{
  "index": "1955-A-6",
  "type": "ALG",
  "tag": [
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "6. Find a necessary and sufficient condition on the positive integer \\( n \\) that the equation\n\\[\nx^{n}+(2+x)^{n}+(2-x)^{n}=0\n\\]\nhave a rational root.",
  "solution": "Solution. There can be no real root if \\( n \\) is even, since for real \\( x \\) each term is non-negative and they cannot vanish simultaneously.\n\nIf \\( n=1 \\), there is obviously a unique root \\( x=-4 \\).\nSuppose \\( n \\) is odd and at least 3 . When the terms of the equation are expanded and collected, the result is monic with all coefficients nonnegative integers and constant term \\( 2^{n+1} \\). The only possible roots therefore are of the form \\( -2^{\\prime} \\). For \\( x=-1 \\), all three terms of the given expression are odd, so -1 is not a root. Putting \\( x=-2 \\), we find ( -2\\( )^{n}+0+ \\) \\( 4^{n} \\neq 0 \\), so -2 is not a root. If we put \\( x=-2^{p+1} \\) where \\( p \\geq 1 \\), the left member of (1) becomes\n\\[\n\\begin{aligned}\n2^{n}\\left[-2^{p n}+\\left(1-2^{p}\\right)^{n}\\right. & \\left.+\\left(1+2^{p}\\right)^{n}\\right] \\\\\n& =2^{n}\\left[-2^{p n}+2\\left\\{1+\\binom{n}{2} 2^{2 p}+\\binom{n}{4} 2^{4 p}+\\cdots\\right\\}\\right] .\n\\end{aligned}\n\\]\n\nThe expression in the brackets is \\( \\equiv 2[\\bmod 4] \\) (recall we are assuming \\( n \\geq 3 \\) ), so \\( -2^{p+1} \\) is not a root for \\( p \\geq 1 \\). So there are no roots if \\( n>1 \\). Summarizing, relation (1) has a rational root if and only if \\( n=1 \\).\n\nRemark. A root with \\( \\boldsymbol{n}>\\mathbf{2} \\) would give a counterexample to the famous Fermat Conjecture.",
  "vars": [
    "x",
    "p"
  ],
  "params": [
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variable",
        "p": "poweridx",
        "n": "exponent"
      },
      "question": "6. Find a necessary and sufficient condition on the positive integer \\( exponent \\) that the equation\n\\[\nvariable^{exponent}+(2+variable)^{exponent}+(2-variable)^{exponent}=0\n\\]\nhave a rational root.",
      "solution": "There can be no real root if \\( exponent \\) is even, since for real \\( variable \\) each term is non-negative and they cannot vanish simultaneously.\n\nIf \\( exponent=1 \\), there is obviously a unique root \\( variable=-4 \\).\nSuppose \\( exponent \\) is odd and at least 3. When the terms of the equation are expanded and collected, the result is monic with all coefficients nonnegative integers and constant term \\( 2^{exponent+1} \\). The only possible roots therefore are of the form \\( -2^{\\prime} \\). For \\( variable=-1 \\), all three terms of the given expression are odd, so -1 is not a root. Putting \\( variable=-2 \\), we find \\( (-2)^{exponent}+0+4^{exponent} \\neq 0 \\), so -2 is not a root. If we put \\( variable=-2^{poweridx+1} \\) where \\( poweridx \\geq 1 \\), the left member of (1) becomes\n\\[\n\\begin{aligned}\n2^{exponent}\\left[-2^{poweridx\\,exponent}+\\left(1-2^{poweridx}\\right)^{exponent}\\right. & \\left.+\\left(1+2^{poweridx}\\right)^{exponent}\\right] \\\\\n& =2^{exponent}\\left[-2^{poweridx\\,exponent}+2\\left\\{1+\\binom{exponent}{2} 2^{2\\,poweridx}+\\binom{exponent}{4} 2^{4\\,poweridx}+\\cdots\\right\\}\\right] .\n\\end{aligned}\n\\]\n\nThe expression in the brackets is \\( \\equiv 2[\\bmod 4] \\) (recall we are assuming \\( exponent \\geq 3 \\) ), so \\( -2^{poweridx+1} \\) is not a root for \\( poweridx \\geq 1 \\). So there are no roots if \\( exponent>1 \\). Summarizing, relation (1) has a rational root if and only if \\( exponent=1 \\).\n\nRemark. A root with \\( \\boldsymbol{exponent}>\\mathbf{2} \\) would give a counterexample to the famous Fermat Conjecture."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "cloudburst",
