{
  "index": "1940-B-5",
  "type": "ALG",
  "tag": [
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "\\begin{array}{l}\n\\text { 13. Determine all rational values for which } a, b, c \\text { are the roots of }\\\\\nx^{3}+a x^{2}+b x+c=0\n\\end{array}",
  "solution": "Solution. The conditions on the roots are equivalent to\n\\[\n\\begin{array}{c}\na+b+c=-a \\\\\na b+b c+c a=b \\\\\na b c=-c\n\\end{array}\n\\]\n\nIf \\( c=0 \\), then \\( a b=b \\) and \\( 2 a+b=0 \\), so either \\( b=0, a=0 \\), or \\( a=1, b=-2 \\).\n\nIf \\( c \\neq 0 \\), then \\( a b=-1 \\). If \\( a+b=0 \\), then (2) becomes \\( a b=b \\) so that \\( a=1, b=-1, c=-1 \\). If \\( a+b \\neq 0 \\), then\n\\[\nc=\\frac{b+1}{a+b}=\\frac{a(b+1)}{a(a+b)}=\\frac{-1+a}{a^{2}-1}=\\frac{1}{a+1}\n\\]\nand (1) becomes\n\\[\n2 a-\\frac{1}{a}+\\frac{1}{a+1}=0\n\\]\nwhence\n\\[\n2 a^{3}+2 a^{2}-1=0\n\\]\n\nThis equation has no rational roots, since the only possibilities are \\( \\pm 1 \\), \\( \\pm 1 / 2 \\), and these are not roots. There are therefore three solutions\n\\[\n\\begin{array}{rrrl}\na & b & c & \\text { corresponding to } \\\\\n0 & 0 & 0 & x^{3}=0 \\\\\n+1 & -2 & 0 & x^{3}+x^{2}-2 x=0 \\\\\n+1 & -1 & -1 & x^{3}+x^{2}-x-1=\\left(x^{2}-1\\right)(x+1)=0 .\n\\end{array}\n\\]",
  "vars": [
    "a",
    "b",
    "c",
    "x"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "rootone",
        "b": "roottwo",
        "c": "rootthree",
        "x": "unknownx"
      },
      "question": "\\begin{array}{l}\n\\text { 13. Determine all rational values for which } rootone, roottwo, rootthree \\text { are the roots of }\\\\\nunknownx^{3}+rootone unknownx^{2}+roottwo unknownx+rootthree=0\n\\end{array}",
      "solution": "Solution. The conditions on the roots are equivalent to\n\\[\n\\begin{array}{c}\nrootone+roottwo+rootthree=-rootone \\\\\nrootone roottwo+roottwo rootthree+rootthree rootone=roottwo \\\\\nrootone roottwo rootthree=-rootthree\n\\end{array}\n\\]\n\nIf \\( rootthree=0 \\), then \\( rootone roottwo=roottwo \\) and \\( 2 rootone+roottwo=0 \\), so either \\( roottwo=0, rootone=0 \\), or \\( rootone=1, roottwo=-2 \\).\n\nIf \\( rootthree \\neq 0 \\), then \\( rootone roottwo=-1 \\). If \\( rootone+roottwo=0 \\), then (2) becomes \\( rootone roottwo=roottwo \\) so that \\( rootone=1, roottwo=-1, rootthree=-1 \\). If \\( rootone+roottwo \\neq 0 \\), then\n\\[\nrootthree=\\frac{roottwo+1}{rootone+roottwo}=\\frac{rootone(roottwo+1)}{rootone(rootone+roottwo)}=\\frac{-1+rootone}{rootone^{2}-1}=\\frac{1}{rootone+1}\n\\]\nand (1) becomes\n\\[\n2 rootone-\\frac{1}{rootone}+\\frac{1}{rootone+1}=0\n\\]\nwhence\n\\[\n2 rootone^{3}+2 rootone^{2}-1=0\n\\]\n\nThis equation has no rational roots, since the only possibilities are \\( \\pm 1 \\), \\( \\pm 1 / 2 \\), and these are not roots. There are therefore three solutions\n\\[\n\\begin{array}{rrrl}\nrootone & roottwo & rootthree & \\text { corresponding to } \\\\\n0 & 0 & 0 & unknownx^{3}=0 \\\\\n+1 & -2 & 0 & unknownx^{3}+unknownx^{2}-2 unknownx=0 \\\\\n+1 & -1 & -1 & unknownx^{3}+unknownx^{2}-unknownx-1=\\left(unknownx^{2}-1\\right)(unknownx+1)=0 .\n\\end{array}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "pinecone",
        "b": "marigold",
