{
  "index": "2021-A-6",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $P(x)$ be a polynomial whose coefficients are all either $0$ or $1$.\nSuppose that $P(x)$ can be written as a product of two nonconstant polynomials with integer coefficients. Does it follow that $P(2)$ is a composite integer?",
  "solution": "Yes, it follows that $P(2)$ is a composite integer. (Note: 1 is neither prime nor composite.)\n\nWrite $P(x) = a_0 + a_1 x + \\cdots + a_n x^n$ with $a_i \\in \\{0,1\\}$ and $a_n = 1$.\nLet $\\alpha$ be an arbitrary root of $P$. Since $P(\\alpha) = 0$, $\\alpha$ cannot be a positive real number.\n%In addition, if $\\alpha \\neq 0$ then\n%\\begin{align*}\n%1 &< |a_{n-1} \\alpha^{-1} + \\cdots + a_0 \\alpha^{-n}| \\\\\n%&\\leq |\\alpha|^{-1} + \\cdots + |\\alpha|^{-n}\n%\\end{align*}\n%and so $|\\alpha| < 2$.\n%\nIn addition, if $\\alpha \\neq 0$ then\n\\begin{align*}\n|1 + a_{n-1} \\alpha^{-1}| &= |a_{n-2} \\alpha^{-2} + \\cdots + a_0 \\alpha^{-n}| \\\\\n&\\leq |\\alpha|^{-2} + \\cdots + |\\alpha|^{-n}.\n\\end{align*}\nIf $\\alpha \\neq 0$ and $\\mathrm{Re}(\\alpha) \\geq 0$, then $\\mathrm{Re}(1 + a_{n-1} \\alpha^{-1}) \\geq 1$\nand \n\\[\n1 \\leq |\\alpha|^{-2} + \\cdots + |\\alpha|^{-n} < \\frac{|\\alpha|^{-2}}{1 - |\\alpha|^{-1}};\n\\]\nthis yields $|\\alpha| < (1 + \\sqrt{5})/2$.\n\nBy the same token, if $\\alpha \\neq 0$ then\n\\[\n|1 + a_{n-1} \\alpha^{-1} + a_{n-2} \\alpha^{-2}| \\leq |\\alpha|^{-3} + \\cdots + |\\alpha|^{-n}.\n\\]\nWe deduce from this that $\\mathrm{Re}(\\alpha) \\leq 3/2$ as follows.\n\\begin{itemize}\n\\item\nThere is nothing to check if $\\mathrm{Re}(\\alpha) \\leq 0$.\n\\item\nIf the argument of $\\alpha$ belongs to $[-\\pi/4, \\pi/4]$, then $\\mathrm{Re}(\\alpha^{-1}), \\mathrm{Re}(\\alpha^{-2}) \\geq 0$, so\n\\[\n1 \\leq |\\alpha|^{-3} + \\cdots + |\\alpha|^{-n} < \\frac{|\\alpha|^{-3}}{1 - |\\alpha|^{-1}}.\n\\]\nHence $|\\alpha|^{-1}$ is greater than the unique positive root of $x^3 + x - 1$, which \nis greater than $2/3$. \n\\item\nOtherwise, $\\alpha$ has argument in $(-\\pi/2,\\pi/4) \\cup (\\pi/4,\\pi/2)$,\nso the bound $|\\alpha| < (1 + \\sqrt{5})/2$ implies that $\\mathrm{Re}(\\alpha) < (1 + \\sqrt{5})/(2 \\sqrt{2}) < 3/2$.\n\\end{itemize}\n\nBy hypothesis, there exists a factorization $P(x) = Q(x)R(x)$ into two nonconstant integer polynomials, which we may assume are monic.\n$Q(x + 3/2)$ is a product of polynomials, each of the form $x - \\alpha$ where $\\alpha$ is a real root of $P$\nor of the form\n\\begin{align*}\n&\\left( x + \\frac{3}{2} - \\alpha\\right) \\left(x + \\frac{3}{2} - \\overline{\\alpha} \\right) \\\\\n&\\quad = x^2 + 2 \\mathrm{Re}\\left(\\frac{3}{2} - \\alpha\\right) x + \\left|\\frac{3}{2} - \\alpha \\right|^2\n\\end{align*}\nwhere $\\alpha$ is a nonreal root of $P$. It follows that $Q(x+3/2)$ has positive coefficients;\ncomparing its values at $x=1/2$ and $x=-1/2$ yields $Q(2) > Q(1)$. We cannot have $Q(1) \\leq 0$, as otherwise the intermediate value theorem would imply that $Q$ has a real root in $[1, \\infty)$; hence $Q(1)  \\geq 1$ and so $Q(2) \\geq 2$.\nSimilarly $R(2) \\geq 2$, so $P(2) = Q(2) R(2)$ is composite.\n\n\\noindent\n\\textbf{Remark.}\nA theorem of Brillhart, Filaseta, and Odlyzko from 1981 states that if a prime $p$ is written as $\\sum_i a_i b^i$ in any base $b \\geq 2$, the polynomial $\\sum_i a_i x^i$ is irreducible.\n(The case $b=10$ is an older result of Cohn.) \nThe solution given above is taken from: Ram Murty, Prime numbers and irreducible polynomials, \\textit{Amer. Math. Monthly} \\textbf{109} (2002), 452--458). The final step is due to P\\'olya and Szeg\\H{o}.",