        "p": "driftwood",
        "n": "starlight"
      },
      "question": "6. Find a necessary and sufficient condition on the positive integer \\( starlight \\) that the equation\n\\[\ncloudburst^{starlight}+(2+cloudburst)^{starlight}+(2-cloudburst)^{starlight}=0\n\\]\nhave a rational root.",
      "solution": "Solution. There can be no real root if \\( starlight \\) is even, since for real \\( cloudburst \\) each term is non-negative and they cannot vanish simultaneously.\n\nIf \\( starlight=1 \\), there is obviously a unique root \\( cloudburst=-4 \\).\nSuppose \\( starlight \\) is odd and at least 3 . When the terms of the equation are expanded and collected, the result is monic with all coefficients nonnegative integers and constant term \\( 2^{starlight+1} \\). The only possible roots therefore are of the form \\( -2^{\\prime} \\). For \\( cloudburst=-1 \\), all three terms of the given expression are odd, so -1 is not a root. Putting \\( cloudburst=-2 \\), we find ( -2\\( )^{starlight}+0+ \\) \\( 4^{starlight} \\neq 0 \\), so -2 is not a root. If we put \\( cloudburst=-2^{driftwood+1} \\) where \\( driftwood \\geq 1 \\), the left member of (1) becomes\n\\[\n\\begin{aligned}\n2^{starlight}\\left[-2^{driftwood\\, starlight}+\\left(1-2^{driftwood}\\right)^{starlight}\\right. & \\left.+\\left(1+2^{driftwood}\\right)^{starlight}\\right] \\\\\n& =2^{starlight}\\left[-2^{driftwood\\, starlight}+2\\left\\{1+\\binom{starlight}{2} 2^{2 driftwood}+\\binom{starlight}{4} 2^{4 driftwood}+\\cdots\\right\\}\\right] .\n\\end{aligned}\n\\]\n\nThe expression in the brackets is \\( \\equiv 2[\\bmod 4] \\) (recall we are assuming \\( starlight \\geq 3 \\) ), so \\( -2^{driftwood+1} \\) is not a root for \\( driftwood \\geq 1 \\). So there are no roots if \\( starlight>1 \\). Summarizing, relation (1) has a rational root if and only if \\( starlight=1 \\).\n\nRemark. A root with \\( \\boldsymbol{starlight}>\\mathbf{2} \\) would give a counterexample to the famous Fermat Conjecture."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "p": "compositefig",
        "n": "fractionalexp"
      },
      "question": "6. Find a necessary and sufficient condition on the positive integer \\( fractionalexp \\) that the equation\n\\[\nconstantval^{fractionalexp}+(2+constantval)^{fractionalexp}+(2-constantval)^{fractionalexp}=0\n\\]\nhave a rational root.",
      "solution": "Solution. There can be no real root if \\( fractionalexp \\) is even, since for real \\( constantval \\) each term is non-negative and they cannot vanish simultaneously.\n\nIf \\( fractionalexp=1 \\), there is obviously a unique root \\( constantval=-4 \\).\nSuppose \\( fractionalexp \\) is odd and at least 3 . When the terms of the equation are expanded and collected, the result is monic with all coefficients nonnegative integers and constant term \\( 2^{fractionalexp+1} \\). The only possible roots therefore are of the form \\( -2^{\\prime} \\). For \\( constantval=-1 \\), all three terms of the given expression are odd, so -1 is not a root. Putting \\( constantval=-2 \\), we find ( -2\\( )^{fractionalexp}+0+ \\) \\( 4^{fractionalexp} \\neq 0 \\), so -2 is not a root. If we put \\( constantval=-2^{compositefig+1} \\) where \\( compositefig \\geq 1 \\), the left member of (1) becomes\n\\[\n\\begin{aligned}\n2^{fractionalexp}\\left[-2^{compositefig fractionalexp}+\\left(1-2^{compositefig}\\right)^{fractionalexp}\\right. & \\left.+\\left(1+2^{compositefig}\\right)^{fractionalexp}\\right] \\\\\n& =2^{fractionalexp}\\left[-2^{compositefig fractionalexp}+2\\left\\{1+\\binom{fractionalexp}{2} 2^{2 compositefig}+\\binom{fractionalexp}{4} 2^{4 compositefig}+\\cdots\\right\\}\\right] .\n\\end{aligned}\n\\]\n\nThe expression in the brackets is \\( \\equiv 2[\\bmod 4] \\) (recall we are assuming \\( fractionalexp \\geq 3 \\) ), so \\( -2^{compositefig+1} \\) is not a root for \\( compositefig \\geq 1 \\). So there are no roots if \\( fractionalexp>1 \\). Summarizing, relation (1) has a rational root if and only if \\( fractionalexp=1 \\).\n\nRemark. A root with \\( \\boldsymbol{fractionalexp}>\\mathbf{2} \\) would give a counterexample to the famous Fermat Conjecture."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "p": "hjgrksla",
