &= (realvarx^{2}+3realvary^{2})(3realvarx^{2}+realvary^{2}) \\\\\n   \\frac{1}{realvarx} - \\frac{1}{2realvary} &= 2(realvary^{4}-realvarx^{4}).\n\\end{align*}",
      "solution": "By adding and subtracting the two given equations, we obtain\nthe equivalent pair of equations\n\\begin{align*}\n2/realvarx &= realvarx^{4} + 10realvarx^{2}realvary^{2} + 5realvary^{4} \\\\\n1/realvary &= 5realvarx^{4} + 10realvarx^{2}realvary^{2} + realvary^{4}.\n\\end{align*}\nMultiplying the former by\n$realvarx$ and the latter by $realvary$, then adding and subtracting the two\nresulting equations, we obtain another pair of equations equivalent\nto the given ones,\n\\[\n3 = (realvarx+realvary)^{5}, \\qquad 1 = (realvarx-realvary)^{5}.\n\\]\nIt follows that\n$realvarx = (3^{1/5}+1)/2$ and $realvary = (3^{1/5}-1)/2$ is the unique solution\nsatisfying the given equations."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lighthouse",
        "y": "tangerine"
      },
      "question": "Find all pairs of real numbers $(lighthouse,tangerine)$ satisfying the system\nof equations\n\\begin{align*}\n \\frac{1}{lighthouse} + \\frac{1}{2tangerine} &= (lighthouse^2+3tangerine^2)(3lighthouse^2+tangerine^2) \\\\\n   \\frac{1}{lighthouse} - \\frac{1}{2tangerine} &= 2(tangerine^4-lighthouse^4).\n\\end{align*}",
      "solution": "By adding and subtracting the two given equations, we obtain\nthe equivalent pair of equations\n\\begin{align*}\n2/lighthouse &= lighthouse^4 + 10lighthouse^2tangerine^2 + 5tangerine^4 \\\\\n1/tangerine &= 5lighthouse^4 + 10lighthouse^2tangerine^2 + tangerine^4.\n\\end{align*}\nMultiplying the former by\n$lighthouse$ and the latter by $tangerine$, then adding and subtracting the two\nresulting equations, we obtain another pair of equations equivalent\nto the given ones,\n\\[\n3 = (lighthouse+tangerine)^5, \\qquad 1 = (lighthouse-tangerine)^5.\n\\]\nIt follows that\n$lighthouse = (3^{1/5}+1)/2$ and $tangerine = (3^{1/5}-1)/2$ is the unique solution\nsatisfying the given equations."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "y": "fixednumber"
      },
      "question": "Find all pairs of real numbers $(constantvalue,fixednumber)$ satisfying the system\nof equations\n\\begin{align*}\n \\frac{1}{constantvalue} + \\frac{1}{2fixednumber} &= (constantvalue^2+3fixednumber^2)(3constantvalue^2+fixednumber^2) \\\\\n   \\frac{1}{constantvalue} - \\frac{1}{2fixednumber} &= 2(fixednumber^4-constantvalue^4).\n\\end{align*}",
