{
  "index": "1952-B-3",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "3. Develop necessary and sufficient conditions that the equation\n\\[\n\\left|\\begin{array}{ccc}\n0 & a_{1}-x & a_{2}-x \\\\\n-a_{1}-x & 0 & a_{3}-x \\\\\n-a_{2}-x & -a_{3}-x & 0\n\\end{array}\\right|=0 \\quad\\left(a_{i} \\neq 0\\right)\n\\]\nshall have a multiple root.",
  "solution": "Solution. The given determinant is\n\\[\n-2 x^{3}+2\\left(a_{1} a_{2}+a_{2} a_{3}-a_{1} a_{3}\\right) x=0 .\n\\]\n\nThe necessary and sufficient condition for a multiple root is\n\\[\na_{1} a_{2}+a_{2} a_{3}-a_{1} a_{3}=0\n\\]\n\nIf \\( a_{1} a_{2} a_{3} \\neq 0 \\), this condition can be expressed in the form\n\\[\n\\frac{1}{a_{1}}+\\frac{1}{a_{3}}=\\frac{1}{a_{2}} .\n\\]\n\nComment. We might consider a slightly more general problem. Let \\( A \\) be a \\( 3 \\times 3 \\) skew-symmetric matrix and \\( S \\) a \\( 3 \\times 3 \\) symmetric matrix. Let \\( f(x)=\\operatorname{det}(A-x S) \\). Then\n\\[\n\\begin{aligned}\nf(-x) & =\\operatorname{det}(A+x S)=\\operatorname{det}\\left(A^{T}+x S^{T}\\right) \\\\\n& =\\operatorname{det}(-A+x S)=-\\operatorname{det}(A-x S)=-f(x) .\n\\end{aligned}\n\\]\n\nSo \\( f \\) must be an odd function, \\( f(x)=\\alpha x^{3}+\\beta x \\). A multiple root exists if abd only if \\( \\beta=0 \\) and \\( \\alpha \\neq 0 \\).",
  "vars": [
    "x"
  ],
  "params": [
    "a_1",
    "a_2",
    "a_3",
    "a_i",
    "A",
    "S",
    "f",
    "\\\\alpha",
    "\\\\beta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "unknownvar",
        "a_1": "firstparam",
        "a_2": "secondparam",
        "a_3": "thirdparam",
        "a_i": "iterparam",
        "A": "skewmatrix",
        "S": "symmatrix",
        "f": "detfunction",
        "\\alpha": "cubecoeff",
        "\\beta": "linecoeff"
      },
      "question": "3. Develop necessary and sufficient conditions that the equation\n\\[\n\\left|\\begin{array}{ccc}\n0 & firstparam-unknownvar & secondparam-unknownvar \\\\\n-firstparam-unknownvar & 0 & thirdparam-unknownvar \\\\\n-secondparam-unknownvar & -thirdparam-unknownvar & 0\n\\end{array}\\right|=0 \\quad\\left(iterparam \\neq 0\\right)\n\\]\nshall have a multiple root.",
      "solution": "Solution. The given determinant is\n\\[\n-2\\, unknownvar^{3}+2\\left(firstparam\\, secondparam+secondparam\\, thirdparam-firstparam\\, thirdparam\\right) unknownvar=0 .\n\\]\n\nThe necessary and sufficient condition for a multiple root is\n\\[\nfirstparam\\, secondparam+secondparam\\, thirdparam-firstparam\\, thirdparam=0\n\\]\n\nIf \\( firstparam\\, secondparam\\, thirdparam \\neq 0 \\), this condition can be expressed in the form\n\\[\n\\frac{1}{firstparam}+\\frac{1}{thirdparam}=\\frac{1}{secondparam} .\n\\]\n\nComment. We might consider a slightly more general problem. Let \\( skewmatrix \\) be a \\( 3 \\times 3 \\) skew-symmetric matrix and \\( symmatrix \\) a \\( 3 \\times 3 \\) symmetric matrix. Let \\( detfunction(unknownvar)=\\operatorname{det}(skewmatrix-unknownvar\\, symmatrix) \\). Then\n\\[\n\\begin{aligned}\ndetfunction(-unknownvar) & =\\operatorname{det}(skewmatrix+unknownvar\\, symmatrix)=\\operatorname{det}\\left(skewmatrix^{T}+unknownvar\\, symmatrix^{T}\\right) \\\\\n& =\\operatorname{det}(-skewmatrix+unknownvar\\, symmatrix)=-\\operatorname{det}(skewmatrix-unknownvar\\, symmatrix)=-detfunction(unknownvar) .\n\\end{aligned}\n\\]\n\nSo \\( detfunction \\) must be an odd function, \\( detfunction(unknownvar)=cubecoeff\\, unknownvar^{3}+linecoeff\\, unknownvar \\). A multiple root exists if and only if \\( linecoeff=0 \\) and \\( cubecoeff \\neq 0 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "dragonfly",