        "c": "lighthouse",
        "x": "waterfall"
      },
      "question": "<<<\n\\begin{array}{l}\n\\text { 13. Determine all rational values for which } pinecone, marigold, lighthouse \\text { are the roots of }\\\\\nwaterfall^{3}+pinecone waterfall^{2}+marigold waterfall+lighthouse=0\n\\end{array}\n>>>",
      "solution": "<<<\nSolution. The conditions on the roots are equivalent to\n\\[\n\\begin{array}{c}\npinecone+marigold+lighthouse=-pinecone \\\\\npinecone marigold+marigold lighthouse+lighthouse pinecone=marigold \\\\\npinecone marigold lighthouse=-lighthouse\n\\end{array}\n\\]\n\nIf \\( lighthouse=0 \\), then \\( pinecone marigold=marigold \\) and \\( 2 pinecone+marigold=0 \\), so either \\( marigold=0, pinecone=0 \\), or \\( pinecone=1, marigold=-2 \\).\n\nIf \\( lighthouse \\neq 0 \\), then \\( pinecone marigold=-1 \\). If \\( pinecone+marigold=0 \\), then (2) becomes \\( pinecone marigold=marigold \\) so that \\( pinecone=1, marigold=-1, lighthouse=-1 \\). If \\( pinecone+marigold \\neq 0 \\), then\n\\[\nlighthouse=\\frac{marigold+1}{pinecone+marigold}=\\frac{pinecone(marigold+1)}{pinecone(pinecone+marigold)}=\\frac{-1+pinecone}{pinecone^{2}-1}=\\frac{1}{pinecone+1}\n\\]\nand (1) becomes\n\\[\n2 pinecone-\\frac{1}{pinecone}+\\frac{1}{pinecone+1}=0\n\\]\nwhence\n\\[\n2 pinecone^{3}+2 pinecone^{2}-1=0\n\\]\n\nThis equation has no rational roots, since the only possibilities are \\( \\pm 1 \\), \\( \\pm 1 / 2 \\), and these are not roots. There are therefore three solutions\n\\[\n\\begin{array}{rrrl}\npinecone & marigold & lighthouse & \\text { corresponding to } \\\\\n0 & 0 & 0 & waterfall^{3}=0 \\\\\n+1 & -2 & 0 & waterfall^{3}+waterfall^{2}-2 waterfall=0 \\\\\n+1 & -1 & -1 & waterfall^{3}+waterfall^{2}-waterfall-1=\\left(waterfall^{2}-1\\right)(waterfall+1)=0 .\n\\end{array}\n\\]\n>>>"
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "terminalpoint",
        "b": "descendingvalue",
        "c": "dynamicvar",
        "x": "knownvalue"
      },
      "question": "\\begin{array}{l}\n\\text { 13. Determine all rational values for which } terminalpoint, descendingvalue, dynamicvar \\text { are the roots of }\\\\\nknownvalue^{3}+terminalpoint knownvalue^{2}+descendingvalue knownvalue+dynamicvar=0\n\\end{array}",
      "solution": "Solution. The conditions on the roots are equivalent to\n\\[\n\\begin{array}{c}\nterminalpoint+descendingvalue+dynamicvar=-terminalpoint \\\\\nterminalpoint descendingvalue+descendingvalue dynamicvar+dynamicvar terminalpoint=descendingvalue \\\\\nterminalpoint descendingvalue dynamicvar=-dynamicvar\n\\end{array}\n\\]\n\nIf \\( dynamicvar=0 \\), then \\( terminalpoint descendingvalue=descendingvalue \\) and \\( 2 terminalpoint+descendingvalue=0 \\), so either \\( descendingvalue=0, terminalpoint=0 \\), or \\( terminalpoint=1, descendingvalue=-2 \\).