{
  "index": "1956-B-7",
  "type": "ALG",
  "tag": [
    "ALG",
    "ANA"
  ],
  "difficulty": "",
  "question": "7. The polynomials \\( P(z) \\) and \\( Q(z) \\) with complex coefficients have the same set of numbers for their zeros but possibly different multiplicities. The same is true of the polynomials\n\\[\nP(z)+1 \\text { and } Q(z)+1\n\\]\n\nProve that \\( P(z) \\equiv Q(z) \\).",
  "solution": "Solution. The desired conclusion is false for polynomials of degree zero, so we shall assume that at least one of the polynomials is not constant. Suppose \\( P \\) has degree \\( m \\) and \\( Q \\) has degree \\( n \\). We may assume by symmetry that \\( m \\geq n \\). Let the distinct zeros of \\( P \\) be \\( \\left\\{\\lambda_{1}, \\lambda_{2}, \\ldots, \\lambda_{r}\\right\\} \\), and let the distinct zeros of \\( P+1 \\) be \\( \\left\\{\\mu_{1}, \\mu_{2}, \\ldots, \\mu_{s}\\right\\} \\). These sets are clearly disjoint. Counting multiplicities, \\( P^{\\prime} \\), the derivative of both \\( P \\) and \\( P+1 \\), must have at least \\( (m-r) \\) zeros in \\( \\left\\{\\lambda_{1}, \\lambda_{2}, \\ldots, \\lambda_{r}\\right\\} \\) and \\( (m-s) \\) zeros in \\( \\left\\{\\mu_{1}, \\mu_{2}, \\ldots, \\mu_{3}\\right\\}: \\) Therefore \\( (m-r)+(m-s) \\leq m-1 \\), the degree of \\( P^{\\prime} \\). (It is here that the assumption \\( m>0 \\) enters.) Hence \\( r+s>m \\). But each of the \\( r+s \\) numbers \\( \\lambda_{1}, \\lambda_{2}, \\ldots, \\lambda_{r}, \\mu_{1}, \\mu_{2}, \\ldots, \\mu_{s} \\) is a zero of \\( P-Q \\), a polynomial of degree at most \\( m \\). It follows that \\( P(z) \\equiv Q(z) \\) as required.\n\nRemark. This problem is discussed in a more general setting by W. W. Adams and E. G. Straus, \"Non-Archimedean Analytic Functions Taking the Same Value at the Same Points,\" Illinois Journal of Mathematics, vol. 15 (1971), pages 418-424.",
  "vars": [
    "z"
  ],
  "params": [
    "P",
    "Q",
    "m",
    "n",
    "r",
    "s",
    "\\\\lambda_1",
    "\\\\lambda_2",
    "\\\\lambda_r",
    "\\\\mu_1",
    "\\\\mu_2",
    "\\\\mu_s"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "z": "variable",
        "P": "polyone",
        "Q": "polytwo",
        "m": "degreep",
        "n": "degreeq",
        "r": "rootcountp",
        "s": "rootcountsum",
        "\\lambda_1": "lambdafirst",
        "\\lambda_2": "lambdasecond",
        "\\lambda_r": "lambdalast",
        "\\mu_1": "mufirst",
        "\\mu_2": "musecond",
        "\\mu_s": "mulast"
      },
      "question": "7. The polynomials \\( polyone(variable) \\) and \\( polytwo(variable) \\) with complex coefficients have the same set of numbers for their zeros but possibly different multiplicities. The same is true of the polynomials\n\\[\npolyone(variable)+1 \\text { and } polytwo(variable)+1\n\\]\n\nProve that \\( polyone(variable) \\equiv polytwo(variable) \\).",
      "solution": "Solution. The desired conclusion is false for polynomials of degree zero, so we shall assume that at least one of the polynomials is not constant. Suppose \\( polyone \\) has degree \\( degreep \\) and \\( polytwo \\) has degree \\( degreeq \\). We may assume by symmetry that \\( degreep \\geq degreeq \\). Let the distinct zeros of \\( polyone \\) be \\( \\left\\{lambdafirst, lambdasecond, \\ldots, lambdalast\\right\\} \\), and let the distinct zeros of \\( polyone+1 \\) be \\( \\left\\{mufirst, musecond, \\ldots, mulast\\right\\} \\). These sets are clearly disjoint. Counting multiplicities, \\( polyone^{\\prime} \\), the derivative of both \\( polyone \\) and \\( polyone+1 \\), must have at least \\( (degreep-rootcountp) \\) zeros in \\( \\left\\{lambdafirst, lambdasecond, \\ldots, lambdalast\\right\\} \\) and \\( (degreep-rootcountsum) \\) zeros in \\( \\left\\{mufirst, musecond, \\ldots, mulast\\right\\}: \\) Therefore \\( (degreep-rootcountp)+(degreep-rootcountsum) \\leq degreep-1 \\), the degree of \\( polyone^{\\prime} \\). (It is here that the assumption \\( degreep>0 \\) enters.) Hence \\( rootcountp+rootcountsum>degreep \\). But each of the \\( rootcountp+rootcountsum \\) numbers \\( lambdafirst, lambdasecond, \\ldots, lambdalast, mufirst, musecond, \\ldots, mulast \\) is a zero of \\( polyone-polytwo \\), a polynomial of degree at most \\( degreep \\). It follows that \\( polyone(variable) \\equiv polytwo(variable) \\) as required.\n\nRemark. This problem is discussed in a more general setting by W. W. Adams and E. G. Straus, \"Non-Archimedean Analytic Functions Taking the Same Value at the Same Points,\" Illinois Journal of Mathematics, vol. 15 (1971), pages 418-424."
    },
    "descriptive_long_confusing": {
      "map": {
        "z": "gingerale",
        "P": "locomotive",
        "Q": "artichoke",
        "m": "firetruck",
        "n": "raspberries",
        "r": "masquerade",
        "s": "backpacker",
        "\\lambda_1": "windmillone",
        "\\lambda_2": "windmilltwo",
        "\\lambda_r": "windmillmask",
        "\\mu_1": "cornflowerone",
        "\\mu_2": "cornflowertwo",
        "\\mu_s": "cornflowerpack"
      },
      "question": "7. The polynomials \\( locomotive(gingerale) \\) and \\( artichoke(gingerale) \\) with complex coefficients have the same set of numbers for their zeros but possibly different multiplicities. The same is true of the polynomials\n\\[\nlocomotive(gingerale)+1 \\text { and } artichoke(gingerale)+1\n\\]\n\nProve that \\( locomotive(gingerale) \\equiv artichoke(gingerale) \\).",
      "solution": "Solution. The desired conclusion is false for polynomials of degree zero, so we shall assume that at least one of the polynomials is not constant. Suppose \\( locomotive \\) has degree \\( firetruck \\) and \\( artichoke \\) has degree \\( raspberries \\). We may assume by symmetry that \\( firetruck \\geq raspberries \\). Let the distinct zeros of \\( locomotive \\) be \\( \\left\\{windmillone, windmilltwo, \\ldots, windmillmask\\right\\} \\), and let the distinct zeros of \\( locomotive+1 \\) be \\( \\left\\{cornflowerone, cornflowertwo, \\ldots, cornflowerpack\\right\\} \\). These sets are clearly disjoint. Counting multiplicities, \\( locomotive^{\\prime} \\), the derivative of both \\( locomotive \\) and \\( locomotive+1 \\), must have at least \\( (firetruck-masquerade) \\) zeros in \\( \\left\\{windmillone, windmilltwo, \\ldots, windmillmask\\right\\} \\) and \\( (firetruck-backpacker) \\) zeros in \\( \\left\\{cornflowerone, cornflowertwo, \\ldots, cornflowerpack\\right\\}: \\) Therefore \\( (firetruck-masquerade)+(firetruck-backpacker) \\leq firetruck-1 \\), the degree of \\( locomotive^{\\prime} \\). (It is here that the assumption \\( firetruck>0 \\) enters.) Hence \\( masquerade+backpacker>firetruck \\). But each of the \\( masquerade+backpacker \\) numbers \\( windmillone, windmilltwo, \\ldots, windmillmask, cornflowerone, cornflowertwo, \\ldots, cornflowerpack \\) is a zero of \\( locomotive-artichoke \\), a polynomial of degree at most \\( firetruck \\). It follows that \\( locomotive(gingerale) \\equiv artichoke(gingerale) \\) as required.\n\nRemark. This problem is discussed in a more general setting by W. W. Adams and E. G. Straus, \"Non-Archimedean Analytic Functions Taking the Same Value at the Same Points,\" Illinois Journal of Mathematics, vol. 15 (1971), pages 418-424."
