{
  "index": "1961-A-6",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "6. If \\( J_{2}=\\{0,1\\} \\) is the field of integers modulo 2 , and if \\( J_{2}[x] \\) is the integral domain of polynomials in one indeterminate with coefficients in \\( J_{2} \\), prove that \\( p(x)=1+x+x^{2}+\\cdots+x^{n} \\) is reducible (factorable) in case \\( n \\) +1 is composite. Is the converse true? That is, if \\( n+1 \\) is prime, is \\( p(x) \\) irreducible?",
  "solution": "Solution. If \\( n+1=r \\cdot s, r>1, s>1 \\) is a factorization of \\( n+1 \\), then\n\\[\n1+x+\\cdots+x^{\\prime \\prime}=\\left(1+x+\\cdots+x^{r-1}\\right)\\left(1+x^{r}+x^{2 r}+\\cdots+x^{(s-1) r}\\right)\n\\]\nis valid over the integers and, a fortiori, when the coefficients are taken modulo 2.\n\nThe converse is not true, since\n\\[\n\\begin{aligned}\n\\left(1+x+x^{3}\\right)\\left(1+x^{2}+x^{3}\\right) & =1+x+x^{2}+3 x^{3}+x^{4}+x^{5}+x^{6} \\\\\n& \\equiv 1+x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}(\\bmod 2)\n\\end{aligned}\n\\]\nand \\( 6+1 \\) is a prime.",
  "vars": [
    "x",
    "n",
    "r",
    "s"
  ],
  "params": [
    "J_2",
    "p"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "indetx",
        "n": "exponent",
        "r": "factorone",
        "s": "factortwo",
        "J_{2}": "fieldtwo",
        "p": "polyfunc"
      },
      "question": "6. If \\( fieldtwo=\\{0,1\\} \\) is the field of integers modulo 2 , and if \\( fieldtwo[indetx] \\) is the integral domain of polynomials in one indeterminate with coefficients in \\( fieldtwo \\), prove that \\( polyfunc(indetx)=1+indetx+indetx^{2}+\\cdots+indetx^{exponent} \\) is reducible (factorable) in case \\( exponent +1 \\) is composite. Is the converse true? That is, if \\( exponent+1 \\) is prime, is \\( polyfunc(indetx) \\) irreducible?",
      "solution": "Solution. If \\( exponent+1=factorone \\cdot factortwo, factorone>1, factortwo>1 \\) is a factorization of \\( exponent+1 \\), then\n\\[\n1+indetx+\\cdots+indetx^{exponent}=\\left(1+indetx+\\cdots+indetx^{factorone-1}\\right)\\left(1+indetx^{factorone}+indetx^{2 factorone}+\\cdots+indetx^{(factortwo-1) factorone}\\right)\n\\]\nis valid over the integers and, a fortiori, when the coefficients are taken modulo 2.\n\nThe converse is not true, since\n\\[\n\\begin{aligned}\n\\left(1+indetx+indetx^{3}\\right)\\left(1+indetx^{2}+indetx^{3}\\right) & =1+indetx+indetx^{2}+3 indetx^{3}+indetx^{4}+indetx^{5}+indetx^{6} \\\\\n& \\equiv 1+indetx+indetx^{2}+indetx^{3}+indetx^{4}+indetx^{5}+indetx^{6}(\\bmod 2)\n\\end{aligned}\n\\]\nand \\( 6+1 \\) is a prime."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lanternfly",
        "n": "buttercup",
        "r": "hummingbird",
        "s": "strawberry",
        "J_2": "raincloud",
        "p": "gemstone"
      },