      "solution": "By adding and subtracting the two given equations, we obtain\nthe equivalent pair of equations\n\\begin{align*}\n2/constantvalue &= constantvalue^4 + 10constantvalue^2fixednumber^2 + 5fixednumber^4 \\\\\n1/fixednumber &= 5constantvalue^4 + 10constantvalue^2fixednumber^2 + fixednumber^4.\n\\end{align*}\nMultiplying the former by\n$constantvalue$ and the latter by $fixednumber$, then adding and subtracting the two\nresulting equations, we obtain another pair of equations equivalent\nto the given ones,\n\\[\n3 = (constantvalue+fixednumber)^5, \\qquad 1 = (constantvalue-fixednumber)^5.\n\\]\nIt follows that\n$constantvalue = (3^{1/5}+1)/2$ and $fixednumber = (3^{1/5}-1)/2$ is the unique solution\nsatisfying the given equations."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla"
      },
      "question": "Find all pairs of real numbers $(qzxwvtnp,hjgrksla)$ satisfying the system\nof equations\n\\begin{align*}\n \\frac{1}{qzxwvtnp} + \\frac{1}{2hjgrksla} &= (qzxwvtnp^2+3hjgrksla^2)(3qzxwvtnp^2+hjgrksla^2) \\\\\n   \\frac{1}{qzxwvtnp} - \\frac{1}{2hjgrksla} &= 2(hjgrksla^4-qzxwvtnp^4).\n\\end{align*}",
      "solution": "By adding and subtracting the two given equations, we obtain\nthe equivalent pair of equations\n\\begin{align*}\n2/qzxwvtnp &= qzxwvtnp^4 + 10qzxwvtnp^2hjgrksla^2 + 5hjgrksla^4 \\\\\n1/hjgrksla &= 5qzxwvtnp^4 + 10qzxwvtnp^2hjgrksla^2 + hjgrksla^4.\n\\end{align*}\nMultiplying the former by\n$qzxwvtnp$ and the latter by $hjgrksla$, then adding and subtracting the two\nresulting equations, we obtain another pair of equations equivalent\nto the given ones,\n\\[\n3 = (qzxwvtnp+hjgrksla)^5, \\qquad 1 = (qzxwvtnp-hjgrksla)^5.\n\\]\nIt follows that\n$qzxwvtnp = (3^{1/5}+1)/2$ and $hjgrksla = (3^{1/5}-1)/2$ is the unique solution\nsatisfying the given equations."
    },
    "kernel_variant": {
      "question": "Determine all ordered triples of positive real numbers  \n\\[\n(x,y,z)\\in(0,\\infty)^{3}\n\\]\nthat satisfy the system of equations  \n\\[\n\\begin{cases}\n\\dfrac1x+\\dfrac1y+\\dfrac1z=(x^{2}+y^{2}+z^{2})\\,(xy+yz+zx),\\\\[6pt]\n\\dfrac1x-\\dfrac1y=\\;(y^{4}-x^{4})+z^{2}(x^{2}-y^{2}),\\\\[6pt]\n\\dfrac1y-\\dfrac1z=\\;(z^{4}-y^{4})+x^{2}(y^{2}-z^{2}).\n\\end{cases}\\tag{$\\star$}\n\\]\n\n────────────────────────────────",
      "solution": "Throughout we assume \\(x,y,z>0\\).\n\nI.  The completely symmetric case \\(x=y=z\\)\n\nLet \\(x=y=z=t>0\\).  \nThe last two equations of \\((\\star)\\) become identities, whereas the first one reduces to\n\\[\n\\frac{3}{t}=9t^{4}\\quad\\Longrightarrow\\quad t^{5}=\\frac13,\n\\]\nhence\n\\[\n\\boxed{(x,y,z)=\\bigl(\\gamma,\\gamma,\\gamma\\bigr)},\\qquad\n\\gamma:=3^{-1/5}\\approx0.80274.\n\\]\n\n────────────────────────────────  \nII.  At least two coordinates are different  \n\nWithout loss of generality assume \\(x\\ne y\\).\n\n─────────────────  \nII a.  Three auxiliary identities  \n\nFrom the second equation of \\((\\star)\\)\n\\[\n\\frac1x-\\frac1y=(y^{4}-x^{4})+z^{2}(x^{2}-y^{2})\n          =(y-x)(y+x)(y^{2}+x^{2}-z^{2}),\n\\]\nand division by \\(y-x\\,( \\neq 0)\\) yields\n\\[\n\\frac1{xy}=(x+y)\\bigl(x^{2}+y^{2}-z^{2}\\bigr).