        "n": "vbmqcrdu"
      },
      "question": "Find a necessary and sufficient condition on the positive integer \\( vbmqcrdu \\) that the equation\n\\[\nqzxwvtnp^{vbmqcrdu}+(2+qzxwvtnp)^{vbmqcrdu}+(2-qzxwvtnp)^{vbmqcrdu}=0\n\\]\nhave a rational root.",
      "solution": "Solution. There can be no real root if \\( vbmqcrdu \\) is even, since for real \\( qzxwvtnp \\) each term is non-negative and they cannot vanish simultaneously.\n\nIf \\( vbmqcrdu=1 \\), there is obviously a unique root \\( qzxwvtnp=-4 \\).\nSuppose \\( vbmqcrdu \\) is odd and at least 3 . When the terms of the equation are expanded and collected, the result is monic with all coefficients nonnegative integers and constant term \\( 2^{vbmqcrdu+1} \\). The only possible roots therefore are of the form \\( -2^{\\prime} \\). For \\( qzxwvtnp=-1 \\), all three terms of the given expression are odd, so -1 is not a root. Putting \\( qzxwvtnp=-2 \\), we find ( -2\\( )^{vbmqcrdu}+0+ \\) \\( 4^{vbmqcrdu} \\neq 0 \\), so -2 is not a root. If we put \\( qzxwvtnp=-2^{hjgrksla+1} \\) where \\( hjgrksla \\geq 1 \\), the left member of (1) becomes\n\\[\n\\begin{aligned}\n2^{vbmqcrdu}\\left[-2^{hjgrksla vbmqcrdu}+\\left(1-2^{hjgrksla}\\right)^{vbmqcrdu}\\right. & \\left.+\\left(1+2^{hjgrksla}\\right)^{vbmqcrdu}\\right] \\\\\n& =2^{vbmqcrdu}\\left[-2^{hjgrksla vbmqcrdu}+2\\left\\{1+\\binom{vbmqcrdu}{2} 2^{2 hjgrksla}+\\binom{vbmqcrdu}{4} 2^{4 hjgrksla}+\\cdots\\right\\}\\right] .\n\\end{aligned}\n\\]\n\nThe expression in the brackets is \\( \\equiv 2[\\bmod 4] \\) (recall we are assuming \\( vbmqcrdu \\geq 3 \\) ), so \\( -2^{hjgrksla+1} \\) is not a root for \\( hjgrksla \\geq 1 \\). So there are no roots if \\( vbmqcrdu>1 \\). Summarizing, relation (1) has a rational root if and only if \\( vbmqcrdu=1 \\).\n\nRemark. A root with \\( \\boldsymbol{vbmqcrdu}>\\mathbf{2} \\) would give a counterexample to the famous Fermat Conjecture."
    },
    "kernel_variant": {
      "question": "Fix the integer  \n\\[\nd=2\n\\]  \nand, for every positive integer \\(n\\), define  \n\\[\nP_{n}(y)=\\bigl(y-2\\bigr)^{n}+y^{\\,n}+\\bigl(y+2\\bigr)^{n}\\in\\mathbf{Q}[y]\\qquad(n\\ge 1).\n\\]\n\n1.  For two \\emph{distinct} positive integers \\(m<n\\), determine explicitly the monic polynomial  \n   \\[\n      \\gcd_{\\mathbf{Q}[y]}\\bigl(P_{m},P_{n}\\bigr).\n   \\]\n\n2.  Find all ordered pairs \\((m,n)\\) of distinct positive integers for which \\(P_{m}\\) and \\(P_{n}\\) possess a common rational root, and list every such root.\n\n3.  Classify (with proof) all non–empty finite index sets  \n   \\[\n      \\mathcal{N}\\subset\\mathbf{N}\n   \\]\n   for which there exists \\(r\\in\\mathbf{Q}\\) satisfying \\(P_{n}(r)=0\\) for \\emph{every} \\(n\\in\\mathcal{N}\\).\n\n(The three questions are strongly coupled; answers obtained independently must nevertheless be proved within a coherent global argument.)\n\n--------------------------------------------------------------------",