      "question": "6. If \\( raincloud_{2}=\\{0,1\\} \\) is the field of integers modulo 2 , and if \\( raincloud_{2}[lanternfly] \\) is the integral domain of polynomials in one indeterminate with coefficients in \\( raincloud_{2} \\), prove that \\( gemstone(lanternfly)=1+lanternfly+lanternfly^{2}+\\cdots+lanternfly^{buttercup} \\) is reducible (factorable) in case \\( buttercup +1 \\) is composite. Is the converse true? That is, if \\( buttercup+1 \\) is prime, is \\( gemstone(lanternfly) \\) irreducible?",
      "solution": "Solution. If \\( buttercup+1=hummingbird \\cdot strawberry, hummingbird>1, strawberry>1 \\) is a factorization of \\( buttercup+1 \\), then\n\\[\n1+lanternfly+\\cdots+lanternfly^{\\prime \\prime}=\\left(1+lanternfly+\\cdots+lanternfly^{hummingbird-1}\\right)\\left(1+lanternfly^{hummingbird}+lanternfly^{2 hummingbird}+\\cdots+lanternfly^{(strawberry-1) hummingbird}\\right)\n\\]\nis valid over the integers and, a fortiori, when the coefficients are taken modulo 2.\n\nThe converse is not true, since\n\\[\n\\begin{aligned}\n\\left(1+lanternfly+lanternfly^{3}\\right)\\left(1+lanternfly^{2}+lanternfly^{3}\\right) & =1+lanternfly+lanternfly^{2}+3 lanternfly^{3}+lanternfly^{4}+lanternfly^{5}+lanternfly^{6} \\\\\n& \\equiv 1+lanternfly+lanternfly^{2}+lanternfly^{3}+lanternfly^{4}+lanternfly^{5}+lanternfly^{6}(\\bmod 2)\n\\end{aligned}\n\\]\nand \\( 6+1 \\) is a prime."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "n": "infiniteindex",
        "r": "multiple",
        "s": "unitvalue",
        "J_2": "nonfield",
        "p": "integerfn"
      },
      "question": "6. If \\( nonfield=\\{0,1\\} \\) is the field of integers modulo 2 , and if \\( nonfield[constantvalue] \\) is the integral domain of polynomials in one indeterminate with coefficients in \\( nonfield \\), prove that \\( integerfn(constantvalue)=1+constantvalue+constantvalue^{2}+\\cdots+constantvalue^{infiniteindex} \\) is reducible (factorable) in case \\( infiniteindex +1 \\) is composite. Is the converse true? That is, if \\( infiniteindex+1 \\) is prime, is \\( integerfn(constantvalue) \\) irreducible?",
      "solution": "Solution. If \\( infiniteindex+1=multiple \\cdot unitvalue, multiple>1, unitvalue>1 \\) is a factorization of \\( infiniteindex+1 \\), then\n\\[\n1+constantvalue+\\cdots+constantvalue^{\\prime \\prime}=\\left(1+constantvalue+\\cdots+constantvalue^{multiple-1}\\right)\\left(1+constantvalue^{multiple}+constantvalue^{2 multiple}+\\cdots+constantvalue^{(unitvalue-1) multiple}\\right)\n\\]\nis valid over the integers and, a fortiori, when the coefficients are taken modulo 2.\n\nThe converse is not true, since\n\\[\n\\begin{aligned}\n\\left(1+constantvalue+constantvalue^{3}\\right)\\left(1+constantvalue^{2}+constantvalue^{3}\\right) & =1+constantvalue+constantvalue^{2}+3 constantvalue^{3}+constantvalue^{4}+constantvalue^{5}+constantvalue^{6} \\\\\n& \\equiv 1+constantvalue+constantvalue^{2}+constantvalue^{3}+constantvalue^{4}+constantvalue^{5}+constantvalue^{6}(\\bmod 2)\n\\end{aligned}\n\\]\nand \\( 6+1 \\) is a prime."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "n": "hjgrksla",
