{
  "index": "1963-B-1",
  "type": "ALG",
  "tag": [
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "\\text { 1. For what integer } a \\text { does } x^{2}-x+a \\text { divide } x^{13}+x+90 \\text { ? (page 577) }",
  "solution": "Solution. Suppose \\( x^{13}+x+90=\\left(x^{2}-x+a\\right) q(x) \\) where \\( a \\) is an integer. Then \\( q \\) is a polynomial with integer coefficients. If \\( a \\leq 0 \\), then \\( x^{2}-x+a \\), and hence also \\( x^{13}+x+90 \\), would have a non-negative zero, which is impossible. So \\( a>0 \\).\n\nSubstituting \\( x=-1,0,1 \\), we find\n\\[\n\\begin{aligned}\n(a+2) q(-1) & =88, \\\\\na q(0) & =90, \\\\\na q(1) & =92 .\n\\end{aligned}\n\\]\n\nThe last two equations show that \\( a \\) divides 2 , so \\( a=1 \\) or 2 . But if \\( a=1 \\), then 3 would divide 88 . So \\( a=2 \\) is the only possibility and, in fact,\n\\[\n\\begin{aligned}\nx^{13}+x+90= & \\left(x^{2}-x+2\\right)\\left(x^{11}+x^{10}-x^{9}-3 x^{8}-x^{7}+5 x^{6}\\right. \\\\\n& \\left.+7 x^{5}-3 x^{4}-7 x^{3}-11 x^{2}+23 x+45\\right) .\n\\end{aligned}\n\\]",
  "vars": [
    "x",
    "q"
  ],
  "params": [
    "a"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variable",
        "q": "quotient",
        "a": "parameter"
      },
      "question": "\\text { 1. For what integer } parameter \\text { does } variable^{2}-variable+parameter \\text { divide } variable^{13}+variable+90 \\text { ? (page 577) }",
      "solution": "Solution. Suppose \\( variable^{13}+variable+90=\\left(variable^{2}-variable+parameter\\right) quotient(variable) \\) where \\( parameter \\) is an integer. Then \\( quotient \\) is a polynomial with integer coefficients. If \\( parameter \\leq 0 \\), then \\( variable^{2}-variable+parameter \\), and hence also \\( variable^{13}+variable+90 \\), would have a non-negative zero, which is impossible. So \\( parameter>0 \\).\n\nSubstituting \\( variable=-1,0,1 \\), we find\n\\[\n\\begin{aligned}\n(parameter+2) \\, quotient(-1) & =88, \\\\\nparameter \\, quotient(0) & =90, \\\\\nparameter \\, quotient(1) & =92 .\n\\end{aligned}\n\\]\n\nThe last two equations show that \\( parameter \\) divides 2 , so \\( parameter=1 \\) or 2 . But if \\( parameter=1 \\), then 3 would divide 88 . So \\( parameter=2 \\) is the only possibility and, in fact,\n\\[\n\\begin{aligned}\nvariable^{13}+variable+90= & \\left(variable^{2}-variable+2\\right)\\left(variable^{11}+variable^{10}-variable^{9}-3 variable^{8}-variable^{7}+5 variable^{6}\\right. \\\\\n& \\left.+7 variable^{5}-3 variable^{4}-7 variable^{3}-11 variable^{2}+23 variable+45\\right) .\n\\end{aligned}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marigold",
        "q": "paintbrush",
        "a": "lighthouse"
      },
      "question": "\\text { 1. For what integer } lighthouse \\text { does } marigold^{2}-marigold+lighthouse \\text { divide } marigold^{13}+marigold+90 \\text { ? (page 577) }",