      "solution": "Throughout we let  \n\\[\nP_{n}(y)=\\bigl(y-2\\bigr)^{n}+y^{\\,n}+\\bigl(y+2\\bigr)^{n}\\qquad(n\\in\\mathbf{N})\n\\]\nand we always choose a greatest common divisor to be monic.\n\n1.  A preliminary analysis of a single polynomial \\(P_{n}\\)\n\n(a)  Parity.  \n\\[\nP_{n}(-y)=(-1)^{\\,n}P_{n}(y)\\qquad(n\\ge 1),\n\\]\nso \\(P_{n}\\) is even when \\(n\\) is even and odd when \\(n\\) is odd.\n\n(b)  A factorisation when \\(n\\) is odd.  \nPut \\(n=2k+1\\) (\\(k\\ge 0\\)).  Expanding \\((y\\pm2)^{2k+1}\\) gives  \n\\[\n\\begin{aligned}\nP_{n}(y)\n      &=(y-2)^{2k+1}+y^{2k+1}+(y+2)^{2k+1}\\\\\n      &=y\\!\\Bigl[\\,3y^{2k}\n        +2\\sum_{j=1}^{k}\\binom{2k+1}{2j}2^{2j}y^{2(k-j)}\\Bigr]\n      =:y\\,Q_{n}(y),\n\\end{aligned}\n\\]\nwhere \\(Q_{n}\\in\\mathbf{Z}[y]\\) has only positive coefficients; hence  \n\\[\nQ_{n}(y)>0\\quad\\forall\\,y\\in\\mathbf{R}. \\tag{1.1}\n\\]\n\n(c)  Real and rational roots.\n\nLemma 1.1.  \nIf \\(n\\) is even, then \\(P_{n}(y)>0\\) for all \\(y\\in\\mathbf{R}\\); consequently \\(P_{n}\\) has no rational root.\n\nProof.  Each of the three summands is non–negative, and at most one of them can vanish; hence the sum is strictly positive. \\(\\square\\)\n\nLemma 1.2.  \nIf \\(n\\) is odd, the only rational root of \\(P_{n}\\) is \\(y=0\\).  This root is simple and  \n\\[\nP_{n}'(0)=\n\\begin{cases}\n3,&n=1,\\\\[4pt]\nn\\,2^{\\,n},&n\\ge 3\\text{ odd}.\n\\end{cases}\n\\]\n\nProof.  By (1.1) we have \\(P_{n}(y)=y\\,Q_{n}(y)\\) with \\(Q_{n}(y)>0\\) on \\(\\mathbf{R}\\), so the sole real (and hence rational) zero is \\(0\\).  Differentiating and evaluating at \\(0\\) gives the claimed formula; for \\(n=1\\) the value is \\(3\\). \\(\\square\\)\n\nCorollary 1.3.  \n\\(P_{n}\\) has a rational root exactly when \\(n\\) is odd; then the unique rational root is \\(y=0\\), which is simple.\n\n--------------------------------------------------------------------\n2.  Uniqueness of a non–zero common root  \n\nFix a non–zero complex number \\(z\\) and set  \n\\[\nu:=\\frac{z-2}{z},\\qquad v:=\\frac{z+2}{z}=2-u,\\qquad\nc:=uv=1-\\frac{4}{z^{2}}\\neq 0.\n\\]\nBecause \\(z\\neq 0\\),  \n\\[\nP_{n}(z)=0\\;\\Longleftrightarrow\\;S_{n}:=u^{\\,n}+v^{\\,n}+1=0. \\tag{2.1}\n\\]\nThe sequence \\(\\bigl(S_{n}\\bigr)_{n\\ge 0}\\) satisfies the inhomogeneous linear recurrence  \n\\[\nS_{n+2}-2\\,S_{n+1}+c\\,S_{n}=c-1\\qquad(n\\ge 0). \\tag{2.2}\n\\]\n\nProposition 2.1 (Uniqueness).  \nFor a fixed non–zero complex number \\(z\\) the equality \\(P_{n}(z)=0\\) can hold for \\emph{at most one} positive integer \\(n\\).\n\nProof.  \nAssume, by contradiction, that \\(P_{m}(z)=P_{n}(z)=0\\) for some integers \\(1\\le m<n\\); put \\(d:=n-m\\ge 1\\).\n\nStep 1.  A basic identity.  \nUsing \\(S_{m}=S_{n}=0\\) we derive\n\\[\nu^{m}\\bigl(u^{d}-1\\bigr)=-v^{m}\\bigl(v^{d}-1\\bigr). \\tag{2.3}\n\\]\n\nStep 2.  The case \\(\\lvert u\\rvert\\neq\\lvert v\\rvert\\).  \nSet \\(r:=\\lvert u/v\\rvert\\neq 1\\).  Taking absolute values in (2.3) yields\n\\[\nr^{\\,m}=\n\\frac{\\lvert v^{d}-1\\rvert}{\\lvert u^{d}-1\\rvert}=:\\kappa(d;u,v),\n\\]\nwhere \\(\\kappa(d;u,v)\\) is independent of \\(m\\).  Because \\(r\\neq 1\\), the left–hand side is a strictly monotone function of \\(m\\); hence equality is impossible for two distinct integers \\(m\\).  Thus the simultaneous vanishing of \\(S_{m}\\) and \\(S_{n}\\) cannot occur.\n\nStep 3.  The critical case \\(\\lvert u\\rvert=\\lvert v\\rvert\\).  \nWrite  \n\\[\nu=r\\,e^{\\,i\\theta},\\quad v=r\\,e^{-i\\theta},\\qquad r>1,\\ 0<\\theta<\\pi, \\tag{2.4}\n\\]\nsince \\(u+v=2\\) forces \\(r\\cos\\theta=1\\) and therefore \\(r>1\\).  \nWith (2.4) we have  \n\\[\nS_{k}=2\\,r^{k}\\cos(k\\theta)+1. \\tag{2.5}\n\\]\n\nSuppose \\(S_{m}=S_{n}=0\\) with \\(m<n\\).  Then (2.5) gives  \n\\[\n\\cos(m\\theta)=-\\frac{1}{2r^{m}},\\qquad\n\\cos(n\\theta)=-\\frac{1}{2r^{n}}. \\tag{2.6}\n\\]\n\nReturning to (2.3) and using \\(v/u=e^{-2i\\theta}\\), we obtain  \n\\[\nu^{d}-1=-e^{-2im\\theta}\\,\\overline{u^{d}-1}. \\tag{2.7}\n\\]\nWrite \\(A:=u^{d}-1\\).  