    },
    "descriptive_long_misleading": {
      "map": {
        "z": "realvalue",
        "P": "constantfn",
        "Q": "fixedvalue",
        "m": "infinitenum",
        "n": "boundless",
        "r": "singleroot",
        "s": "uniqueroot",
        "\\lambda_1": "nonrootone",
        "\\lambda_2": "nonroottwo",
        "\\lambda_r": "nonrootmax",
        "\\mu_1": "nonshiftone",
        "\\mu_2": "nonshifttwo",
        "\\mu_s": "nonshifts"
      },
      "question": "7. The polynomials \\( constantfn(realvalue) \\) and \\( fixedvalue(realvalue) \\) with complex coefficients have the same set of numbers for their zeros but possibly different multiplicities. The same is true of the polynomials\n\\[\nconstantfn(realvalue)+1 \\text { and } fixedvalue(realvalue)+1\n\\]\n\nProve that \\( constantfn(realvalue) \\equiv fixedvalue(realvalue) \\).",
      "solution": "Solution. The desired conclusion is false for polynomials of degree zero, so we shall assume that at least one of the polynomials is not constant. Suppose \\( constantfn \\) has degree \\( infinitenum \\) and \\( fixedvalue \\) has degree \\( boundless \\). We may assume by symmetry that \\( infinitenum \\geq boundless \\). Let the distinct zeros of \\( constantfn \\) be \\( \\left\\{nonrootone, nonroottwo, \\ldots, nonrootmax\\right\\} \\), and let the distinct zeros of \\( constantfn+1 \\) be \\( \\left\\{nonshiftone, nonshifttwo, \\ldots, nonshifts\\right\\} \\). These sets are clearly disjoint. Counting multiplicities, \\( constantfn^{\\prime} \\), the derivative of both \\( constantfn \\) and \\( constantfn+1 \\), must have at least \\( (infinitenum-singleroot) \\) zeros in \\( \\left\\{nonrootone, nonroottwo, \\ldots, nonrootmax\\right\\} \\) and \\( (infinitenum-uniqueroot) \\) zeros in \\( \\left\\{nonshiftone, nonshifttwo, \\ldots, \\mu_{3}\\right\\}: \\) Therefore \\( (infinitenum-singleroot)+(infinitenum-uniqueroot) \\leq infinitenum-1 \\), the degree of \\( constantfn^{\\prime} \\). (It is here that the assumption \\( infinitenum>0 \\) enters.) Hence \\( singleroot+uniqueroot>infinitenum \\). But each of the \\( singleroot+uniqueroot \\) numbers \\( nonrootone, nonroottwo, \\ldots, nonrootmax, nonshiftone, nonshifttwo, \\ldots, nonshifts \\) is a zero of \\( constantfn-fixedvalue \\), a polynomial of degree at most \\( infinitenum \\). It follows that \\( constantfn(realvalue) \\equiv fixedvalue(realvalue) \\) as required.\n\nRemark. This problem is discussed in a more general setting by W. W. Adams and E. G. Straus, \"Non-Archimedean Analytic Functions Taking the Same Value at the Same Points,\" Illinois Journal of Mathematics, vol. 15 (1971), pages 418-424."