  "vars": [
    "x",
    "a_0",
    "a_1",
    "a_n",
    "a_i",
    "n",
    "\\\\alpha",
    "Q",
    "R",
    "b",
    "p"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "indepvar",
        "a_0": "coeffzero",
        "a_1": "coeffone",
        "a_n": "coeffhigh",
        "a_i": "coeffgen",
        "n": "degreeval",
        "\\alpha": "rootvar",
        "Q": "factorone",
        "R": "factortwo",
        "b": "baseval",
        "p": "primeval"
      },
      "question": "Let $P(indepvar)$ be a polynomial whose coefficients are all either $0$ or $1$. Suppose that $P(indepvar)$ can be written as a product of two nonconstant polynomials with integer coefficients. Does it follow that $P(2)$ is a composite integer?",
      "solution": "Yes, it follows that $P(2)$ is a composite integer. (Note: 1 is neither prime nor composite.)\n\nWrite $P(indepvar) = coeffzero + coeffone\\,indepvar + \\cdots + coeffhigh\\,indepvar^{degreeval}$ with $coeffgen \\in \\{0,1\\}$ and $coeffhigh = 1$.\n\nLet $rootvar$ be an arbitrary root of $P$. Since $P(rootvar) = 0$, $rootvar$ cannot be a positive real number.\n\nIn addition, if $rootvar \\neq 0$ then\n\\begin{align*}\n|1 + a_{degreeval-1}\\,rootvar^{-1}| &= |a_{degreeval-2}\\,rootvar^{-2} + \\cdots + coeffzero\\,rootvar^{-degreeval}| \\\\\n&\\leq |rootvar|^{-2} + \\cdots + |rootvar|^{-degreeval}.\n\\end{align*}\nIf $rootvar \\neq 0$ and $\\mathrm{Re}(rootvar) \\geq 0$, then $\\mathrm{Re}(1 + a_{degreeval-1}\\,rootvar^{-1}) \\geq 1$ and\n\\[\n1 \\leq |rootvar|^{-2} + \\cdots + |rootvar|^{-degreeval} < \\frac{|rootvar|^{-2}}{1 - |rootvar|^{-1}};\n\\]\nthis yields $|rootvar| < (1 + \\sqrt{5})/2$.\n\nBy the same token, if $rootvar \\neq 0$ then\n\\[\n|1 + a_{degreeval-1}\\,rootvar^{-1} + a_{degreeval-2}\\,rootvar^{-2}| \\leq |rootvar|^{-3} + \\cdots + |rootvar|^{-degreeval}.\n\\]\nWe deduce from this that $\\mathrm{Re}(rootvar) \\leq 3/2$ as follows.\n\\begin{itemize}\n\\item There is nothing to check if $\\mathrm{Re}(rootvar) \\leq 0$.\n\\item If the argument of $rootvar$ belongs to $[-\\pi/4, \\pi/4]$, then $\\mathrm{Re}(rootvar^{-1}), \\mathrm{Re}(rootvar^{-2}) \\geq 0$, so\n\\[\n1 \\leq |rootvar|^{-3} + \\cdots + |rootvar|^{-degreeval} < \\frac{|rootvar|^{-3}}{1 - |rootvar|^{-1}}.\n\\]\nHence $|rootvar|^{-1}$ is greater than the unique positive root of $indepvar^3 + indepvar - 1$, which is greater than $2/3$.\n\\item Otherwise, $rootvar$ has argument in $(-\\pi/2,\\pi/4) \\cup (\\pi/4,\\pi/2)$, so the bound $|rootvar| < (1 + \\sqrt{5})/2$ implies that $\\mathrm{Re}(rootvar) < (1 + \\sqrt{5})/(2 \\sqrt{2}) < 3/2$.\n\\end{itemize}\n\nBy hypothesis, there exists a factorization $P(indepvar) = factorone(indepvar)\\,factortwo(indepvar)$ into two nonconstant integer polynomials, which we may assume are monic.\n$factorone(indepvar + 3/2)$ is a product of polynomials, each of the form $indepvar - rootvar$ where $rootvar$ is a real root of $P$ or of the form\n\\begin{align*}\n&\\left( indepvar + \\frac{3}{2} - rootvar\\right) \\left(indepvar + \\frac{3}{2} - \\overline{rootvar} \\right) \\\\\n&\\quad = indepvar^2 + 2\\,\\mathrm{Re}\\!