{
  "index": "1940-A-1",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "1. Prove that if \\( f(x) \\) is a polynomial with integral coefficients, and there exists an integer \\( k \\) such that none of the integers \\( f(1), f(2), \\ldots, f(k) \\) is divisible by \\( k \\), then \\( f(x) \\) has no integral root.",
  "solution": "Solution. Suppose \\( f \\) has an integral root \\( r \\). Then \\( f(x)=(x-r) g(x) \\) where \\( g(x) \\) is also a polynomial with integral coefficients. Then there are integers \\( p \\) and \\( q \\) such that \\( r=p+k q \\) and \\( 1 \\leq p \\leq k \\). But \\( f(p)= \\) \\( (p-r) g(p)=-k q g(p) \\) and hence \\( f(p) \\) is divisible by \\( k \\) contrary to the hypothesis. This contradiction shows that \\( f(x) \\) has no integral root.",
  "vars": [
    "x",
    "f",
    "r",
    "g",
    "p",
    "q"
  ],
  "params": [
    "k"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variable",
        "f": "polynomial",
        "r": "rootval",
        "g": "factorpoly",
        "p": "residue",
        "q": "quotient",
        "k": "modulus"
      },
      "question": "1. Prove that if \\( polynomial(variable) \\) is a polynomial with integral coefficients, and there exists an integer \\( modulus \\) such that none of the integers \\( polynomial(1), polynomial(2), \\ldots, polynomial(modulus) \\) is divisible by \\( modulus \\), then \\( polynomial(variable) \\) has no integral root.",
      "solution": "Solution. Suppose \\( polynomial \\) has an integral root \\( rootval \\). Then \\( polynomial(variable)=(variable-rootval) factorpoly(variable) \\) where \\( factorpoly \\) is also a polynomial with integral coefficients. Then there are integers \\( residue \\) and \\( quotient \\) such that \\( rootval=residue+modulus\\, quotient \\) and \\( 1 \\leq residue \\leq modulus \\). But \\( polynomial(residue)=(residue-rootval) factorpoly(residue)=-modulus\\, quotient\\, factorpoly(residue) \\) and hence \\( polynomial(residue) \\) is divisible by \\( modulus \\) contrary to the hypothesis. This contradiction shows that \\( polynomial(variable) \\) has no integral root."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pebblestone",
        "f": "compassrose",
        "r": "windvessel",
        "g": "lanternbeam",
        "p": "harvestmoon",
        "q": "driftwood",
        "k": "rainshadow"
      },
      "question": "Prove that if \\( compassrose(pebblestone) \\) is a polynomial with integral coefficients, and there exists an integer \\( rainshadow \\) such that none of the integers \\( compassrose(1), compassrose(2), \\ldots, compassrose(rainshadow) \\) is divisible by \\( rainshadow \\), then \\( compassrose(pebblestone) \\) has no integral root.",
      "solution": "Suppose \\( compassrose \\) has an integral root \\( windvessel \\). Then \\( compassrose(pebblestone)=(pebblestone-windvessel) lanternbeam(pebblestone) \\) where \\( lanternbeam(pebblestone) \\) is also a polynomial with integral coefficients. Then there are integers \\( harvestmoon \\) and \\( driftwood \\) such that \\( windvessel=harvestmoon+rainshadow driftwood \\) and \\( 1 \\leq harvestmoon \\leq rainshadow \\). But \\( compassrose(harvestmoon)= \\) \\( (harvestmoon-windvessel) lanternbeam(harvestmoon)=-rainshadow driftwood lanternbeam(harvestmoon) \\) and hence \\( compassrose(harvestmoon) \\) is divisible by \\( rainshadow \\) contrary to the hypothesis. This contradiction shows that \\( compassrose(pebblestone) \\) has no integral root."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "f": "staticnumber",
        "r": "peakvalue",
        "g": "constantterm",
        "p": "floatingreal",
        "q": "fractionalpart",
        "k": "irrational"
      },
      "question": "1. Prove that if \\( staticnumber(constantvalue) \\) is a polynomial with integral coefficients, and there exists an integer \\( irrational \\) such that none of the integers \\( staticnumber(1), staticnumber(2), \\ldots, staticnumber(irrational) \\) is divisible by \\( irrational \\), then \\( staticnumber(constantvalue) \\) has no integral root.",
      "solution": "Solution. Suppose \\( staticnumber \\) has an integral root \\( peakvalue \\). Then \\( staticnumber(constantvalue)=(constantvalue-peakvalue) constantterm(constantvalue) \\) where \\( constantterm(constantvalue) \\) is also a polynomial with integral coefficients. Then there are integers \\( floatingreal \\) and \\( fractionalpart \\) such that \\( peakvalue=floatingreal+irrational fractionalpart \\) and \\( 1 \\leq floatingreal \\leq irrational \\). But \\( staticnumber(floatingreal)= \\) \\( (floatingreal-peakvalue) constantterm(floatingreal)=-irrational fractionalpart constantterm(floatingreal) \\) and hence \\( staticnumber(floatingreal) \\) is divisible by \\( irrational \\) contrary to the hypothesis. This contradiction shows that \\( staticnumber(constantvalue) \\) has no integral root."