\\tag{A}\n\\]\n\nBy cyclic permutation of \\((x,y,z)\\) we analogously obtain  \n\\[\n\\frac1{yz}=(y+z)\\bigl(y^{2}+z^{2}-x^{2}\\bigr),\\tag{B}\n\\qquad\n\\frac1{zx}=(z+x)\\bigl(z^{2}+x^{2}-y^{2}\\bigr).\\tag{C}\n\\]\n\n─────────────────  \nII b.  All three coordinates pairwise different  \n\nAssume for the moment that \\(x,y,z\\) are mutually distinct, so that (A)–(C) are all valid.\n\n1.  Multiplying (A)–(C) yields\n\\[\n\\frac1{(xyz)^{2}}\n   =(x+y)(y+z)(z+x)\\!\n     \\prod_{\\text{cyc}}\\bigl(x^{2}+y^{2}-z^{2}\\bigr).\\tag{1}\n\\]\n\n2.  From the first equation of \\((\\star)\\) we obtain, after multiplication by \\(xyz\\) and division by the positive factor \\(xy+yz+zx\\),\n\\[\n\\boxed{x^{2}+y^{2}+z^{2}=\\frac1{xyz}}.\\tag{2}\n\\]\n\n3.  A crucial identity (properly derived).  \n   Divide (A), (B) and (C) by \\(x+y,\\;y+z,\\;z+x\\), respectively:\n   \\[\n   \\frac1{xy(x+y)}=x^{2}+y^{2}-z^{2},\\quad\n   \\frac1{yz(y+z)}=y^{2}+z^{2}-x^{2},\\quad\n   \\frac1{zx(z+x)}=z^{2}+x^{2}-y^{2}.\n   \\]\n   Adding these three equalities gives\n   \\[\n   \\sum_{\\text{cyc}}\\frac1{xy(x+y)}\n            =x^{2}+y^{2}+z^{2}=\\frac1{xyz}\\quad\\text{by (2).}\n   \\]\n   Multiplying by \\(xyz\\) we arrive at the *exact* identity\n   \\[\n   \\boxed{\\frac{z}{x+y}+\\frac{x}{y+z}+\\frac{y}{z+x}=1.}\\tag{3}\n   \\]\n\n4.  Contradiction via Nesbitt’s inequality.  \n   For all positive reals \\(x,y,z\\),\n   \\[\n   \\frac{x}{y+z}+\\frac{y}{z+x}+\\frac{z}{x+y}\\ge\\frac32,\n   \\]\n   with equality if and only if \\(x=y=z\\).  \n   Yet (3) asserts that the same sum equals \\(1<\\tfrac32\\).  \n   This contradiction shows that\n\n no solution exists with three distinct coordinates.\n\n─────────────────  \nII c.  Exactly two coordinates coincide  \n\nSuppose \\(x=y=:a\\) and \\(z=:b\\) with \\(a\\neq b\\)\n(the other possibilities follow by symmetry).\n\nBecause \\(x=y\\), the second equation of \\((\\star)\\) is an identity, while the third becomes\n\\[\n\\frac1a-\\frac1b=(b^{2}-a^{2})b^{2}\n\\quad\\Longrightarrow\\quad\n\\frac1{ab}=(a+b)b^{2}.\\tag{4}\n\\]\n\nWith \\(x=y=a\\), the first equation of \\((\\star)\\) reads\n\\[\n\\frac2a+\\frac1b=(2a^{2}+b^{2})(a^{2}+2ab).\\tag{5}\n\\]\n\nIntroduce the ratio \\(t:=\\dfrac{a}{b}\\;(t>0,\\;t\\neq1)\\).\nEquation (4) gives\n\\[\nb^{5}=\\frac{1}{t(t+1)}.\\tag{6}\n\\]\nPut \\(a=tb\\) and substitute (6) into (5); after cancellation one obtains the quartic\n\\[\n2t^{4}+4t^{3}-t-2=0.\\tag{7}\n\\]\nFactorising,\n\\[\n(t+2)(2t^{3}-1)=0.\\tag{8}\n\\]\nPositivity rules out \\(t=-2\\); consequently\n\\[\nt_{0}=2^{-1/3}.\\tag{9}\n\\]\n\nInsert \\(t_{0}\\) into (6):\n\\[\nb=\\Bigl(\\frac{2^{\\,2/3}}{1+2^{\\,1/3}}\\Bigr)^{1/5}\\! =: \\beta,\n\\qquad\na= t_{0}b = 2^{-1/3}\\beta =: \\alpha.\\tag{10}\n\\]\nNumerically\n\\[\n\\beta\\approx0.93180,\\qquad\n\\alpha\\approx0.73909.\n\\]\n\nHence \\((a,a,b)=(\\alpha,\\alpha,\\beta)\\) is a solution, and by cyclic permutation so are \\((\\alpha,\\beta,\\alpha)\\) and \\((\\beta,\\alpha,\\alpha)\\).