\\left(\\frac{3}{2} - rootvar\\right) indepvar + \\left|\\frac{3}{2} - rootvar \\right|^2\n\\end{align*}\nwhere $rootvar$ is a nonreal root of $P$. It follows that $factorone(indepvar+3/2)$ has positive coefficients; comparing its values at $indepvar=1/2$ and $indepvar=-1/2$ yields $factorone(2) > factorone(1)$. We cannot have $factorone(1) \\leq 0$, as otherwise the intermediate value theorem would imply that $factorone$ has a real root in $[1, \\infty)$; hence $factorone(1) \\geq 1$ and so $factorone(2) \\geq 2$.\nSimilarly $factortwo(2) \\geq 2$, so $P(2) = factorone(2)\\,factortwo(2)$ is composite.\n\nRemark.\nA theorem of Brillhart, Filaseta, and Odlyzko from 1981 states that if a prime $primeval$ is written as $\\sum_i a_i baseval^i$ in any base $baseval \\geq 2$, the polynomial $\\sum_i a_i indepvar^i$ is irreducible.\n(The case $baseval=10$ is an older result of Cohn.)\nThe solution given above is taken from: Ram Murty, Prime numbers and irreducible polynomials, \\textit{Amer. Math. Monthly} \\textbf{109} (2002), 452--458). The final step is due to P\\'olya and Szeg\\H{o}."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineapple",
        "a_0": "waterfall",
        "a_1": "locomotive",
        "a_n": "blueberry",
        "a_i": "sailboat",
        "n": "kangaroo",
        "\\alpha": "sunflower",
        "Q": "microscope",
        "R": "adventure",
        "b": "photograph",
        "p": "bookshelf"
      },
      "question": "Let $P(pineapple)$ be a polynomial whose coefficients are all either $0$ or $1$.\nSuppose that $P(pineapple)$ can be written as a product of two nonconstant polynomials with integer coefficients. Does it follow that $P(2)$ is a composite integer?",
      "solution": "Yes, it follows that $P(2)$ is a composite integer. (Note: 1 is neither prime nor composite.)\n\nWrite $P(pineapple) = waterfall + locomotive\\, pineapple + \\cdots + blueberry\\, pineapple^{kangaroo}$ with $sailboat \\in \\{0,1\\}$ and $blueberry = 1$.\nLet $sunflower$ be an arbitrary root of $P$. Since $P(sunflower) = 0$, $sunflower$ cannot be a positive real number.\n%In addition, if $sunflower \\neq 0$ then\n%\\begin{align*}\n%1 &< |a_{kangaroo-1} sunflower^{-1} + \\cdots + waterfall sunflower^{-kangaroo}| \\\\\n%&\\leq |sunflower|^{-1} + \\cdots + |sunflower|^{-kangaroo}\n%\\end{align*}\n%and so $|sunflower| < 2$.\n%\nIn addition, if $sunflower \\neq 0$ then\n\\begin{align*}\n|1 + a_{kangaroo-1} sunflower^{-1}| &= |a_{kangaroo-2} sunflower^{-2} + \\cdots + waterfall sunflower^{-kangaroo}| \\\\\n&\\leq |sunflower|^{-2} + \\cdots + |sunflower|^{-kangaroo}.\n\\end{align*}\nIf $sunflower \\neq 0$ and $\\mathrm{Re}(sunflower) \\geq 0$, then $\\mathrm{Re}(1 + a_{kangaroo-1} sunflower^{-1}) \\geq 1$ and\n\\[\n1 \\leq |sunflower|^{-2} + \\cdots + |sunflower|^{-kangaroo} < \\frac{|sunflower|^{-2}}{1 - |sunflower|^{-1}};\n\\]\nthis yields $|sunflower| < (1 + \\sqrt{5})/2$.\n\nBy the same token, if $sunflower \\neq 0$ then\n\\[\n|1 + a_{kangaroo-1} sunflower^{-1} + a_{kangaroo-2} sunflower^{-2}| \\leq |sunflower|^{-3} + \\cdots + |sunflower|^{-kangaroo}.\n\\]\nWe deduce from this that $\\mathrm{Re}(sunflower) \\leq 3/2$ as follows.\n\\begin{itemize}\n\\item\nThere is nothing to check if $\\mathrm{Re}(sunflower) \\leq 0$.\n\\item\nIf the argument of $sunflower$ belongs to $[-\\pi/4, \\pi/4]$, then $\\mathrm{Re}(sunflower^{-1}), \\mathrm{Re}(sunflower^{-2}) \\geq 0$, so\n\\[\n1 \\leq |sunflower|^{-3} + \\cdots + |sunflower|^{-kangaroo} < \\frac{|sunflower|^{-3}}{1 - |sunflower|^{-1}}.\n\\]\nHence $|sunflower|^{-1}$ is greater than the unique positive root of $pineapple^3 + pineapple - 1$, which is greater than $2/3$.