        "a_1": "raspberry",
        "a_2": "blueberry",
        "a_3": "blackberry",
        "a_i": "elderberry",
        "A": "pineapple",
        "S": "watermelon",
        "f": "gooseberry",
        "\\alpha": "pomegranate",
        "\\beta": "boysenberry"
      },
      "question": "3. Develop necessary and sufficient conditions that the equation\n\\[\n\\left|\\begin{array}{ccc}\n0 & raspberry-dragonfly & blueberry-dragonfly \\\\\n-raspberry-dragonfly & 0 & blackberry-dragonfly \\\\\n-blueberry-dragonfly & -blackberry-dragonfly & 0\n\\end{array}\\right|=0 \\quad\\left(elderberry \\neq 0\\right)\n\\]\nshall have a multiple root.",
      "solution": "Solution. The given determinant is\n\\[\n-2 dragonfly^{3}+2\\left(raspberry\\,blueberry+blueberry\\,blackberry-raspberry\\,blackberry\\right) dragonfly=0 .\n\\]\n\nThe necessary and sufficient condition for a multiple root is\n\\[\nraspberry\\,blueberry+blueberry\\,blackberry-raspberry\\,blackberry=0\n\\]\n\nIf \\( raspberry\\,blueberry\\,blackberry \\neq 0 \\), this condition can be expressed in the form\n\\[\n\\frac{1}{raspberry}+\\frac{1}{blackberry}=\\frac{1}{blueberry} .\n\\]\n\nComment. We might consider a slightly more general problem. Let \\( pineapple \\) be a \\( 3 \\times 3 \\) skew-symmetric matrix and \\( watermelon \\) a \\( 3 \\times 3 \\) symmetric matrix. Let \\( gooseberry(dragonfly)=\\operatorname{det}(pineapple-dragonfly\\,watermelon) \\). Then\n\\[\n\\begin{aligned}\ngooseberry(-dragonfly) & =\\operatorname{det}(pineapple+dragonfly\\,watermelon)=\\operatorname{det}\\left(pineapple^{T}+dragonfly\\,watermelon^{T}\\right) \\\\\n& =\\operatorname{det}(-pineapple+dragonfly\\,watermelon)=-\\operatorname{det}(pineapple-dragonfly\\,watermelon)=-gooseberry(dragonfly) .\n\\end{aligned}\n\\]\n\nSo \\( gooseberry \\) must be an odd function, \\( gooseberry(dragonfly)=pomegranate\\,dragonfly^{3}+boysenberry\\,dragonfly \\). A multiple root exists if abd only if \\( boysenberry=0 \\) and \\( pomegranate \\neq 0 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "knownvalue",
        "a_1": "variableone",
        "a_2": "variabletwo",
        "a_3": "variablethree",
        "a_i": "variableindex",
        "A": "nonsymmetric",
        "S": "antisymmetric",
        "f": "constant",
        "\\alpha": "nonfactor",
        "\\beta": "nonparameter"
      },
      "question": "3. Develop necessary and sufficient conditions that the equation\n\\[\n\\left|\\begin{array}{ccc}\n0 & variableone-knownvalue & variabletwo-knownvalue \\\\\n-variableone-knownvalue & 0 & variablethree-knownvalue \\\\\n-variabletwo-knownvalue & -variablethree-knownvalue & 0\n\\end{array}\\right|=0 \\quad\\left(variableindex \\neq 0\\right)\n\\]\nshall have a multiple root.",