\n\n────────────────────────────────  \nIII.  Exhaustiveness of the list  \n\n•  Section I furnished the symmetric solution  \n  \\((\\gamma,\\gamma,\\gamma)\\).\n\n•  Section II b rigorously ruled out the possibility that all three coordinates are different.\n\n•  Section II c produced exactly three solutions with two equal coordinates.\n\nTherefore the complete set of positive solutions to \\((\\star)\\) is\n\\[\n\\boxed{(\\gamma,\\gamma,\\gamma)},\\qquad\n\\boxed{(\\alpha,\\alpha,\\beta)},\\quad\n\\boxed{(\\alpha,\\beta,\\alpha)},\\quad\n\\boxed{(\\beta,\\alpha,\\alpha)},\n\\]\nwhere\n\\[\n\\gamma=3^{-1/5},\\qquad\n\\beta=\\Bigl(\\frac{2^{\\,2/3}}{1+2^{\\,1/3}}\\Bigr)^{1/5},\\qquad\n\\alpha=2^{-1/3}\\beta.\n\\]\n\nApproximate decimal values:\n\\[\n(0.80274,0.80274,0.80274),\\;\n(0.73909,0.73909,0.93180),\\;\n(0.73909,0.93180,0.73909),\\;\n(0.93180,0.73909,0.73909).\n\\]\n\nThe problem is thus completely solved.\n\n────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.773934",
        "was_fixed": false,
        "difficulty_analysis": "• The number of variables has been raised from two to three, creating a fully-coupled **system of three nonlinear equations** instead of two.  \n• Each equation now mixes different symmetric polynomials (power sums and elementary symmetric polynomials of degree 2 and 4), so direct pattern‐matching is impossible.  \n• Factorising the last two equations reveals hidden common factors whose vanishing demands a careful case analysis; one must prove that the only admissible case forces equality of all three variables.  \n• After reducing to the diagonal line $x=y=z$, the remaining equation is a quintic, preserving the quintic flavour of the original but embedded in a higher-dimensional setting.  \n• The solution requires several layers of reasoning: algebraic factorisation, contradiction via sign analysis, elimination, and finally solving a one-variable quintic.  \nTogether these ingredients make the enhanced variant substantially harder and longer than both the original and the previous kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Determine all ordered triples of positive real numbers  \n\\[\n(x,y,z)\\in(0,\\infty)^{3}\n\\]\nthat satisfy the system of equations  \n\\[\n\\begin{cases}\n\\dfrac1x+\\dfrac1y+\\dfrac1z=(x^{2}+y^{2}+z^{2})\\,(xy+yz+zx),\\\\[6pt]\n\\dfrac1x-\\dfrac1y=\\;(y^{4}-x^{4})+z^{2}(x^{2}-y^{2}),\\\\[6pt]\n\\dfrac1y-\\dfrac1z=\\;(z^{4}-y^{4})+x^{2}(y^{2}-z^{2}).\n\\end{cases}\\tag{$\\star$}\n\\]\n\n────────────────────────────────",