    },
    "garbled_string": {
      "map": {
        "z": "vxmpeojq",
        "P": "qzxwvtnp",
        "Q": "hjgrksla",
        "m": "lkzprmva",
        "n": "vdoyxgtr",
        "r": "cmpqznea",
        "s": "fhtbglow",
        "\\lambda_1": "arnsleven",
        "\\lambda_2": "ythopkui",
        "\\lambda_r": "gquzsmec",
        "\\mu_1": "xfwpphri",
        "\\mu_2": "dlcqazno",
        "\\mu_s": "zemkoati"
      },
      "question": "7. The polynomials \\( qzxwvtnp(vxmpeojq) \\) and \\( hjgrksla(vxmpeojq) \\) with complex coefficients have the same set of numbers for their zeros but possibly different multiplicities. The same is true of the polynomials\n\\[\nqzxwvtnp(vxmpeojq)+1 \\text { and } hjgrksla(vxmpeojq)+1\n\\]\n\nProve that \\( qzxwvtnp(vxmpeojq) \\equiv hjgrksla(vxmpeojq) \\).",
      "solution": "Solution. The desired conclusion is false for polynomials of degree zero, so we shall assume that at least one of the polynomials is not constant. Suppose \\( qzxwvtnp \\) has degree \\( lkzprmva \\) and \\( hjgrksla \\) has degree \\( vdoyxgtr \\). We may assume by symmetry that \\( lkzprmva \\geq vdoyxgtr \\). Let the distinct zeros of \\( qzxwvtnp \\) be \\( \\left\\{arnsleven, ythopkui, \\ldots, gquzsmec\\right\\} \\), and let the distinct zeros of \\( qzxwvtnp+1 \\) be \\( \\left\\{xfwpphri, dlcqazno, \\ldots, zemkoati\\right\\} \\). These sets are clearly disjoint. Counting multiplicities, \\( qzxwvtnp^{\\prime} \\), the derivative of both \\( qzxwvtnp \\) and \\( qzxwvtnp+1 \\), must have at least \\( (lkzprmva-cmpqznea) \\) zeros in \\( \\left\\{arnsleven, ythopkui, \\ldots, gquzsmec\\right\\} \\) and \\( (lkzprmva-fhtbglow) \\) zeros in \\( \\left\\{xfwpphri, dlcqazno, \\ldots, \\mu_{3}\\right\\}: \\) Therefore \\( (lkzprmva-cmpqznea)+(lkzprmva-fhtbglow) \\leq lkzprmva-1 \\), the degree of \\( qzxwvtnp^{\\prime} \\). (It is here that the assumption \\( lkzprmva>0 \\) enters.) Hence \\( cmpqznea+fhtbglow>lkzprmva \\). But each of the \\( cmpqznea+fhtbglow \\) numbers \\( arnsleven, ythopkui, \\ldots, gquzsmec, xfwpphri, dlcqazno, \\ldots, zemkoati \\) is a zero of \\( qzxwvtnp-hjgrksla \\), a polynomial of degree at most \\( lkzprmva \\). It follows that \\( qzxwvtnp(vxmpeojq) \\equiv hjgrksla(vxmpeojq) \\) as required.\n\nRemark. This problem is discussed in a more general setting by W. W. Adams and E. G. Straus, \"Non-Archimedean Analytic Functions Taking the Same Value at the Same Points,\" Illinois Journal of Mathematics, vol. 15 (1971), pages 418-424."