Since \\(r>1\\), \\(u^{d}\\neq 1\\) and \\(A\\neq 0\\).  \nTaking complex conjugates in (2.7) and multiplying the two equalities gives  \n\\[\n\\bigl(1-e^{-2im\\theta}\\bigr)\\,|A|^{2}=0.\n\\]\nHence \\(e^{-2im\\theta}=1\\); that is,  \n\\[\n2m\\theta=2\\pi k\\quad(k\\in\\mathbf{Z}). \\tag{2.8}\n\\]\n\nConsequently \\(\\cos(m\\theta)=1\\).  Plugging this into the first relation in (2.6) yields  \n\\[\n1=-\\frac{1}{2r^{m}},\n\\]\ncontradicting \\(r>1\\).  Therefore the assumption \\(S_{m}=S_{n}=0\\) is impossible.\n\nCombining Steps 2 and 3 completes the proof. \\(\\square\\)\n\n--------------------------------------------------------------------\n3.  The greatest common divisor  \n\nProposition 3.1.  For \\(m<n\\) we have  \n\\[\n\\gcd_{\\mathbf{Q}[y]}\\bigl(P_{m},P_{n}\\bigr)=\n\\begin{cases}\ny,&\\text{if }m,n\\text{ are both odd},\\\\[6pt]\n1,&\\text{otherwise}.\n\\end{cases}\n\\]\n\nProof.  \n\n(i)  Assume at least one exponent, say \\(m\\), is even.  \nThen \\(P_{m}\\) possesses no rational root (Lemma 1.1), so \\(y\\nmid P_{m}\\).  \nSuppose, towards a contradiction, that a non–constant \\(D\\in\\mathbf{Q}[y]\\) divides both \\(P_{m}\\) and \\(P_{n}\\).  Let \\(z\\) be a root of \\(D\\).  \n\n• If \\(z=0\\) then \\(P_{m}(0)=2^{\\,m}\\neq 0\\), impossible.  \n\n• If \\(z\\neq 0\\) then \\(P_{m}(z)=P_{n}(z)=0\\), contradicting Proposition 2.1.  \n\nHence \\(\\gcd=1\\).\n\n(ii)  Now suppose both exponents are odd.  \nBy Corollary 1.3, \\(y\\mid P_{m}\\) and \\(y\\mid P_{n}\\), so \\(y\\) divides the gcd.  \nThe zero at \\(y=0\\) is simple in each polynomial (Lemma 1.2), implying \\(y^{2}\\nmid P_{m}\\) and therefore \\(y^{2}\\nmid\\gcd\\).  \nIf the gcd possessed another irreducible factor, it would yield a non–zero common root, impossible by Proposition 2.1.  Thus \\(\\gcd=y\\).  \\(\\square\\)\n\n--------------------------------------------------------------------\n4.  Common rational roots of two members  \n\nTheorem 4.1.  \nThe polynomials \\(P_{m}\\) and \\(P_{n}\\) share a rational root precisely when both \\(m\\) and \\(n\\) are odd; their only common rational root is \\(y=0\\).\n\nProof.  \nIf one exponent is even, the corresponding polynomial has no rational root (Lemma 1.1); hence no common rational root exists.  \nIf both exponents are odd, Corollary 1.3 gives the common root \\(0\\), and Proposition 3.1 shows there is no other.  \\(\\square\\)\n\n--------------------------------------------------------------------\n5.  Simultaneous vanishing on a finite set of indices  \n\nLet \\(\\varnothing\\neq\\mathcal{N}\\subset\\mathbf{N}\\) be finite.  \nBy Corollary 1.3, \\(P_{n}\\) has a rational root iff \\(n\\) is odd, in which case the only rational root is \\(0\\).  Consequently  \n\\[\n\\exists\\,r\\in\\mathbf{Q}\\;\n   \\forall\\,n\\in\\mathcal{N}:P_{n}(r)=0\n\\quad\\Longleftrightarrow\\quad\n\\mathcal{N}\\subset 2\\mathbf{N}-1,\n\\]\nand then necessarily \\(r=0\\).\n\n--------------------------------------------------------------------\nAnswer summary  \n\n1.  \\(\\displaystyle\n       \\gcd_{\\mathbf{Q}[y]}\\bigl(P_{m},P_{n}\\bigr)=\n       \\begin{cases}\n         y,&\\text{if }m,n\\text{ are odd},\\\\\n         1,&\\text{otherwise}.\n       \\end{cases}\\)\n\n2.  \\(P_{m}\\) and \\(P_{n}\\) possess a common rational root iff both indices are odd; that root is \\(y=0\\) (simple in each polynomial).\n\n3.  A finite non–empty set \\(\\mathcal{N}\\subset\\mathbf{N}\\) admits a rational \\(r\\) with \\(P_{n}(r)=0\\) for all \\(n\\in\\mathcal{N}\\) precisely when every element of \\(\\mathcal{N}\\) is odd; then \\(r=0\\) is forced.