\n\\item\nOtherwise, $sunflower$ has argument in $(-\\pi/2,\\pi/4) \\cup (\\pi/4,\\pi/2)$, so the bound $|sunflower| < (1 + \\sqrt{5})/2$ implies that $\\mathrm{Re}(sunflower) < (1 + \\sqrt{5})/(2 \\sqrt{2}) < 3/2$.\n\\end{itemize}\n\nBy hypothesis, there exists a factorization $P(pineapple) = microscope(pineapple) adventure(pineapple)$ into two nonconstant integer polynomials, which we may assume are monic.\n$microscope(pineapple + 3/2)$ is a product of polynomials, each of the form $pineapple - sunflower$ where $sunflower$ is a real root of $P$ or of the form\n\\begin{align*}\n&\\left( pineapple + \\frac{3}{2} - sunflower \\right) \\left( pineapple + \\frac{3}{2} - \\overline{sunflower} \\right) \\\\\n&\\quad = pineapple^2 + 2 \\, \\mathrm{Re}\\left( \\frac{3}{2} - sunflower \\right) pineapple + \\left| \\frac{3}{2} - sunflower \\right|^2\n\\end{align*}\nwhere $sunflower$ is a nonreal root of $P$. It follows that $microscope(pineapple+3/2)$ has positive coefficients; comparing its values at $pineapple=1/2$ and $pineapple=-1/2$ yields $microscope(2) > microscope(1)$. We cannot have $microscope(1) \\leq 0$, as otherwise the intermediate value theorem would imply that $microscope$ has a real root in $[1, \\infty)$; hence $microscope(1) \\geq 1$ and so $microscope(2) \\geq 2$.\nSimilarly $adventure(2) \\geq 2$, so $P(2) = microscope(2) adventure(2)$ is composite.\n\n\\noindent\n\\textbf{Remark.} A theorem of Brillhart, Filaseta, and Odlyzko from 1981 states that if a prime $bookshelf$ is written as $\\sum_i sailboat photograph^i$ in any base $photograph \\geq 2$, the polynomial $\\sum_i sailboat pineapple^i$ is irreducible.\n(The case $photograph=10$ is an older result of Cohn.) The solution given above is taken from: Ram Murty, Prime numbers and irreducible polynomials, \\textit{Amer. Math. Monthly} \\textbf{109} (2002), 452--458). The final step is due to P\\'olya and Szeg\\H{o}."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "a_0": "mutablezero",
        "a_1": "mutableone",
        "a_n": "mutableend",
        "a_i": "mutableith",
        "n": "beginning",
        "\\alpha": "endpoint",
        "Q": "antifactor",
        "R": "counterpart",
        "b": "apexvalue",
        "p": "composite"
      },
      "question": "Let $P(constantval)$ be a polynomial whose coefficients are all either $0$ or $1$.\nSuppose that $P(constantval)$ can be written as a product of two nonconstant polynomials with integer coefficients. Does it follow that $P(2)$ is a composite integer?",
      "solution": "Yes, it follows that $P(2)$ is a composite integer. (Note: 1 is neither prime nor composite.)\n\nWrite $P(constantval) = mutablezero + mutableone\\, constantval + \\cdots + mutableend\\, constantval^{beginning}$ with $mutableith \\in \\{0,1\\}$ and $mutableend = 1$.\nLet endpoint be an arbitrary root of $P$. Since $P(endpoint) = 0$, endpoint cannot be a positive real number.\n\nIn addition, if $endpoint \\neq 0$ then\n\\begin{align*}\n|1 + a_{beginning-1} endpoint^{-1}| &= |a_{beginning-2} endpoint^{-2} + \\cdots + mutablezero\\, endpoint^{-beginning}| \\\\\n&\\le |endpoint|^{-2} + \\cdots + |endpoint|^{-beginning}.\n\\end{align*}\nIf $endpoint \\neq 0$ and $\\mathrm{Re}(endpoint) \\ge 0$, then $\\mathrm{Re}(1 + a_{beginning-1} endpoint^{-1}) \\ge 1$\nand \n\\[\n1 \\le |endpoint|^{-2} + \\cdots + |endpoint|^{-beginning} < \\frac{|endpoint|^{-2}}{1 - |endpoint|^{-1}};\n\\]\nthis yields $|endpoint| < (1 + \\sqrt{5})/2$.