      "solution": "Solution. The given determinant is\n\\[\n-2 knownvalue^{3}+2\\left(variableone variabletwo+variabletwo variablethree-variableone variablethree\\right) knownvalue=0 .\n\\]\n\nThe necessary and sufficient condition for a multiple root is\n\\[\nvariableone variabletwo+variabletwo variablethree-variableone variablethree=0\n\\]\n\nIf \\( variableone variabletwo variablethree \\neq 0 \\), this condition can be expressed in the form\n\\[\n\\frac{1}{variableone}+\\frac{1}{variablethree}=\\frac{1}{variabletwo} .\n\\]\n\nComment. We might consider a slightly more general problem. Let \\( nonsymmetric \\) be a \\( 3 \\times 3 \\) skew-symmetric matrix and \\( antisymmetric \\) a \\( 3 \\times 3 \\) symmetric matrix. Let \\( constant(knownvalue)=\\operatorname{det}(nonsymmetric-knownvalue antisymmetric) \\). Then\n\\[\n\\begin{aligned}\nconstant(-knownvalue) & =\\operatorname{det}(nonsymmetric+knownvalue antisymmetric)=\\operatorname{det}\\left(nonsymmetric^{T}+knownvalue antisymmetric^{T}\\right) \\\\\n& =\\operatorname{det}(-nonsymmetric+knownvalue antisymmetric)=-\\operatorname{det}(nonsymmetric-knownvalue antisymmetric)=-constant(knownvalue) .\n\\end{aligned}\n\\]\n\nSo \\( constant \\) must be an odd function, \\( constant(knownvalue)=nonfactor knownvalue^{3}+nonparameter knownvalue \\). A multiple root exists if abd only if \\( nonparameter=0 \\) and \\( nonfactor \\neq 0 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "pldkjmas",
        "a_1": "qzxwvtnp",
        "a_2": "hjgrksla",
        "a_3": "bqtmnsdf",
        "a_i": "lyptzsnc",
        "A": "ugotrhps",
        "S": "mnlkjhgf",
        "f": "rtyuiopa",
        "\\alpha": "asdfghjk",
        "\\beta": "zxcvbnml"
      },
      "question": "3. Develop necessary and sufficient conditions that the equation\n\\[\n\\left|\\begin{array}{ccc}\n0 & qzxwvtnp-pldkjmas & hjgrksla-pldkjmas \\\\\n-qzxwvtnp-pldkjmas & 0 & bqtmnsdf-pldkjmas \\\\\n-hjgrksla-pldkjmas & -bqtmnsdf-pldkjmas & 0\n\\end{array}\\right|=0 \\quad\\left(lyptzsnc \\neq 0\\right)\n\\]\nshall have a multiple root.",
      "solution": "Solution. The given determinant is\n\\[\n-2\\,pldkjmas^{3}+2\\left(qzxwvtnp hjgrksla+hjgrksla bqtmnsdf-qzxwvtnp bqtmnsdf\\right)pldkjmas=0 .\n\\]\n\nThe necessary and sufficient condition for a multiple root is\n\\[\nqzxwvtnp hjgrksla+hjgrksla bqtmnsdf-qzxwvtnp bqtmnsdf=0\n\\]\n\nIf \\( qzxwvtnp hjgrksla bqtmnsdf \\neq 0 \\), this condition can be expressed in the form\n\\[\n\\frac{1}{qzxwvtnp}+\\frac{1}{bqtmnsdf}=\\frac{1}{hjgrksla} .\n\\]\n\nComment. We might consider a slightly more general problem. Let \\( ugotrhps \\) be a \\( 3 \\times 3 \\) skew-symmetric matrix and \\( mnlkjhgf \\) a \\( 3 \\times 3 \\) symmetric matrix. Let \\( rtyuiopa(pldkjmas)=\\operatorname{det}(ugotrhps-pldkjmas mnlkjhgf) \\). Then\n\\[\n\\begin{aligned}\nrtyuiopa(-pldkjmas) & =\\operatorname{det}(ugotrhps+pldkjmas mnlkjhgf)=\\operatorname{det}\\left(ugotrhps^{T}+pldkjmas mnlkjhgf^{T}\\right) \\\\\n& =\\operatorname{det}(-ugotrhps+pldkjmas mnlkjhgf)=-\\operatorname{det}(ugotrhps-pldkjmas mnlkjhgf)=-rtyuiopa(pldkjmas) .\n\\end{aligned}\n\\]\n\nSo \\( rtyuiopa \\) must be an odd function, \\( rtyuiopa(pldkjmas)=asdfghjk\\,pldkjmas^{3}+zxcvbnml\\,pldkjmas \\). A multiple root exists if and only if \\( zxcvbnml=0 \\) and \\( asdfghjk \\neq 0 \\)."