    },
    "garbled_string": {
      "map": {
        "x": "zdinqsom",
        "f": "plorxvne",
        "r": "hqmtzlwa",
        "g": "vpyksdre",
        "p": "wlenjazs",
        "q": "ybrudkoc",
        "k": "smenqfut"
      },
      "question": "Prove that if \\( plorxvne(zdinqsom) \\) is a polynomial with integral coefficients, and there exists an integer \\( smenqfut \\) such that none of the integers \\( plorxvne(1), plorxvne(2), \\ldots, plorxvne(smenqfut) \\) is divisible by \\( smenqfut \\), then \\( plorxvne(zdinqsom) \\) has no integral root.",
      "solution": "Suppose \\( plorxvne \\) has an integral root \\( hqmtzlwa \\). Then \\( plorxvne(zdinqsom)=(zdinqsom-hqmtzlwa) vpyksdre(zdinqsom) \\) where \\( vpyksdre(zdinqsom) \\) is also a polynomial with integral coefficients. Then there are integers \\( wlenjazs \\) and \\( ybrudkoc \\) such that \\( hqmtzlwa=wlenjazs+smenqfut ybrudkoc \\) and \\( 1 \\leq wlenjazs \\leq smenqfut \\). But \\( plorxvne(wlenjazs)= (wlenjazs-hqmtzlwa) vpyksdre(wlenjazs)=-smenqfut ybrudkoc vpyksdre(wlenjazs) \\) and hence \\( plorxvne(wlenjazs) \\) is divisible by \\( smenqfut \\) contrary to the hypothesis. This contradiction shows that \\( plorxvne(zdinqsom) \\) has no integral root."
    },
    "kernel_variant": {
      "question": "Let $m>1$ be an odd integer and let $f(x)\\in\\mathbb{Z}[x]$.  Assume that none of the $m$ numbers\n\\[\n f(2),\\;f(4),\\;f(6),\\;\\dots,\\;f(2m)\n\\]\nis divisible by $m$.  Prove that $f(x)$ has no integral root.",
      "solution": "Assume, toward a contradiction, that f has an integral root r; then\n   f(x)=(x−r)g(x) with g(x)∈ℤ[x].\n\nBecause m is odd, gcd(2,m)=1, so the map s↦2s (mod m) is a permutation of the residue classes modulo m. Hence there is a unique s∈{1,2,…,m} such that\n   r≡2s (mod m).\nPut\n   p:=2s∈{2,4,…,2m},  so r=p+mq for some q∈ℤ.\n\nEvaluate f at x=p:\n   f(p)=f(r−mq) = (p−r)g(p) = −m q·g(p),\nso m divides f(p). But p is one of 2,4,…,2m, and by hypothesis none of those values is divisible by m. This contradiction shows that f cannot have an integral root.\n\nTherefore f(x) has no integer zero.",
      "_meta": {
        "core_steps": [
          "Assume, for contradiction, that f has an integer root r and factor f(x) = (x−r)g(x) with g ∈ ℤ[x].",
          "Express r modulo k:  r = p + kq with p chosen from a complete residue system {1,…,k}.",
          "Evaluate at that residue:  f(p) = (p−r) g(p) = −kq·g(p), so k divides f(p).",
          "This divisibility contradicts the hypothesis that none of f(1),…,f(k) is divisible by k; hence f has no integer root."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The particular residue representatives used; any complete residue system modulo k would suffice.",
            "original": "the consecutive integers 1,2,…,k"
          },
          "slot2": {
            "description": "The modulus involved in the divisibility condition; any integer m>1 works with the same argument.",
            "original": "k"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}