\n\nBy the same token, if $endpoint \\neq 0$ then\n\\[\n|1 + a_{beginning-1} endpoint^{-1} + a_{beginning-2} endpoint^{-2}| \\le |endpoint|^{-3} + \\cdots + |endpoint|^{-beginning}.\n\\]\nWe deduce from this that $\\mathrm{Re}(endpoint) \\le 3/2$ as follows.\n\\begin{itemize}\n\\item\nThere is nothing to check if $\\mathrm{Re}(endpoint) \\le 0$.\n\\item\nIf the argument of $endpoint$ belongs to $[-\\pi/4, \\pi/4]$, then $\\mathrm{Re}(endpoint^{-1}), \\mathrm{Re}(endpoint^{-2}) \\ge 0$, so\n\\[\n1 \\le |endpoint|^{-3} + \\cdots + |endpoint|^{-beginning} < \\frac{|endpoint|^{-3}}{1 - |endpoint|^{-1}}.\n\\]\nHence $|endpoint|^{-1}$ is greater than the unique positive root of $constantval^{3} + constantval - 1$, which \nis greater than $2/3$. \n\\item\nOtherwise, $endpoint$ has argument in $(-\\pi/2,\\pi/4) \\cup (\\pi/4,\\pi/2)$,\nso the bound $|endpoint| < (1 + \\sqrt{5})/2$ implies that $\\mathrm{Re}(endpoint) < (1 + \\sqrt{5})/(2 \\sqrt{2}) < 3/2$.\n\\end{itemize}\n\nBy hypothesis, there exists a factorization $P(constantval) = antifactor(constantval)\\, counterpart(constantval)$ into two nonconstant integer polynomials, which we may assume are monic.\n$antifactor(constantval + 3/2)$ is a product of polynomials, each of the form $constantval - endpoint$ where $endpoint$ is a real root of $P$\nor of the form\n\\begin{align*}\n&\\left( constantval + \\frac{3}{2} - endpoint\\right) \\left( constantval + \\frac{3}{2} - \\overline{endpoint} \\right) \\\\\n&\\quad = constantval^{2} + 2 \\, \\mathrm{Re}\\left(\\frac{3}{2} - endpoint\\right) \\, constantval + \\left|\\frac{3}{2} - endpoint \\right|^{2}\n\\end{align*}\nwhere $endpoint$ is a nonreal root of $P$. It follows that $antifactor(constantval+3/2)$ has positive coefficients;\ncomparing its values at $constantval=1/2$ and $constantval=-1/2$ yields $antifactor(2) > antifactor(1)$. We cannot have $antifactor(1) \\le 0$, as otherwise the intermediate value theorem would imply that $antifactor$ has a real root in $[1, \\infty)$; hence $antifactor(1)  \\ge 1$ and so $antifactor(2) \\ge 2$.\nSimilarly $counterpart(2) \\ge 2$, so $P(2) = antifactor(2)\\, counterpart(2)$ is composite.\n\nRemark.\nA theorem of Brillhart, Filaseta, and Odlyzko from 1981 states that if a prime $composite$ is written as $\\sum_i mutableith\\, apexvalue^{i}$ in any base $apexvalue \\ge 2$, the polynomial $\\sum_i mutableith\\, constantval^{i}$ is irreducible.\n(The case $apexvalue=10$ is an older result of Cohn.) \nThe solution given above is taken from: Ram Murty, Prime numbers and irreducible polynomials, \\textit{Amer. Math. Monthly} \\textbf{109} (2002), 452--458). The final step is due to P\\'olya and Szeg\\H{o}."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "a_0": "hjgrksla",
        "a_1": "mnctrpqo",
        "a_n": "vblsdpzi",
        "a_i": "fqsrlkdu",
        "n": "wjthpzre",
        "\\\\alpha": "zqmvbnde",
        "Q": "kdrnvsqe",
        "R": "pjwzclta",
        "b": "clhzsmfu",
        "p": "lgxkdtne"
      },
      "question": "Let $P(qzxwvtnp)$ be a polynomial whose coefficients are all either $0$ or $1$. Suppose that $P(qzxwvtnp)$ can be written as a product of two nonconstant polynomials with integer coefficients. Does it follow that $P(2)$ is a composite integer?",