    },
    "kernel_variant": {
      "question": "Let  \n\n          ⎛   0    c₁   c₂   c₃ ⎞  \n      K = ⎜ −c₁   0    c₄   c₅ ⎟ ,    (c₁,…,c₆ ∈ ℝ, not all zero),  \n          ⎜ −c₂ −c₄   0    c₆ ⎟  \n          ⎝ −c₃ −c₅ −c₆   0  ⎠  \n\nbe an arbitrary real 4 × 4 skew–symmetric matrix.  \nFor every real parameter x put  \n\n          M(x) = K − x I₄ ,  I₄ = 4 × 4 identity.\n\nDefine the quartic polynomial  \n\n          f(x) := det M(x).  (⋆)\n\n(a)  Show that f is an even polynomial of degree 4 and therefore can be written in the bi-quadratic form  \n\n          f(x) = x⁴ + σ x² + π,                               (1)\n\n     and express the two scalar invariants σ and π solely in terms of the six entries c₁,…,c₆.\n\n(b)  Give a necessary and sufficient condition, written only with c₁,…,c₆, for the quartic (⋆) to possess at least one multiple (possibly complex) root.\n\n(c)  Under that condition classify the multiplicities of the roots and write down explicitly every repeated root, distinguishing the three possibilities  \n\n     • x = 0 is a repeated root,  \n     • x = ± i√(σ / 2) are repeated roots,  \n     • all four roots coincide.  \n\n     Prove in particular that the last alternative can occur only when K = 0 and is therefore ruled out by the standing assumption “not all cᵢ are zero”.\n\n(The problem is the 4 × 4 analogue of the classical 3 × 3 cubic that involves the Pfaffian and tr K².)\n\n--------------------------------------------------------------------",
      "solution": "Throughout we write  \n\n          S := c₁² + c₂² + c₃² + c₄² + c₅² + c₆² ≥ 0,   \n          P := c₁c₆ − c₂c₅ + c₃c₄  (the Pfaffian of K).\n\nThe Pfaffian is real, so P² ≥ 0.\n\nStep 1.  Reduction to the bi-quadratic form.  \nFor every 4 × 4 skew matrix K one has the well–known identity\n\n          det(λ I₄ − K) = λ⁴ − ½ tr K² λ² + pf(K)².          (2)\n\n(An elementary derivation uses the block form of K and the Cayley–Hamilton\ntheorem.)\n\nReplacing λ by x and changing the overall sign (|K − xI| = | xI − K|) gives   \n\n          f(x)=det(K − xI₄)=x⁴ − ½ tr K² x² + pf(K)².         (3)\n\nHence  \n\n          f(x)=x⁴+σx²+π                                        (1)\n\nwith the invariants  \n\n          σ := −½ tr K²,  π := pf(K)².                       (4)\n\nBecause K is skew, tr K = 0, hence f is even.  \nNote that σ ≥ 0 and π ≥ 0.\n\nStep 2.  σ and π in terms of c₁,…,c₆.  \n\n2.1  The trace term.  \nFor a skew matrix, (K²)_{ii}=Σ_j K_{ij}K_{ji}=−Σ_j K_{ij}².\nTherefore  \n\n          tr K² = −2 S, so σ = −½(−2S) = S.                   (5)\n\nThus σ is simply the sum of the squares of the six independent entries and\nvanishes only when K = 0.\n\n2.2  The Pfaffian.  \nA direct expansion gives  \n\n          pf(K)= P = c₁c₆ − c₂c₅ + c₃c₄,                       (6)\n          π = P².                                              (7)\n\nStep 3.  Necessary and sufficient condition for a multiple root.  \n\nWith f(x)=x⁴+σx²+π we have  \n\n          f′(x)=4x³+2σx = 2x(2x²+σ).                           (8)\n\nA number x₀ is a multiple root of f  ⇔  f(x₀)=0 and f′(x₀)=0.\n\n(i)  x₀ = 0.  Then f′(0)=0 automatically, and 0 is a root ⇔ π=0.   (9)\n\n(ii) x₀ ≠ 0.  