        "r": "plmnbvcx",
        "s": "oiuytrew",
        "J_2": "asdfghjk",
        "p": "zxcvbnml"
      },
      "question": "6. If \\( asdfghjk=\\{0,1\\} \\) is the field of integers modulo 2 , and if \\( asdfghjk[qzxwvtnp] \\) is the integral domain of polynomials in one indeterminate with coefficients in \\( asdfghjk \\), prove that \\( zxcvbnml(qzxwvtnp)=1+qzxwvtnp+qzxwvtnp^{2}+\\cdots+qzxwvtnp^{hjgrksla} \\) is reducible (factorable) in case \\( hjgrksla+1 \\) is composite. Is the converse true? That is, if \\( hjgrksla+1 \\) is prime, is \\( zxcvbnml(qzxwvtnp) \\) irreducible?",
      "solution": "Solution. If \\( hjgrksla+1=plmnbvcx \\cdot oiuytrew, plmnbvcx>1, oiuytrew>1 \\) is a factorization of \\( hjgrksla+1 \\), then\n\\[\n1+qzxwvtnp+\\cdots+qzxwvtnp^{\\prime \\prime}=\\left(1+qzxwvtnp+\\cdots+qzxwvtnp^{plmnbvcx-1}\\right)\\left(1+qzxwvtnp^{plmnbvcx}+qzxwvtnp^{2 plmnbvcx}+\\cdots+qzxwvtnp^{(oiuytrew-1) plmnbvcx}\\right)\n\\]\nis valid over the integers and, a fortiori, when the coefficients are taken modulo 2.\n\nThe converse is not true, since\n\\[\n\\begin{aligned}\n\\left(1+qzxwvtnp+qzxwvtnp^{3}\\right)\\left(1+qzxwvtnp^{2}+qzxwvtnp^{3}\\right) & =1+qzxwvtnp+qzxwvtnp^{2}+3 qzxwvtnp^{3}+qzxwvtnp^{4}+qzxwvtnp^{5}+qzxwvtnp^{6} \\\n& \\equiv 1+qzxwvtnp+qzxwvtnp^{2}+qzxwvtnp^{3}+qzxwvtnp^{4}+qzxwvtnp^{5}+qzxwvtnp^{6}(\\bmod 2)\n\\end{aligned}\n\\]\nand \\( 6+1 \\) is a prime."
    },
    "kernel_variant": {
      "question": "Let $\\mathbb F_{2}=\\{0,1\\}$ be the field with two elements and let $\\mathbb F_{2}[x]$ be the polynomial ring in one indeterminate over $\\mathbb F_{2}$.  \nFor a positive integer $n$ put  \n\\[\nf_{n}(x)=1+x+x^{2}+\\dots+x^{n}\\;\\in\\mathbb F_{2}[x]\\qquad(\\deg f_{n}=n).\n\\]\n\nFor a positive integer $m$ write $\\Phi_{m}(x)\\in\\mathbb Z[x]$ for the $m$-th cyclotomic polynomial and denote its reduction modulo $2$ by  \n\\[\n\\varphi_{2,m}(x):=\\bigl(\\Phi_{m}(x)\\bigr)\\pmod 2 \\;\\in\\mathbb F_{2}[x].\n\\]\n\nThroughout Parts (B)–(E) we assume  \n\\[\nn\\text{ is even}\\quad\\Longleftrightarrow\\quad n+1\\text{ is an odd integer }\\ge 3.\\tag{$\\star$}\n\\]  \n(Consequently $\\gcd(2,d)=1$ for every divisor $d>1$ of $n+1$, so the multiplicative order $\\operatorname{ord}_{d}(2)$ is well defined.)\n\nAnswer the following questions (all rings and polynomials are taken over $\\mathbb F_{2}$).\n\n(A)  (Cyclotomic factorisation)  \n Prove that  \n\\[\nf_{n}(x)=\\prod_{\\substack{d\\mid (n+1)\\\\ d>1}}\\varphi_{2,d}(x).\n\\]\n\n(B)  (Degrees of irreducible factors)  \n Show that every irreducible factor of $f_{n}(x)$ has degree equal to $\\operatorname{ord}_{d}(2)$ for some (necessarily odd) divisor $d>1$ of $n+1$.\n\n(C)  (Reducibility criterion)  \n Under the standing assumption $(\\star)$ prove the equivalence  \n\\[\n\\bigl(f_{n}(x)\\text{ reducible in }\\mathbb F_{2}[x]\\bigr)\n\\;\\Longleftrightarrow\\;\n\\bigl((n+1)\\text{ composite}\\bigr)\\ \\text{ or }\\ \\bigl(2\\text{ not primitive modulo }n+1\\bigr).