      "solution": "Yes, it follows that $P(2)$ is a composite integer. (Note: 1 is neither prime nor composite.)\n\nWrite $P(qzxwvtnp) = hjgrksla + mnctrpqo qzxwvtnp + \\cdots + vblsdpzi qzxwvtnp^{wjthpzre}$ with $fqsrlkdu \\in \\{0,1\\}$ and $vblsdpzi = 1$. Let $zqmvbnde$ be an arbitrary root of $P$. Since $P(zqmvbnde) = 0$, $zqmvbnde$ cannot be a positive real number.\n%In addition, if $\\alpha \\neq 0$ then\n%\\begin{align*}\n%1 &< |a_{n-1} \\alpha^{-1} + \\cdots + a_0 \\alpha^{-n}| \\\\\n%&\\leq |\\alpha|^{-1} + \\cdots + |\\alpha|^{-n}\n%\\end{align*}\n%and so $|\\alpha| < 2$.\n%\nIn addition, if $zqmvbnde \\neq 0$ then\n\\begin{align*}\n|1 + a_{wjthpzre-1} zqmvbnde^{-1}| &= |a_{wjthpzre-2} zqmvbnde^{-2} + \\cdots + hjgrksla zqmvbnde^{-wjthpzre}| \\\\\n&\\leq |zqmvbnde|^{-2} + \\cdots + |zqmvbnde|^{-wjthpzre}.\n\\end{align*}\nIf $zqmvbnde \\neq 0$ and $\\mathrm{Re}(zqmvbnde) \\geq 0$, then $\\mathrm{Re}(1 + a_{wjthpzre-1} zqmvbnde^{-1}) \\geq 1$ and \n\\[\n1 \\leq |zqmvbnde|^{-2} + \\cdots + |zqmvbnde|^{-wjthpzre} < \\frac{|zqmvbnde|^{-2}}{1 - |zqmvbnde|^{-1}};\n\\]\nthis yields $|zqmvbnde| < (1 + \\sqrt{5})/2$.\n\nBy the same token, if $zqmvbnde \\neq 0$ then\n\\[\n|1 + a_{wjthpzre-1} zqmvbnde^{-1} + a_{wjthpzre-2} zqmvbnde^{-2}| \\leq |zqmvbnde|^{-3} + \\cdots + |zqmvbnde|^{-wjthpzre}.\n\\]\nWe deduce from this that $\\mathrm{Re}(zqmvbnde) \\leq 3/2$ as follows.\n\\begin{itemize}\n\\item\nThere is nothing to check if $\\mathrm{Re}(zqmvbnde) \\leq 0$.\n\\item\nIf the argument of $zqmvbnde$ belongs to $[-\\pi/4, \\pi/4]$, then $\\mathrm{Re}(zqmvbnde^{-1}), \\mathrm{Re}(zqmvbnde^{-2}) \\geq 0$, so\n\\[\n1 \\leq |zqmvbnde|^{-3} + \\cdots + |zqmvbnde|^{-wjthpzre} < \\frac{|zqmvbnde|^{-3}}{1 - |zqmvbnde|^{-1}}.\n\\]\nHence $|zqmvbnde|^{-1}$ is greater than the unique positive root of $x^3 + x - 1$, which is greater than $2/3$.\n\\item\nOtherwise, $zqmvbnde$ has argument in $(-\\pi/2,\\pi/4) \\cup (\\pi/4,\\pi/2)$, so the bound $|zqmvbnde| < (1 + \\sqrt{5})/2$ implies that $\\mathrm{Re}(zqmvbnde) < (1 + \\sqrt{5})/(2 \\sqrt{2}) < 3/2$.\n\\end{itemize}\n\nBy hypothesis, there exists a factorization $P(qzxwvtnp) = kdrnvsqe(qzxwvtnp)pjwzclta(qzxwvtnp)$ into two nonconstant integer polynomials, which we may assume are monic. $kdrnvsqe(qzxwvtnp + 3/2)$ is a product of polynomials, each of the form $qzxwvtnp - zqmvbnde$ where $zqmvbnde$ is a real root of $P$ or of the form\n\\begin{align*}\n&\\left( qzxwvtnp + \\frac{3}{2} - zqmvbnde\\right) \\left(qzxwvtnp + \\frac{3}{2} - \\overline{zqmvbnde} \\right) \\\\\n&\\quad = qzxwvtnp^2 + 2 \\,\\mathrm{Re}\\left(\\frac{3}{2} - zqmvbnde\\right) qzxwvtnp + \\left|\\frac{3}{2} - zqmvbnde \\right|^2\n\\end{align*}\nwhere $zqmvbnde$ is a nonreal root of $P$. It follows that $kdrnvsqe(qzxwvtnp+3/2)$ has positive coefficients; comparing its values at $qzxwvtnp=1/2$ and $qzxwvtnp=-1/2$ yields $kdrnvsqe(2) > kdrnvsqe(1)$. We cannot have $kdrnvsqe(1) \\leq 0$, as otherwise the intermediate value theorem would imply that $kdrnvsqe$ has a real root in $[1, \\infty)$; hence $kdrnvsqe(1) \\geq 1$ and so $kdrnvsqe(2) \\geq 2$. Similarly $pjwzclta(2) \\geq 2$, so $P(2) = kdrnvsqe(2) pjwzclta(2)$ is composite.\n\n\\noindent\n\\textbf{Remark.} A theorem of Brillhart, Filaseta, and Odlyzko from 1981 states that if a prime $lgxkdtne$ is written as $\\sum_i a_i clhzsmfu^i$ in any base $clhzsmfu \\geq 2$, the polynomial $\\sum_i a_i qzxwvtnp^i$ is irreducible. (The case $clhzsmfu=10$ is an older result of Cohn.) The solution given above is taken from: Ram Murty, Prime numbers and irreducible polynomials, \\textit{Amer. Math. Monthly} \\textbf{109} (2002), 452--458). The final step is due to P\\'olya and Szeg\\H{o}."