Condition f′(x₀)=0 forces 2x₀²+σ=0 ⇔ x₀²=−σ/2.\nInserting into f(x₀)=0 yields\n\n        (−σ/2)² + σ(−σ/2) + π = π − σ²/4 = 0.                  (10)\n\nHence there exists a non–zero multiple root (necessarily imaginary unless σ=0)  \n\n          ⇔  σ² = 4π and σ ≠ 0.                               (11)\n\nCombining (9) and (11):\n\n          f has a multiple root ⇔  π = 0   or   σ² = 4π.       (12)\n\nStep 4.  Translation into the entries of K.  \nUsing (5) and (7),\n\n          σ = S, π = P², σ² = S².  \n\nCondition (12) becomes  \n\n   P·(4P² − S²) = 0.                                          (13)\n\nEquivalently  \n\n   (I)  P = 0, or (II)  S² = 4P².                              (14)\n\nBecause S and |P| are rotation–invariant quantities, (14) is a concise\nentry–wise description.\n\nStep 5.  Classification of multiplicities.  \n\nRecall σ = S ≥ 0.\n\n• Case (I) only (P = 0, S > 0).  \n  Then π = 0, so  \n\n          f(x)=x²(x²+S).                                       (15)\n\n  The root x = 0 has multiplicity 2, whereas  \n\n          x = ± i√S                                            (16)\n\n  are distinct and simple purely imaginary roots.\n\n• Case (II) only (P ≠ 0, S²=4P², hence S>0).  \n  Write σ=S and set ρ² := S/2>0.  Then π = σ²/4 = ρ⁴ and  \n\n          f(x)=x⁴+σx²+σ²/4 = (x²+ρ²)².                         (17)\n\n  The two purely imaginary numbers  \n\n          x = ± iρ = ± i√(σ/2)                                 (18)\n\n  are double roots.\n\n• Both (I) and (II) (S=0=σ, P=0=π).  \n  Here f(x)=x⁴, so x=0 would be a quadruple root.  \n  However S=0 forces c₁=…=c₆=0 by (5), contradicting the hypothesis\n  “not all cᵢ are zero”.  Hence this scenario cannot actually occur.\n\nConsequently, for any non-zero 4 × 4 skew matrix K the only possible\nmultiplicity patterns are:\n\n   – a double root at 0 and two simple purely imaginary roots (Pfaffian 0);  \n   – two conjugate double purely imaginary roots (σ² = 4π ≠ 0).\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.446608",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: The problem moves from a 3 × 3 to a 4 × 4 matrix, raising the characteristic polynomial from cubic to quartic.  \n2. Additional mathematical structure: The solution exploits deep properties of skew–symmetric matrices—Pfaffians, trace identities and the even-degree nature of the characteristic polynomial.  \n3. Deeper algebra: Detecting repeated roots now requires manipulating quartic discriminants and interpreting them through invariants σ, π, demanding facility with resultants and discriminants rather than the linear observation used in the original cubic.  \n4. Multiple interacting concepts: Linear-algebraic invariants (tr K², Pfaffian), polynomial algebra (biquadratic reduction, discriminant computation) and root-multiplicity classification all interact.  \n5. More steps: One must (i) establish the even quartic form, (ii) compute σ, π from traces and Pfaffians, (iii) derive the discriminant, (iv) translate its vanishing into explicit conditions, and (v) analyse each resulting case—considerably lengthier and conceptually denser than the original single-line criterion a₁a₂+a₂a₃−a₁a₃=0."