      "solution": "Solution. Suppose \\( marigold^{13}+marigold+90=\\left(marigold^{2}-marigold+lighthouse\\right) paintbrush(marigold) \\) where \\( lighthouse \\) is an integer. Then \\( paintbrush \\) is a polynomial with integer coefficients. If \\( lighthouse \\leq 0 \\), then \\( marigold^{2}-marigold+lighthouse \\), and hence also \\( marigold^{13}+marigold+90 \\), would have a non-negative zero, which is impossible. So \\( lighthouse>0 \\).\n\nSubstituting \\( marigold=-1,0,1 \\), we find\n\\[\n\\begin{aligned}\n(lighthouse+2) paintbrush(-1) & =88, \\\\\nlighthouse paintbrush(0) & =90, \\\\\nlighthouse paintbrush(1) & =92 .\n\\end{aligned}\n\\]\n\nThe last two equations show that \\( lighthouse \\) divides 2 , so \\( lighthouse=1 \\) or 2 . But if \\( lighthouse=1 \\), then 3 would divide 88 . So \\( lighthouse=2 \\) is the only possibility and, in fact,\n\\[\n\\begin{aligned}\nmarigold^{13}+marigold+90= & \\left(marigold^{2}-marigold+2\\right)\\left(marigold^{11}+marigold^{10}-marigold^{9}-3 marigold^{8}-marigold^{7}+5 marigold^{6}\\right. \\\\\n& \\left.+7 marigold^{5}-3 marigold^{4}-7 marigold^{3}-11 marigold^{2}+23 marigold+45\\right) .\n\\end{aligned}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "q": "nonquotient",
        "a": "irrational"
      },
      "question": "\\text { 1. For what integer } irrational \\text { does } constantval^{2}-constantval+irrational \\text { divide } constantval^{13}+constantval+90 \\text { ? (page 577) }",
      "solution": "Solution. Suppose \\( constantval^{13}+constantval+90=\\left(constantval^{2}-constantval+irrational\\right) nonquotient(constantval) \\) where \\( irrational \\) is an integer. Then \\( nonquotient \\) is a polynomial with integer coefficients. If \\( irrational \\leq 0 \\), then \\( constantval^{2}-constantval+irrational \\), and hence also \\( constantval^{13}+constantval+90 \\), would have a non-negative zero, which is impossible. So \\( irrational>0 \\).\n\nSubstituting \\( constantval=-1,0,1 \\), we find\n\\[\n\\begin{aligned}\n(irrational+2) nonquotient(-1) & =88, \\\\\nirrational\\, nonquotient(0) & =90, \\\\\nirrational\\, nonquotient(1) & =92 .\n\\end{aligned}\n\\]\n\nThe last two equations show that \\( irrational \\) divides 2 , so \\( irrational=1 \\) or 2 . But if \\( irrational=1 \\), then 3 would divide 88 . So \\( irrational=2 \\) is the only possibility and, in fact,\n\\[\n\\begin{aligned}\nconstantval^{13}+constantval+90= & \\left(constantval^{2}-constantval+2\\right)\\left(constantval^{11}+constantval^{10}-constantval^{9}-3 constantval^{8}-constantval^{7}+5 constantval^{6}\\right. \\\\\n& \\left.+7 constantval^{5}-3 constantval^{4}-7 constantval^{3}-11 constantval^{2}+23 constantval+45\\right) .\n\\end{aligned}\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "q": "hbvcnqwe",
        "a": "zjrnksla"
      },
      "question": "\\text { 1. For what integer } zjrnksla \\text { does } qzxwvtnp^{2}-qzxwvtnp+zjrnksla \\text { divide } qzxwvtnp^{13}+qzxwvtnp+90 \\text { ? (page 577) }",