    },
    "kernel_variant": {
      "question": "Let  P(z),Q(z) \\in \\overline{\\mathbb Q}[z]  be two polynomials which are NOT BOTH constant.  Assume\n1.  P and Q have the same set of (distinct) zeros in \\overline{\\mathbb Q}  (multiplicities need not agree), and\n2.  the polynomials  P(z)+2  and  Q(z)+2  also have the same set of (distinct) zeros.\n\nProve that  P \\equiv Q.",
      "solution": "We work over the algebraic closure  \\overline{\\Q}.  Throughout, equality of zero-sets is meant \"as sets\", i.e. multiplicities are ignored.\n\nStep 1.  Both polynomials have positive degree.\nThe hypothesis excludes the possibility that **both** polynomials are constant, but a priori one of them might be constant.  Suppose, for instance, that  P is constant.  If P = 0 then its zero-set is all of  \\overline{\\Q}, whereas Q, being a non–zero polynomial, has only finitely many zeros – contradiction.  If P = c \\neq 0 then P has no zeros, so Q would have to have no zeros either.  A non–constant polynomial over an algebraically closed field always has at least one zero, so Q would also be constant, again contradicting the hypothesis that P and Q are not both constant.  Hence both polynomials are non-constant and therefore of positive degree.\n\nLet  m = \\deg P  and  n = \\deg Q.  By symmetry we may assume  n \\ge m > 0.\n\nStep 2.  Notation for the distinct zeros.\nWrite\n    {\\lambda_1,\\dots,\\lambda_r}   for the distinct zeros of  Q,\n    {\\mu_1,\\dots,\\mu_s}           for the distinct zeros of  Q+2.\nBecause  Q(\\lambda_i)=0  implies  Q(\\lambda_i)+2=2\\neq0, and  Q(\\mu_j)+2=0  implies  Q(\\mu_j)=-2\\neq0, the two sets \\{\\lambda_i\\} and \\{\\mu_j\\} are disjoint.\n\nStep 3.  A bound coming from the derivative.\nEach multiple root of  Q  (respectively  Q+2) is also a root of the common derivative  Q'.  Counting multiplicities one gets\n   Q' has at least  (n-r)  zeros among the \\lambda_i\n   Q' has at least  (n-s)  zeros among the \\mu_j.\nTherefore\n   (n-r) + (n-s) \\le \\deg Q' = n-1,\nwhence  r+s \\ge n+1, i.e.  r+s > n.   (1)\n\nStep 4.  The polynomial  Q-P.\nFor every \\lambda_i we have  P(\\lambda_i)=0  because P and Q share their zero-set.  For every \\mu_j we have  P(\\mu_j)+2=0 (by the second hypothesis) and hence  P(\\mu_j)=Q(\\mu_j).  Thus each \\lambda_i and each \\mu_j is a zero of  Q-P.  Altogether  Q-P  has at least  r+s  distinct zeros.\n\nBut  \\deg(Q-P) \\le \\max(n,m)=n, while (1) says that  r+s>n.  A non-zero polynomial of degree \\le n cannot have more than n distinct zeros, so the only possibility is  Q-P\\equiv0.  Hence  P \\equiv Q.\n\nThe proof is complete.",
      "_meta": {
        "core_steps": [
          "Assume at least one polynomial is non-constant; let m = deg P ≥ deg Q.",
          "Let λᵢ be the distinct zeros of P and μⱼ those of P+1; these two sets are disjoint.",
          "Because every multiple root of P or P+1 is also a root of P′, P′ has ≥(m−r)+(m−s) zeros, so (m−r)+(m−s) ≤ m−1 ⇒ r+s>m.",
          "Both λᵢ and μⱼ are zeros of P−Q, which has degree ≤ m; since r+s>m, P−Q is identically zero.",
          "Conclude P ≡ Q."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The fixed constant added to the polynomials.",
            "original": "+1"
          },
          "slot2": {
            "description": "The field over which the polynomials are taken (requires algebraic closure so every root is counted).",
            "original": "complex numbers ℂ"
          },
          "slot3": {
            "description": "The designation of which polynomial is assumed to have larger (or equal) degree.",
            "original": "Assume m = deg P ≥ deg Q by symmetry"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}