\n\n\\[\n\\square\n\\]\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.473698",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original single-equation question, the enhanced\nvariant demands:\n\n• Working simultaneously with two (and in part 3, arbitrarily many)\ndifferent exponents, so one must analyse a whole family of polynomials\ninstead of a single one.\n\n• Determining the exact greatest common divisor of two high-degree\ntrinomials in ℚ[y].  This obliges the solver to mix structural\nparity arguments with polynomial factorisation over ℚ—concepts that do\nnot appear in the original exercise.\n\n• Proving coprimality of quotient polynomials, which requires a\nnon-trivial combination of the Rational Root Theorem, parity\nconsiderations, modular arithmetic and multiplicity arguments.\n\n• Classifying all finite index sets that admit a common rational root,\nwhich converts the problem from an isolated Diophantine question into\na global description invoking results from the previous parts.\n\nEach of these layers adds substantial algebraic complexity and forces\nthe use of several advanced tools (factorisation in polynomial rings,\ngcd computations, root multiplicities, and parity-based modular\nobstructions), making the enhanced variant markedly more intricate\nthan both the original problem and the current kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Fix the integer  \n\\[\nd=2\n\\]  \nand, for every positive integer \\(n\\), define  \n\\[\nP_{n}(y)=\\bigl(y-2\\bigr)^{n}+y^{\\,n}+\\bigl(y+2\\bigr)^{n}\\in\\mathbf{Q}[y]\\qquad(n\\ge 1).\n\\]\n\n1.  For two \\emph{distinct} positive integers \\(m<n\\), determine explicitly the monic polynomial  \n   \\[\n      \\gcd_{\\mathbf{Q}[y]}\\bigl(P_{m},P_{n}\\bigr).\n   \\]\n\n2.  Find all ordered pairs \\((m,n)\\) of distinct positive integers for which \\(P_{m}\\) and \\(P_{n}\\) possess a common rational root, and list every such root.\n\n3.  Classify (with proof) all non–empty finite index sets  \n   \\[\n      \\mathcal{N}\\subset\\mathbf{N}\n   \\]\n   for which there exists \\(r\\in\\mathbf{Q}\\) satisfying \\(P_{n}(r)=0\\) for \\emph{every} \\(n\\in\\mathcal{N}\\).\n\n(The three questions are strongly coupled; answers obtained independently must nevertheless be proved within a coherent global argument.)\n\n--------------------------------------------------------------------",
      "solution": "Throughout we let  \n\\[\nP_{n}(y)=\\bigl(y-2\\bigr)^{n}+y^{\\,n}+\\bigl(y+2\\bigr)^{n}\\qquad(n\\in\\mathbf{N})\n\\]\nand we always choose a greatest common divisor to be monic.\n\n1.  A preliminary analysis of a single polynomial \\(P_{n}\\)\n\n(a)  Parity.  \n\\[\nP_{n}(-y)=(-1)^{\\,n}P_{n}(y)\\qquad(n\\ge 1),\n\\]\nso \\(P_{n}\\) is even when \\(n\\) is even and odd when \\(n\\) is odd.\n\n(b)  A factorisation when \\(n\\) is odd.  \nPut \\(n=2k+1\\) (\\(k\\ge 0\\)).  Expanding \\((y\\pm2)^{2k+1}\\) gives  \n\\[\n\\begin{aligned}\nP_{n}(y)\n      &=(y-2)^{2k+1}+y^{2k+1}+(y+2)^{2k+1}\\\\\n      &=y\\!\\Bigl[\\,3y^{2k}\n        +2\\sum_{j=1}^{k}\\binom{2k+1}{2j}2^{2j}y^{2(k-j)}\\Bigr]\n      =:y\\,Q_{n}(y),\n\\end{aligned}\n\\]\nwhere \\(Q_{n}\\in\\mathbf{Z}[y]\\) has only positive coefficients; hence  \n\\[\nQ_{n}(y)>0\\quad\\forall\\,y\\in\\mathbf{R}. \\tag{1.1}\n\\]\n\n(c)  Real and rational roots.\n\nLemma 1.1.  \nIf \\(n\\) is even, then \\(P_{n}(y)>0\\) for all \\(y\\in\\mathbf{R}\\); consequently \\(P_{n}\\) has no rational root.\n\nProof.  Each of the three summands is non–negative, and at most one of them can vanish; hence the sum is strictly positive. \\(\\square\\)\n\nLemma 1.2.  \nIf \\(n\\) is odd, the only rational root of \\(P_{n}\\) is \\(y=0\\).  This root is simple and  \n\\[\nP_{n}'(0)=\n\\begin{cases}\n3,&n=1,\\\\[4pt]\nn\\,2^{\\,n},&n\\ge 3\\text{ odd}.\n\\end{cases}\n\\]\n\nProof.  By (1.1) we have \\(P_{n}(y)=y\\,Q_{n}(y)\\) with \\(Q_{n}(y)>0\\) on \\(\\mathbf{R}\\), so the sole real (and hence rational) zero is \\(0\\).  