      "solution": "Throughout we assume \\(x,y,z>0\\).\n\nI.  The completely symmetric case \\(x=y=z\\)\n\nLet \\(x=y=z=t>0\\).  \nThe last two equations of \\((\\star)\\) become identities, whereas the first one reduces to\n\\[\n\\frac{3}{t}=9t^{4}\\quad\\Longrightarrow\\quad t^{5}=\\frac13,\n\\]\nhence\n\\[\n\\boxed{(x,y,z)=\\bigl(\\gamma,\\gamma,\\gamma\\bigr)},\\qquad\n\\gamma:=3^{-1/5}\\approx0.80274.\n\\]\n\n────────────────────────────────  \nII.  At least two coordinates are different  \n\nWithout loss of generality assume \\(x\\ne y\\).\n\n─────────────────  \nII a.  Three auxiliary identities  \n\nFrom the second equation of \\((\\star)\\)\n\\[\n\\frac1x-\\frac1y=(y^{4}-x^{4})+z^{2}(x^{2}-y^{2})\n          =(y-x)(y+x)(y^{2}+x^{2}-z^{2}),\n\\]\nand division by \\(y-x\\,( \\neq 0)\\) yields\n\\[\n\\frac1{xy}=(x+y)\\bigl(x^{2}+y^{2}-z^{2}\\bigr).\\tag{A}\n\\]\n\nBy cyclic permutation of \\((x,y,z)\\) we analogously obtain  \n\\[\n\\frac1{yz}=(y+z)\\bigl(y^{2}+z^{2}-x^{2}\\bigr),\\tag{B}\n\\qquad\n\\frac1{zx}=(z+x)\\bigl(z^{2}+x^{2}-y^{2}\\bigr).\\tag{C}\n\\]\n\n─────────────────  \nII b.  All three coordinates pairwise different  \n\nAssume for the moment that \\(x,y,z\\) are mutually distinct, so that (A)–(C) are all valid.\n\n1.  Multiplying (A)–(C) yields\n\\[\n\\frac1{(xyz)^{2}}\n   =(x+y)(y+z)(z+x)\\!\n     \\prod_{\\text{cyc}}\\bigl(x^{2}+y^{2}-z^{2}\\bigr).\\tag{1}\n\\]\n\n2.  From the first equation of \\((\\star)\\) we obtain, after multiplication by \\(xyz\\) and division by the positive factor \\(xy+yz+zx\\),\n\\[\n\\boxed{x^{2}+y^{2}+z^{2}=\\frac1{xyz}}.\\tag{2}\n\\]\n\n3.  A crucial identity (properly derived).  \n   Divide (A), (B) and (C) by \\(x+y,\\;y+z,\\;z+x\\), respectively:\n   \\[\n   \\frac1{xy(x+y)}=x^{2}+y^{2}-z^{2},\\quad\n   \\frac1{yz(y+z)}=y^{2}+z^{2}-x^{2},\\quad\n   \\frac1{zx(z+x)}=z^{2}+x^{2}-y^{2}.\n   \\]\n   Adding these three equalities gives\n   \\[\n   \\sum_{\\text{cyc}}\\frac1{xy(x+y)}\n            =x^{2}+y^{2}+z^{2}=\\frac1{xyz}\\quad\\text{by (2).}\n   \\]\n   Multiplying by \\(xyz\\) we arrive at the *exact* identity\n   \\[\n   \\boxed{\\frac{z}{x+y}+\\frac{x}{y+z}+\\frac{y}{z+x}=1.}\\tag{3}\n   \\]\n\n4.  Contradiction via Nesbitt’s inequality.  \n   For all positive reals \\(x,y,z\\),\n   \\[\n   \\frac{x}{y+z}+\\frac{y}{z+x}+\\frac{z}{x+y}\\ge\\frac32,\n   \\]\n   with equality if and only if \\(x=y=z\\).  \n   Yet (3) asserts that the same sum equals \\(1<\\tfrac32\\).  \n   This contradiction shows that\n\n no solution exists with three distinct coordinates.\n\n─────────────────  \nII c.  Exactly two coordinates coincide  \n\nSuppose \\(x=y=:a\\) and \\(z=:b\\) with \\(a\\neq b\\)\n(the other possibilities follow by symmetry).\n\nBecause \\(x=y\\), the second equation of \\((\\star)\\) is an identity, while the third becomes\n\\[\n\\frac1a-\\frac1b=(b^{2}-a^{2})b^{2}\n\\quad\\Longrightarrow\\quad\n\\frac1{ab}=(a+b)b^{2}.\\tag{4}\n\\]\n\nWith \\(x=y=a\\), the first equation of \\((\\star)\\) reads\n\\[\n\\frac2a+\\frac1b=(2a^{2}+b^{2})(a^{2}+2ab).\\tag{5}\n\\]\n\nIntroduce the ratio \\(t:=\\dfrac{a}{b}\\;(t>0,\\;t\\neq1)\\).