\n\nIf \\( dynamicvar \\neq 0 \\), then \\( terminalpoint descendingvalue=-1 \\). If \\( terminalpoint+descendingvalue=0 \\), then (2) becomes \\( terminalpoint descendingvalue=descendingvalue \\) so that \\( terminalpoint=1, descendingvalue=-1, dynamicvar=-1 \\). If \\( terminalpoint+descendingvalue \\neq 0 \\), then\n\\[\ndynamicvar=\\frac{descendingvalue+1}{terminalpoint+descendingvalue}=\\frac{terminalpoint(descendingvalue+1)}{terminalpoint(terminalpoint+descendingvalue)}=\\frac{-1+terminalpoint}{terminalpoint^{2}-1}=\\frac{1}{terminalpoint+1}\n\\]\nand (1) becomes\n\\[\n2 terminalpoint-\\frac{1}{terminalpoint}+\\frac{1}{terminalpoint+1}=0\n\\]\nwhence\n\\[\n2 terminalpoint^{3}+2 terminalpoint^{2}-1=0\n\\]\n\nThis equation has no rational roots, since the only possibilities are \\( \\pm 1 \\), \\( \\pm 1 / 2 \\), and these are not roots. There are therefore three solutions\n\\[\n\\begin{array}{rrrl}\nterminalpoint & descendingvalue & dynamicvar & \\text { corresponding to } \\\\\n0 & 0 & 0 & knownvalue^{3}=0 \\\\\n+1 & -2 & 0 & knownvalue^{3}+knownvalue^{2}-2 knownvalue=0 \\\\\n+1 & -1 & -1 & knownvalue^{3}+knownvalue^{2}-knownvalue-1=\\left(knownvalue^{2}-1\\right)(knownvalue+1)=0 .\n\\end{array}\n\\]"
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "mfdqnibe",
        "x": "roplqsev"
      },
      "question": "\\begin{array}{l}\n\\text { 13. Determine all rational values for which } qzxwvtnp, hjgrksla, mfdqnibe \\text { are the roots of }\\\\\nroplqsev^{3}+qzxwvtnp roplqsev^{2}+hjgrksla roplqsev+mfdqnibe=0\n\\end{array}",
      "solution": "Solution. The conditions on the roots are equivalent to\n\\[\n\\begin{array}{c}\nqzxwvtnp+hjgrksla+mfdqnibe=-qzxwvtnp \\\\\nqzxwvtnp hjgrksla+hjgrksla mfdqnibe+mfdqnibe qzxwvtnp=hjgrksla \\\\\nqzxwvtnp hjgrksla mfdqnibe=-mfdqnibe\n\\end{array}\n\\]\n\nIf \\( mfdqnibe=0 \\), then \\( qzxwvtnp hjgrksla=hjgrksla \\) and \\( 2 qzxwvtnp+hjgrksla=0 \\), so either \\( hjgrksla=0, qzxwvtnp=0 \\), or \\( qzxwvtnp=1, hjgrksla=-2 \\).\n\nIf \\( mfdqnibe \\neq 0 \\), then \\( qzxwvtnp hjgrksla=-1 \\). If \\( qzxwvtnp+hjgrksla=0 \\), then (2) becomes \\( qzxwvtnp hjgrksla=hjgrksla \\) so that \\( qzxwvtnp=1, hjgrksla=-1, mfdqnibe=-1 \\). If \\( qzxwvtnp+hjgrksla \\neq 0 \\), then\n\\[\nmfdqnibe=\\frac{hjgrksla+1}{qzxwvtnp+hjgrksla}=\\frac{qzxwvtnp(hjgrksla+1)}{qzxwvtnp(qzxwvtnp+hjgrksla)}=\\frac{-1+qzxwvtnp}{qzxwvtnp^{2}-1}=\\frac{1}{qzxwvtnp+1}\n\\]\nand (1) becomes\n\\[\n2 qzxwvtnp-\\frac{1}{qzxwvtnp}+\\frac{1}{qzxwvtnp+1}=0\n\\]\nwhence\n\\[\n2 qzxwvtnp^{3}+2 qzxwvtnp^{2}-1=0\n\\]\n\nThis equation has no rational roots, since the only possibilities are \\( \\pm 1 \\), \\( \\pm 1 / 2 \\), and these are not roots. There are therefore three solutions\n\\[\n\\begin{array}{rrrl}\nqzxwvtnp & hjgrksla & mfdqnibe & \\text { corresponding to } \\\\\n0 & 0 & 0 & roplqsev^{3}=0 \\\\\n+1 & -2 & 0 & roplqsev^{3}+roplqsev^{2}-2 roplqsev=0 \\\\\n+1 & -1 & -1 & roplqsev^{3}+roplqsev^{2}-roplqsev-1=\\left(roplqsev^{2}-1\\right)(roplqsev+1)=0 .\n\\end{array}\n\\]"
    },
    "kernel_variant": {
      "question": "Let a, b, c, d be integers.  \nDetermine all ordered quadruples (a,b,c,d) ∈ ℤ⁴ for which the four numbers a, b, c, d themselves are (with multiplicity) the four roots of the monic quartic polynomial  \n\n  Q(x)=x⁴ + a x³ + c x² + d x + b.  \n\n(Notice that, as in the cubic kernel, the constant term and the x²– and x–coefficients are deliberately “scrambled’’.)",