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\n          ⎛   0    c₁   c₂   c₃ ⎞  \n      K = ⎜ −c₁   0    c₄   c₅ ⎟ ,    (c₁,…,c₆ ∈ ℝ, not all zero),  \n          ⎜ −c₂ −c₄   0    c₆ ⎟  \n          ⎝ −c₃ −c₅ −c₆   0  ⎠  \n\nbe an arbitrary real 4 × 4 skew–symmetric matrix.  \nFor every real parameter x put  \n\n          M(x) = K − x I₄ ,  I₄ = 4 × 4 identity.\n\nDefine the quartic polynomial  \n\n          f(x) := det M(x).  (⋆)\n\n(a)  Show that f is an even polynomial of degree 4 and therefore can be written in the bi-quadratic form  \n\n          f(x) = x⁴ + σ x² + π,                               (1)\n\n     and express the two scalar invariants σ and π solely in terms of the six entries c₁,…,c₆.\n\n(b)  Give a necessary and sufficient condition, written only with c₁,…,c₆, for the quartic (⋆) to possess at least one multiple (possibly complex) root.\n\n(c)  Under that condition classify the multiplicities of the roots and write down explicitly every repeated root, distinguishing the three possibilities  \n\n     • x = 0 is a repeated root,  \n     • x = ± i√(σ / 2) are repeated roots,  \n     • all four roots coincide.  \n\n     Prove in particular that the last alternative can occur only when K = 0 and is therefore ruled out by the standing assumption “not all cᵢ are zero”.\n\n(The problem is the 4 × 4 analogue of the classical 3 × 3 cubic that involves the Pfaffian and tr K².)\n\n--------------------------------------------------------------------",
      "solution": "Throughout we write  \n\n          S := c₁² + c₂² + c₃² + c₄² + c₅² + c₆² ≥ 0,   \n          P := c₁c₆ − c₂c₅ + c₃c₄  (the Pfaffian of K).\n\nThe Pfaffian is real, so P² ≥ 0.\n\nStep 1.  Reduction to the bi-quadratic form.  \nFor every 4 × 4 skew matrix K one has the well–known identity\n\n          det(λ I₄ − K) = λ⁴ − ½ tr K² λ² + pf(K)².          (2)\n\n(An elementary derivation uses the block form of K and the Cayley–Hamilton\ntheorem.)\n\nReplacing λ by x and changing the overall sign (|K − xI| = | xI − K|) gives   \n\n          f(x)=det(K − xI₄)=x⁴ − ½ tr K² x² + pf(K)².         (3)\n\nHence  \n\n          f(x)=x⁴+σx²+π                                        (1)\n\nwith the invariants  \n\n          σ := −½ tr K²,  π := pf(K)².                       (4)\n\nBecause K is skew, tr K = 0, hence f is even.  \nNote that σ ≥ 0 and π ≥ 0.\n\nStep 2.  σ and π in terms of c₁,…,c₆.  \n\n2.1  The trace term.  \nFor a skew matrix, (K²)_{ii}=Σ_j K_{ij}K_{ji}=−Σ_j K_{ij}².\nTherefore  \n\n          tr K² = −2 S, so σ = −½(−2S) = S.                   (5)\n\nThus σ is simply the sum of the squares of the six independent entries and\nvanishes only when K = 0.\n\n2.2  The Pfaffian.  \nA direct expansion gives  \n\n          pf(K)= P = c₁c₆ − c₂c₅ + c₃c₄,                       (6)\n          π = P².                                              (7)\n\nStep 3.  Necessary and sufficient condition for a multiple root.  \n\nWith f(x)=x⁴+σx²+π we have  \n\n          f′(x)=4x³+2σx = 2x(2x²+σ).                           (8)\n\nA number x₀ is a multiple root of f  ⇔  f(x₀)=0 and f′(x₀)=0.\n\n(i)  x₀ = 0.  Then f′(0)=0 automatically, and 0 is a root ⇔ π=0.   (9)\n\n(ii) x₀ ≠ 0.  Condition f′(x₀)=0 forces 2x₀²+σ=0 ⇔ x₀²=−σ/2.