\nEquation (4) gives\n\\[\nb^{5}=\\frac{1}{t(t+1)}.\\tag{6}\n\\]\nPut \\(a=tb\\) and substitute (6) into (5); after cancellation one obtains the quartic\n\\[\n2t^{4}+4t^{3}-t-2=0.\\tag{7}\n\\]\nFactorising,\n\\[\n(t+2)(2t^{3}-1)=0.\\tag{8}\n\\]\nPositivity rules out \\(t=-2\\); consequently\n\\[\nt_{0}=2^{-1/3}.\\tag{9}\n\\]\n\nInsert \\(t_{0}\\) into (6):\n\\[\nb=\\Bigl(\\frac{2^{\\,2/3}}{1+2^{\\,1/3}}\\Bigr)^{1/5}\\! =: \\beta,\n\\qquad\na= t_{0}b = 2^{-1/3}\\beta =: \\alpha.\\tag{10}\n\\]\nNumerically\n\\[\n\\beta\\approx0.93180,\\qquad\n\\alpha\\approx0.73909.\n\\]\n\nHence \\((a,a,b)=(\\alpha,\\alpha,\\beta)\\) is a solution, and by cyclic permutation so are \\((\\alpha,\\beta,\\alpha)\\) and \\((\\beta,\\alpha,\\alpha)\\).\n\n────────────────────────────────  \nIII.  Exhaustiveness of the list  \n\n•  Section I furnished the symmetric solution  \n  \\((\\gamma,\\gamma,\\gamma)\\).\n\n•  Section II b rigorously ruled out the possibility that all three coordinates are different.\n\n•  Section II c produced exactly three solutions with two equal coordinates.\n\nTherefore the complete set of positive solutions to \\((\\star)\\) is\n\\[\n\\boxed{(\\gamma,\\gamma,\\gamma)},\\qquad\n\\boxed{(\\alpha,\\alpha,\\beta)},\\quad\n\\boxed{(\\alpha,\\beta,\\alpha)},\\quad\n\\boxed{(\\beta,\\alpha,\\alpha)},\n\\]\nwhere\n\\[\n\\gamma=3^{-1/5},\\qquad\n\\beta=\\Bigl(\\frac{2^{\\,2/3}}{1+2^{\\,1/3}}\\Bigr)^{1/5},\\qquad\n\\alpha=2^{-1/3}\\beta.\n\\]\n\nApproximate decimal values:\n\\[\n(0.80274,0.80274,0.80274),\\;\n(0.73909,0.73909,0.93180),\\;\n(0.73909,0.93180,0.73909),\\;\n(0.93180,0.73909,0.73909).\n\\]\n\nThe problem is thus completely solved.\n\n────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.592609",
        "was_fixed": false,
        "difficulty_analysis": "• The number of variables has been raised from two to three, creating a fully-coupled **system of three nonlinear equations** instead of two.  \n• Each equation now mixes different symmetric polynomials (power sums and elementary symmetric polynomials of degree 2 and 4), so direct pattern‐matching is impossible.  \n• Factorising the last two equations reveals hidden common factors whose vanishing demands a careful case analysis; one must prove that the only admissible case forces equality of all three variables.  \n• After reducing to the diagonal line $x=y=z$, the remaining equation is a quintic, preserving the quintic flavour of the original but embedded in a higher-dimensional setting.  \n• The solution requires several layers of reasoning: algebraic factorisation, contradiction via sign analysis, elimination, and finally solving a one-variable quintic.  \nTogether these ingredients make the enhanced variant substantially harder and longer than both the original and the previous kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}