      "solution": "Step 1.  Translate the root–coefficient condition via Viète.  \nBecause a, b, c, d are the roots of Q we have  \n\n(1) a + b + c + d = –a,  \n\n(2) ab + ac + ad + bc + bd + cd =  c,  \n\n(3) abc + abd + acd + bcd = –d,  \n\n(4) abcd =  b.  \n\nStep 2.  Treat the two main cases b = 0 and b ≠ 0.\n\n────────────────────────────────────────  \nCase 1.  b = 0  \n\nEquation (4) is automatic.  The remaining relations become  \n\n(1′) c + d = –2a,  \n\n(2′) ac + ad + cd = c,  \n\n(3′) acd = –d.  \n\n–––––  Sub-case 1.1: d = 0  \nThen (3′) is automatic and (2′) reduces to ac = c, i.e. c(a – 1)=0.\n\n•  If c = 0, then (1′) gives a = 0, yielding the solution (0,0,0,0).  \n•  If a = 1, then (1′) gives c = –2, giving (1,0,–2,0).\n\n–––––  Sub-case 1.2: d ≠ 0  \nFrom (3′) we get ac = –1  ⇒  c = –1/a and a ≠ 0.\n\nInsert c and d = –2a – c = –2a + 1/a into (2′):\n\n ac + ad + cd = c  \n –1 + a d + c d = –1/a.\n\nAfter clearing denominators one obtains\n\n 2a⁴ – 2a² – a + 1 = 0.                     (★)\n\nFactorising,   \n 2a⁴ – 2a² – a + 1 = (a – 1)(2a³ + 2a² – 1).\n\nThe cubic factor has no integer (indeed no rational) root by the Rational-Root Test, so the only integral solution of (★) is a = 1.  \nThus c = –1 and d = –2·1 + 1 = –1, giving the third solution (1,0,–1,–1).\n\n────────────────────────────────────────  \nCase 2.  b ≠ 0  \n\nNow (4) gives acd = 1.  Hence a, c, d are all divisors of 1, so each is ±1.\n\nList the four possibilities for (a,c,d):\n\n (i) (1, 1, 1) (ii) (1, –1, –1) (iii) (–1, 1, –1) (iv) (–1, –1, 1).\n\nFor each, compute b from (1) (b = –2a – c – d) and test (2).\n\n(i) (1,1,1): b = –4; (2) gives –9 = 1 ✘  \n(ii) (1,–1,–1): b = 0 (not allowed in Case 2)  \n(iii) (–1,1,–1): b = 2; (2) gives –3 = 1 ✘  \n(iv) (–1,–1,1): b = 2; (2) gives –3 = –1 ✘  \n\nNo choice satisfies all equations, so Case 2 yields no solutions.\n\n────────────────────────────────────────  \nStep 3.  Collect the solutions.\n\nThe only integral quadruples satisfying (1)–(4), hence the only ones for which a, b, c, d are the roots of Q, are  \n\n (0, 0, 0, 0), (1, 0, –2, 0), (1, 0, –1, –1).\n\nDirect substitution confirms that each quadruple indeed makes Q(x) vanish at x = a, b, c, d.\n\nTherefore, the complete answer is  \n {(0,0,0,0), (1,0,–2,0), (1,0,–1,–1)}.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.377348",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: The unknown-set grew from triples to quadruples, turning three Viète equations into four intertwined ones, and forcing a two-level case-analysis.\n\n2. Additional constraints: Scrambling the coefficients (constant term = b, not d) prevents naively re-using the cubic argument and introduces the non-trivial product condition abcd = b.\n\n3. Deeper theory:  \n   • Rational-Root Test on a quartic and its cubic factor;  \n   • Complete enumeration of sign patterns using divisibility (acd = 1);  \n   • Factorisation of a quartic to isolate the only admissible integral root.\n\n4. Multi-concept interaction: Viète symmetries, divisibility, rational-root techniques and systematic case-splitting all interact; omitting any part leaves the problem unsolved.\n\nAltogether the enhanced variant demands substantially more algebraic manipulation and finer number-theoretic reasoning than either the original or the current kernel cubic, hence is significantly harder."