    },
    "kernel_variant": {
      "question": "Let\n\nP(x)=a_0+a_1x+\\dots +a_nx^{n}\\qquad(a_i\\in\\{0,1\\},\\;a_n=1)\n\nbe a polynomial whose coefficients are all 0 or 1.  Assume that P admits a non-trivial factorisation over the integers,\n\nP(x)=Q(x)\\,R(x),\\qquad Q,R\\in\\mathbb Z[x],\\;\\deg Q,\\deg R\\ge 1.\n\nProve that the integer P(3) is composite (i.e. it is neither 1 nor a prime).",
      "solution": "Write\n\\[\nP(x)=a_0+a_1x+\\dots +a_nx^{n},\\qquad a_i\\in\\{0,1\\},\\;a_n=1,\\;n\\ge 1,\n\\]\nand suppose there is a non–trivial factorisation over \\(\\mathbb Z\\)\n\\[\nP(x)=Q(x)R(x),\\qquad \\deg Q,\\deg R\\ge 1 .\n\\]\nOur goal is to show that \\(P(3)\\) is not a prime.\n\nPre-liminaries: arranging monic factors.\n------------------------------------------------\nBecause the leading coefficient of \\(P\\) is 1, the product of the leading coefficients of \\(Q\\) and \\(R\\) is 1.  Hence each leading coefficient is either \\(+1\\) or \\(-1\\); moreover they are equal.  If both are \\(-1\\) we replace \\(Q,R\\) by \\(-Q,-R\\).  The new pair still satisfies \\(P=QR\\) and both polynomials are now *monic*.  From now on we assume\n\\[\nQ,R\\text{ are monic.}\n\\]\nThis fact will be used twice, once in Step 2 and once in Step 3.\n\nStep 0.  The easy case \\(a_0=0\\).\n----------------------------------\nIf \\(a_0=0\\) we can write \\(P(x)=x\\,P_1(x)\\) with\n\\(\nP_1(x)=a_1+a_2x+\\dots +a_nx^{n-1}\\;(a_n=1).\n\\)\nBecause the given factorisation of \\(P\\) is non-trivial, \\(n\\ge 2\\) and so \\(\\deg P_1\\ge 1\\).  The smallest possible value of \\(P_1(3)\\) occurs when the only non-zero coefficient is the leading one; then \\(P_1(3)=3^{n-1}\\ge 3\\).  Hence\n\\[\nP(3)=3\\,P_1(3)\\ge 3\\times 3=9,\n\\]\nwhich is composite.  Therefore we may (and do) assume from now on\n\\[\n\\boxed{\\;a_0=1\\;}.  \\qquad(1)\n\\]\n\nStep 1.  Where can the roots of \\(P\\) lie?\n-------------------------------------------\n(a)  No positive real roots.  For every real \\(x>0\\), all summands in \\(P(x)\\) are non-negative and at least one is positive, so \\(P(x)>0\\).  Consequently neither \\(P\\) nor its factors \\(Q,R\\) have positive real roots.\n\n(b)  A bound for the moduli.  Let \\(\\alpha\\) be any root of \\(P\\).  If \\(|\\alpha|>2\\) then\n\\[\n|\\alpha|^{n}=|\\alpha^n|>\\sum_{j=0}^{n-1}|\\alpha|^j\\ge\\Bigl|\\sum_{j=0}^{n-1}a_j \\alpha^j\\Bigr|=|P(\\alpha)-\\alpha^n|=|0-\\alpha^n|=|\\alpha|^n,\n\\]\na contradiction.  Hence every root satisfies \\(|\\alpha|\\le 2\\); in particular \\(\\operatorname{Re}(\\alpha)\\le 2\\).\n\nStep 2.  Shifting the factors so that all coefficients are positive.\n-------------------------------------------------------------------\nWrite \\(Q\\) and \\(R\\) over \\(\\mathbb R\\) as products of linear factors for real roots and quadratic factors for non-real conjugate pairs.  Replace \\(x\\) by \\(x+2\\).  \n\n• For a real root \\(\\alpha\\) we obtain the factor \\(x+2-\\alpha\\) whose coefficients are \\(1\\) and \\(2-\\alpha\\); since \\(\\alpha<0\\) (no non-negative real roots) we have \\(2-\\alpha>0\\).