      "solution": "Solution. Suppose \\( qzxwvtnp^{13}+qzxwvtnp+90=\\left(qzxwvtnp^{2}-qzxwvtnp+zjrnksla\\right) hbvcnqwe(qzxwvtnp) \\) where \\( zjrnksla \\) is an integer. Then \\( hbvcnqwe \\) is a polynomial with integer coefficients. If \\( zjrnksla \\leq 0 \\), then \\( qzxwvtnp^{2}-qzxwvtnp+zjrnksla \\), and hence also \\( qzxwvtnp^{13}+qzxwvtnp+90 \\), would have a non-negative zero, which is impossible. So \\( zjrnksla>0 \\).\n\nSubstituting \\( qzxwvtnp=-1,0,1 \\), we find\n\\[\n\\begin{aligned}\n(zjrnksla+2) hbvcnqwe(-1) & =88, \\\\\nzjrnksla hbvcnqwe(0) & =90, \\\\\nzjrnksla hbvcnqwe(1) & =92 .\n\\end{aligned}\n\\]\n\nThe last two equations show that \\( zjrnksla \\) divides 2 , so \\( zjrnksla=1 \\) or 2 . But if \\( zjrnksla=1 \\), then 3 would divide 88 . So \\( zjrnksla=2 \\) is the only possibility and, in fact,\n\\[\n\\begin{aligned}\nqzxwvtnp^{13}+qzxwvtnp+90= & \\left(qzxwvtnp^{2}-qzxwvtnp+2\\right)\\left(qzxwvtnp^{11}+qzxwvtnp^{10}-qzxwvtnp^{9}-3 qzxwvtnp^{8}-qzxwvtnp^{7}+5 qzxwvtnp^{6}\\right. \\\\\n& \\left.+7 qzxwvtnp^{5}-3 qzxwvtnp^{4}-7 qzxwvtnp^{3}-11 qzxwvtnp^{2}+23 qzxwvtnp+45\\right) .\n\\end{aligned}\n\\]"
    },
    "kernel_variant": {
      "question": "For which integer values of $a$ does the quadratic polynomial\n\\[\n    x^{2}-x+a\n\\]\ndivide the polynomial\n\\[\n    P(x)=x^{17}+x+100\\ ?\n\\]",
      "solution": "Suppose x^2–x+a divides P(x)=x^17+x+100 in ℤ[x], so\n   P(x)=(x^2–x+a)Q(x),  Q∈ℤ[x].\n1.  If a≤0 then Δ=1–4a≥1 so x^2–x+a has a real root r≥1.  But P(r)=r^17+r+100>0, contradiction.  Hence a>0.\n2.  Plug in x=0,1:  aQ(0)=P(0)=100,  aQ(1)=P(1)=1+1+100=102.  Thus a|100 and a|102 ⇒ a|gcd(100,102)=2 ⇒ a=1 or 2.\n3.  (i)  If a=1 then x^2–x+1 has roots ω=e^{±iπ/3}, and ω^6=1 ⇒ ω^17=ω^5.  Thus P(ω)=ω^5+ω+100=(ω+ω^5)+100=1+100=101≠0, so no.\n    (ii)  If a=2 then x^2–x+2 has roots α with α^2=α–2.  One checks by the recurrence α^n=α·α^{n–1}–2α^{n–2} that α^17+α+100=272α–86≠0, so again no.\nTherefore neither a=1 nor a=2 works, and no other integer is possible.  Conclusion: there is no integer a for which x^2–x+a divides x^17+x+100.",
      "_meta": {
        "core_steps": [
          "Assume (x^2 − x + a) divides the given polynomial and let the quotient have integer coefficients.",
          "Observe that a > 0; otherwise the quadratic factor would have a non-negative real root, contradicting the positive value of the dividend there.",
          "Evaluate at x = 0 and x = 1 to get a q(0)=c and a q(1)=c+2, so a divides both c and c+2 ⇒ a | 2 ⇒ a ∈ {1,2}.",
          "Evaluate at x = −1 to obtain (a+2) q(−1)=c−2; use this to rule out a=1, leaving a=2 as the only possibility."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Exponent of the highest-degree term in the dividend polynomial (currently 13 in x^13). It only needs to be odd so that (−1)^n = −1 in step 4.",
            "original": 13
          },
          "slot2": {
            "description": "Constant term of the dividend polynomial (currently 90). It can be replaced by any even integer not congruent to 2 mod 3, so that gcd(c, c+2)=2 and 3∤(c−2) in steps 3–4.",
            "original": 90
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}