\nInserting into f(x₀)=0 yields\n\n        (−σ/2)² + σ(−σ/2) + π = π − σ²/4 = 0.                  (10)\n\nHence there exists a non–zero multiple root (necessarily imaginary unless σ=0)  \n\n          ⇔  σ² = 4π and σ ≠ 0.                               (11)\n\nCombining (9) and (11):\n\n          f has a multiple root ⇔  π = 0   or   σ² = 4π.       (12)\n\nStep 4.  Translation into the entries of K.  \nUsing (5) and (7),\n\n          σ = S, π = P², σ² = S².  \n\nCondition (12) becomes  \n\n   P·(4P² − S²) = 0.                                          (13)\n\nEquivalently  \n\n   (I)  P = 0, or (II)  S² = 4P².                              (14)\n\nBecause S and |P| are rotation–invariant quantities, (14) is a concise\nentry–wise description.\n\nStep 5.  Classification of multiplicities.  \n\nRecall σ = S ≥ 0.\n\n• Case (I) only (P = 0, S > 0).  \n  Then π = 0, so  \n\n          f(x)=x²(x²+S).                                       (15)\n\n  The root x = 0 has multiplicity 2, whereas  \n\n          x = ± i√S                                            (16)\n\n  are distinct and simple purely imaginary roots.\n\n• Case (II) only (P ≠ 0, S²=4P², hence S>0).  \n  Write σ=S and set ρ² := S/2>0.  Then π = σ²/4 = ρ⁴ and  \n\n          f(x)=x⁴+σx²+σ²/4 = (x²+ρ²)².                         (17)\n\n  The two purely imaginary numbers  \n\n          x = ± iρ = ± i√(σ/2)                                 (18)\n\n  are double roots.\n\n• Both (I) and (II) (S=0=σ, P=0=π).  \n  Here f(x)=x⁴, so x=0 would be a quadruple root.  \n  However S=0 forces c₁=…=c₆=0 by (5), contradicting the hypothesis\n  “not all cᵢ are zero”.  Hence this scenario cannot actually occur.\n\nConsequently, for any non-zero 4 × 4 skew matrix K the only possible\nmultiplicity patterns are:\n\n   – a double root at 0 and two simple purely imaginary roots (Pfaffian 0);  \n   – two conjugate double purely imaginary roots (σ² = 4π ≠ 0).\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.385098",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: The problem moves from a 3 × 3 to a 4 × 4 matrix, raising the characteristic polynomial from cubic to quartic.  \n2. Additional mathematical structure: The solution exploits deep properties of skew–symmetric matrices—Pfaffians, trace identities and the even-degree nature of the characteristic polynomial.  \n3. Deeper algebra: Detecting repeated roots now requires manipulating quartic discriminants and interpreting them through invariants σ, π, demanding facility with resultants and discriminants rather than the linear observation used in the original cubic.  \n4. Multiple interacting concepts: Linear-algebraic invariants (tr K², Pfaffian), polynomial algebra (biquadratic reduction, discriminant computation) and root-multiplicity classification all interact.  \n5. More steps: One must (i) establish the even quartic form, (ii) compute σ, π from traces and Pfaffians, (iii) derive the discriminant, (iv) translate its vanishing into explicit conditions, and (v) analyse each resulting case—considerably lengthier and conceptually denser than the original single-line criterion a₁a₂+a₂a₃−a₁a₃=0."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}