      }
    },
    "original_kernel_variant": {
      "question": "Let a, b, c, d be integers.  \nDetermine all ordered quadruples (a,b,c,d) ∈ ℤ⁴ for which the four numbers a, b, c, d themselves are (with multiplicity) the four roots of the monic quartic polynomial  \n\n  Q(x)=x⁴ + a x³ + c x² + d x + b.  \n\n(Notice that, as in the cubic kernel, the constant term and the x²– and x–coefficients are deliberately “scrambled’’.)",
      "solution": "Step 1.  Translate the root–coefficient condition via Viète.  \nBecause a, b, c, d are the roots of Q we have  \n\n(1) a + b + c + d = –a,  \n\n(2) ab + ac + ad + bc + bd + cd =  c,  \n\n(3) abc + abd + acd + bcd = –d,  \n\n(4) abcd =  b.  \n\nStep 2.  Treat the two main cases b = 0 and b ≠ 0.\n\n────────────────────────────────────────  \nCase 1.  b = 0  \n\nEquation (4) is automatic.  The remaining relations become  \n\n(1′) c + d = –2a,  \n\n(2′) ac + ad + cd = c,  \n\n(3′) acd = –d.  \n\n–––––  Sub-case 1.1: d = 0  \nThen (3′) is automatic and (2′) reduces to ac = c, i.e. c(a – 1)=0.\n\n•  If c = 0, then (1′) gives a = 0, yielding the solution (0,0,0,0).  \n•  If a = 1, then (1′) gives c = –2, giving (1,0,–2,0).\n\n–––––  Sub-case 1.2: d ≠ 0  \nFrom (3′) we get ac = –1  ⇒  c = –1/a and a ≠ 0.\n\nInsert c and d = –2a – c = –2a + 1/a into (2′):\n\n ac + ad + cd = c  \n –1 + a d + c d = –1/a.\n\nAfter clearing denominators one obtains\n\n 2a⁴ – 2a² – a + 1 = 0.                     (★)\n\nFactorising,   \n 2a⁴ – 2a² – a + 1 = (a – 1)(2a³ + 2a² – 1).\n\nThe cubic factor has no integer (indeed no rational) root by the Rational-Root Test, so the only integral solution of (★) is a = 1.  \nThus c = –1 and d = –2·1 + 1 = –1, giving the third solution (1,0,–1,–1).\n\n────────────────────────────────────────  \nCase 2.  b ≠ 0  \n\nNow (4) gives acd = 1.  Hence a, c, d are all divisors of 1, so each is ±1.\n\nList the four possibilities for (a,c,d):\n\n (i) (1, 1, 1) (ii) (1, –1, –1) (iii) (–1, 1, –1) (iv) (–1, –1, 1).\n\nFor each, compute b from (1) (b = –2a – c – d) and test (2).\n\n(i) (1,1,1): b = –4; (2) gives –9 = 1 ✘  \n(ii) (1,–1,–1): b = 0 (not allowed in Case 2)  \n(iii) (–1,1,–1): b = 2; (2) gives –3 = 1 ✘  \n(iv) (–1,–1,1): b = 2; (2) gives –3 = –1 ✘  \n\nNo choice satisfies all equations, so Case 2 yields no solutions.\n\n────────────────────────────────────────  \nStep 3.  Collect the solutions.\n\nThe only integral quadruples satisfying (1)–(4), hence the only ones for which a, b, c, d are the roots of Q, are  \n\n (0, 0, 0, 0), (1, 0, –2, 0), (1, 0, –1, –1).\n\nDirect substitution confirms that each quadruple indeed makes Q(x) vanish at x = a, b, c, d.\n\nTherefore, the complete answer is  \n {(0,0,0,0), (1,0,–2,0), (1,0,–1,–1)}.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.325101",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: The unknown-set grew from triples to quadruples, turning three Viète equations into four intertwined ones, and forcing a two-level case-analysis.\n\n2. Additional constraints: Scrambling the coefficients (constant term = b, not d) prevents naively re-using the cubic argument and introduces the non-trivial product condition abcd = b.\n\n3. Deeper theory:  \n   • Rational-Root Test on a quartic and its cubic factor;  \n   • Complete enumeration of sign patterns using divisibility (acd = 1);  \n   • Factorisation of a quartic to isolate the only admissible integral root.\n\n4. Multi-concept interaction: Viète symmetries, divisibility, rational-root techniques and systematic case-splitting all interact; omitting any part leaves the problem unsolved.\n\nAltogether the enhanced variant demands substantially more algebraic manipulation and finer number-theoretic reasoning than either the original or the current kernel cubic, hence is significantly harder."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}