\n\n• For a non-real root \\(\\alpha\\) the quadratic factor becomes\n\\[\n(x+2-\\alpha)(x+2-\\overline{\\alpha})=x^2+2(2-\\operatorname{Re}\\alpha)\\,x+\\bigl((2-\\operatorname{Re}\\alpha)^2+(\\operatorname{Im}\\alpha)^2\\bigr),\n\\]\nall of whose coefficients are strictly positive because \\(\\operatorname{Re}\\alpha<2\\).\n\nAs \\(Q\\) and \\(R\\) are monic, the leading coefficient of each shifted factor remains \\(+1\\).  Hence\n\\[\n\\widetilde Q(x):=Q(x+2),\\qquad \\widetilde R(x):=R(x+2)\n\\]\nare monic polynomials *with strictly positive integer coefficients*.\n\nStep 3.  Comparing the values at \\(x=\\pm1\\).\n-------------------------------------------\nFor any polynomial \\(F\\) with strictly positive coefficients we have\n\\[\nF(1)=\\sum_{j}c_j, \\qquad F(-1)=\\sum_{j}(-1)^j c_j, \\qquad F(1)-F(-1)=2\\sum_{j\\,\\text{odd}}c_j>0.\n\\]\nThus \\(F(1)>F(-1)\\).  Applying this to \\(F=\\widetilde Q,\\widetilde R\\) gives\n\\[\n\\widetilde Q(1)>\\widetilde Q(-1),\\qquad \\widetilde R(1)>\\widetilde R(-1).\n\\]\nBecause \\(\\widetilde Q(-1)=Q(1)\\) and \\(\\widetilde Q(1)=Q(3)\\), we deduce\n\\[\nQ(3)>Q(1).\\qquad(2)\n\\]\nIf \\(Q(1)\\le 0\\) then, by continuity and the fact that \\(Q(x)\\to+\\infty\\) as \\(x\\to+\\infty\\) (\\(Q\\) is monic), the intermediate value theorem would force a real root of \\(Q\\) in \\([1,\\infty)\\), contradicting Step 1(a).  Therefore \\(Q(1)\\ge 1\\), and from (2) we get\n\\[\nQ(3)\\ge Q(1)+1\\ge 2.\n\\]\nExactly the same reasoning gives\n\\[\nR(3)\\ge 2.\n\\]\n\nStep 4.  Finishing up.\n----------------------\nEvaluating \\(P=QR\\) at \\(x=3\\) yields\n\\[\nP(3)=Q(3)R(3),\\qquad Q(3),R(3)\\ge 2.\n\\]\nHence \\(P(3)\\) is a product of two integers each at least 2, so it is composite. \\(\\square\\)",
      "_meta": {
        "core_steps": [
          "Estimate the location of every root α of P, obtaining an absolute bound M on Re(α).",
          "Shift the variable by M (consider Q(x+M) and R(x+M)); every linear or quadratic factor now has strictly positive coefficients, so the whole product has positive coefficients.",
          "Compare the shifted factor at two nearby real points (one on each side of 0) to deduce 1 ≤ Q(1) < Q(2) (and likewise for R), hence Q(2), R(2) ≥ 2.",
          "Multiply:  P(2)=Q(2)·R(2) with each factor ≥2 ⇒ P(2) is composite."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The integer value at which P is finally evaluated and shown composite (any base ≥2 works).",
            "original": 2
          },
          "slot2": {
            "description": "The right-hand bound M for Re(α) and the corresponding shift x → x+M; it just has to exceed every root’s real part.",
            "original": 0.3
          },
          "slot3": {
            "description": "The two symmetric test points ±δ used after the shift to compare values; only their ordering (|δ| small, Q(δ) > Q(–δ)) matters.",
            "original": [
              0.5,
              -0.5
            ]
          },
          "slot4": {
            "description": "Any numerical bound used inside the root-location argument (e.g. (1+√5)/2); its exact value is irrelevant once it is < slot2).",
            "original": "(1+√5)/2 ≈ 1.618"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}