Differentiating and evaluating at \\(0\\) gives the claimed formula; for \\(n=1\\) the value is \\(3\\). \\(\\square\\)\n\nCorollary 1.3.  \n\\(P_{n}\\) has a rational root exactly when \\(n\\) is odd; then the unique rational root is \\(y=0\\), which is simple.\n\n--------------------------------------------------------------------\n2.  Uniqueness of a non–zero common root  \n\nFix a non–zero complex number \\(z\\) and set  \n\\[\nu:=\\frac{z-2}{z},\\qquad v:=\\frac{z+2}{z}=2-u,\\qquad\nc:=uv=1-\\frac{4}{z^{2}}\\neq 0.\n\\]\nBecause \\(z\\neq 0\\),  \n\\[\nP_{n}(z)=0\\;\\Longleftrightarrow\\;S_{n}:=u^{\\,n}+v^{\\,n}+1=0. \\tag{2.1}\n\\]\nThe sequence \\(\\bigl(S_{n}\\bigr)_{n\\ge 0}\\) satisfies the inhomogeneous linear recurrence  \n\\[\nS_{n+2}-2\\,S_{n+1}+c\\,S_{n}=c-1\\qquad(n\\ge 0). \\tag{2.2}\n\\]\n\nProposition 2.1 (Uniqueness).  \nFor a fixed non–zero complex number \\(z\\) the equality \\(P_{n}(z)=0\\) can hold for \\emph{at most one} positive integer \\(n\\).\n\nProof.  \nAssume, by contradiction, that \\(P_{m}(z)=P_{n}(z)=0\\) for some integers \\(1\\le m<n\\); put \\(d:=n-m\\ge 1\\).\n\nStep 1.  A basic identity.  \nUsing \\(S_{m}=S_{n}=0\\) we derive\n\\[\nu^{m}\\bigl(u^{d}-1\\bigr)=-v^{m}\\bigl(v^{d}-1\\bigr). \\tag{2.3}\n\\]\n\nStep 2.  The case \\(\\lvert u\\rvert\\neq\\lvert v\\rvert\\).  \nSet \\(r:=\\lvert u/v\\rvert\\neq 1\\).  Taking absolute values in (2.3) yields\n\\[\nr^{\\,m}=\n\\frac{\\lvert v^{d}-1\\rvert}{\\lvert u^{d}-1\\rvert}=:\\kappa(d;u,v),\n\\]\nwhere \\(\\kappa(d;u,v)\\) is independent of \\(m\\).  Because \\(r\\neq 1\\), the left–hand side is a strictly monotone function of \\(m\\); hence equality is impossible for two distinct integers \\(m\\).  Thus the simultaneous vanishing of \\(S_{m}\\) and \\(S_{n}\\) cannot occur.\n\nStep 3.  The critical case \\(\\lvert u\\rvert=\\lvert v\\rvert\\).  \nWrite  \n\\[\nu=r\\,e^{\\,i\\theta},\\quad v=r\\,e^{-i\\theta},\\qquad r>1,\\ 0<\\theta<\\pi, \\tag{2.4}\n\\]\nsince \\(u+v=2\\) forces \\(r\\cos\\theta=1\\) and therefore \\(r>1\\).  \nWith (2.4) we have  \n\\[\nS_{k}=2\\,r^{k}\\cos(k\\theta)+1. \\tag{2.5}\n\\]\n\nSuppose \\(S_{m}=S_{n}=0\\) with \\(m<n\\).  Then (2.5) gives  \n\\[\n\\cos(m\\theta)=-\\frac{1}{2r^{m}},\\qquad\n\\cos(n\\theta)=-\\frac{1}{2r^{n}}. \\tag{2.6}\n\\]\n\nReturning to (2.3) and using \\(v/u=e^{-2i\\theta}\\), we obtain  \n\\[\nu^{d}-1=-e^{-2im\\theta}\\,\\overline{u^{d}-1}. \\tag{2.7}\n\\]\nWrite \\(A:=u^{d}-1\\).  Since \\(r>1\\), \\(u^{d}\\neq 1\\) and \\(A\\neq 0\\).  \nTaking complex conjugates in (2.7) and multiplying the two equalities gives  \n\\[\n\\bigl(1-e^{-2im\\theta}\\bigr)\\,|A|^{2}=0.\n\\]\nHence \\(e^{-2im\\theta}=1\\); that is,  \n\\[\n2m\\theta=2\\pi k\\quad(k\\in\\mathbf{Z}). \\tag{2.8}\n\\]\n\nConsequently \\(\\cos(m\\theta)=1\\).  Plugging this into the first relation in (2.6) yields  \n\\[\n1=-\\frac{1}{2r^{m}},\n\\]\ncontradicting \\(r>1\\).  Therefore the assumption \\(S_{m}=S_{n}=0\\) is impossible.\n\nCombining Steps 2 and 3 completes the proof. \\(\\square\\)\n\n--------------------------------------------------------------------\n3.  The greatest common divisor  \n\nProposition 3.1.  For \\(m<n\\) we have  \n\\[\n\\gcd_{\\mathbf{Q}[y]}\\bigl(P_{m},P_{n}\\bigr)=\n\\begin{cases}\ny,&\\text{if }m,n\\text{ are both odd},\\\\[6pt]\n1,&\\text{otherwise}.\n\\end{cases}\n\\]\n\nProof.  \n\n(i)  Assume at least one exponent, say \\(m\\), is even.  \nThen \\(P_{m}\\) possesses no rational root (Lemma 1.1), so \\(y\\nmid P_{m}\\).  \nSuppose, towards a contradiction, that a non–constant \\(D\\in\\mathbf{Q}[y]\\) divides both \\(P_{m}\\) and \\(P_{n}\\).  Let \\(z\\) be a root of \\(D\\).  \n\n• If \\(z=0\\) then \\(P_{m}(0)=2^{\\,m}\\neq 0\\), impossible.  \n\n• If \\(z\\neq 0\\) then \\(P_{m}(z)=P_{n}(z)=0\\), contradicting Proposition 2.1.  \n\nHence \\(\\gcd=1\\).\n\n(ii)  Now suppose both exponents are odd.  \nBy Corollary 1.3, \\(y\\mid P_{m}\\) and \\(y\\mid P_{n}\\), so \\(y\\) divides the gcd.  \nThe zero at \\(y=0\\) is simple in each polynomial (Lemma 1.2), implying \\(y^{2}\\nmid P_{m}\\) and therefore \\(y^{2}\\nmid\\gcd\\).  \nIf the gcd possessed another irreducible factor, it would yield a non–zero common root, impossible by Proposition 2.1.  Thus \\(\\gcd=y\\).  \\(\\square\\)\n\n--------------------------------------------------------------------\n4.  Common rational roots of two members  \n\nTheorem 4.1.  \nThe polynomials \\(P_{m}\\) and \\(P_{n}\\) share a rational root precisely when both \\(m\\) and \\(n\\) are odd; their only common rational root is \\(y=0\\).\n\nProof.  \nIf one exponent is even, the corresponding polynomial has no rational root (Lemma 1.1); hence no common rational root exists.  \nIf both exponents are odd, Corollary 1.3 gives the common root \\(0\\), and Proposition 3.1 shows there is no other.  \\(\\square\\)\n\n--------------------------------------------------------------------\n5.  Simultaneous vanishing on a finite set of indices  \n\nLet \\(\\varnothing\\neq\\mathcal{N}\\subset\\mathbf{N}\\) be finite.  \nBy Corollary 1.3, \\(P_{n}\\) has a rational root iff \\(n\\) is odd, in which case the only rational root is \\(0\\).  Consequently  \n\\[\n\\exists\\,r\\in\\mathbf{Q}\\;\n   \\forall\\,n\\in\\mathcal{N}:P_{n}(r)=0\n\\quad\\Longleftrightarrow\\quad\n\\mathcal{N}\\subset 2\\mathbf{N}-1,\n\\]\nand then necessarily \\(r=0\\).\n\n--------------------------------------------------------------------\nAnswer summary  \n\n1.  \\(\\displaystyle\n       \\gcd_{\\mathbf{Q}[y]}\\bigl(P_{m},P_{n}\\bigr)=\n       \\begin{cases}\n         y,&\\text{if }m,n\\text{ are odd},\\\\\n         1,&\\text{otherwise}.\n       \\end{cases}\\)\n\n2.  \\(P_{m}\\) and \\(P_{n}\\) possess a common rational root iff both indices are odd; that root is \\(y=0\\) (simple in each polynomial).\n\n3.  A finite non–empty set \\(\\mathcal{N}\\subset\\mathbf{N}\\) admits a rational \\(r\\) with \\(P_{n}(r)=0\\) for all \\(n\\in\\mathcal{N}\\) precisely when every element of \\(\\mathcal{N}\\) is odd; then \\(r=0\\) is forced.\n\n\\[\n\\square\n\\]\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.398405",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original single-equation question, the enhanced\nvariant demands:\n\n• Working simultaneously with two (and in part 3, arbitrarily many)\ndifferent exponents, so one must analyse a whole family of polynomials\ninstead of a single one.\n\n• Determining the exact greatest common divisor of two high-degree\ntrinomials in ℚ[y].  This obliges the solver to mix structural\nparity arguments with polynomial factorisation over ℚ—concepts that do\nnot appear in the original exercise.\n\n• Proving coprimality of quotient polynomials, which requires a\nnon-trivial combination of the Rational Root Theorem, parity\nconsiderations, modular arithmetic and multiplicity arguments.\n\n• Classifying all finite index sets that admit a common rational root,\nwhich converts the problem from an isolated Diophantine question into\na global description invoking results from the previous parts.\n\nEach of these layers adds substantial algebraic complexity and forces\nthe use of several advanced tools (factorisation in polynomial rings,\ngcd computations, root multiplicities, and parity-based modular\nobstructions), making the enhanced variant markedly more intricate\nthan both the original problem and the current kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}