\n\\]\n\n(D)  (Prime index)  \n Let $p\\ge 3$ be prime.  Show that  \n\\[\nf_{p-1}(x)\\text{ irreducible in }\\mathbb F_{2}[x]\\;\\Longleftrightarrow\\;2\\text{ is a primitive root modulo }p.\n\\]\n\n(E)  (Explicit examples)  \n Determine the complete factorisations in $\\mathbb F_{2}[x]$ as well as the degrees of all irreducible factors for  \n\n (i) $f_{10}(x)=1+x+\\dots+x^{10}$, and  \n\n (ii) $f_{30}(x)=1+x+\\dots+x^{30}$.",
      "solution": "Fix an algebraic closure $\\overline{\\mathbb F}_{2}$ of $\\mathbb F_{2}$ and write $\\operatorname{ord}_{m}(2)$ for the multiplicative order of $2$ modulo the (odd) integer $m$.\n\n-------------------------------------------------\n(A)  Cyclotomic factorisation of $f_{n}$\n-------------------------------------------------\nIn characteristic $2$ we have\n\\[\nx^{\\,n+1}+1=(x+1)\\bigl(1+x+\\dots+x^{n}\\bigr)=(x+1)f_{n}(x).\\tag{1}\n\\]\n\nOver $\\mathbb Z$ the classical cyclotomic factorisation reads\n\\[\nx^{\\,n+1}-1=\\prod_{d\\mid(n+1)}\\Phi_{d}(x).\\tag{2}\n\\]\n\nReplacing “$-1$’’ by “$+1$’’ makes no difference modulo $2$, so upon reducing (2) we obtain\n\\[\nx^{\\,n+1}+1=\\prod_{d\\mid(n+1)}\\varphi_{2,d}(x)\\qquad\\text{in }\\mathbb F_{2}[x].\\tag{3}\n\\]\n\nBecause $\\Phi_{1}(x)=x-1$,  \n\\[\n\\varphi_{2,1}(x)=\\bigl(\\Phi_{1}(x)\\bigr)\\pmod 2 = x+1 .\n\\]\nDividing (3) by the linear factor $x+1$ and invoking (1) yields\n\\[\nf_{n}(x)=\\prod_{\\substack{d\\mid(n+1)\\\\ d>1}}\\varphi_{2,d}(x).\\qquad\\Box\n\\]\n\n-------------------------------------------------\n(B)  Degrees of the irreducible factors of $f_{n}$\n-------------------------------------------------\nLet $d>1$ be an odd divisor of $n+1$ and let $\\zeta\\in\\overline{\\mathbb F}_{2}$ be a primitive $d$-th root of unity.  \nThe Frobenius automorphism $F:\\alpha\\mapsto\\alpha^{2}$ acts on the set $\\mu_{d}$ of $d$-th roots of unity; the orbit through $\\zeta$ has length\n\\[\nk=\\operatorname{ord}_{d}(2).\n\\]\nHence the minimal polynomial of $\\zeta$ over $\\mathbb F_{2}$ is\n\\[\nM_{\\zeta}(x)=\\prod_{j=0}^{k-1}\\bigl(x-F^{j}(\\zeta)\\bigr),\n\\]\nwhence $\\deg M_{\\zeta}=k$.  Because $\\zeta$ is a root of $\\varphi_{2,d}(x)$, every irreducible factor of $\\varphi_{2,d}(x)$, and therefore of $f_{n}(x)$, has degree $\\operatorname{ord}_{d}(2).\\;\\Box$\n\n-------------------------------------------------\n(C)  Reducibility criterion ($n$ even, $n+1$ odd $\\ge 3$)\n-------------------------------------------------\n“$\\Leftarrow$’’  If $n+1$ is composite, choose a proper divisor $d>1$.  \nBy (A) $\\varphi_{2,d}(x)$ is a non-constant factor of $f_{n}$, so $f_{n}$ is reducible.\n\nAssume now that $n+1=p$ is prime but $2$ is not primitive modulo $p$.  \nPut $k=\\operatorname{ord}_{p}(2)<p-1=n$.  \nBy (A) and (B) every irreducible factor of $f_{n}(x)=\\varphi_{2,p}(x)$ has degree $k<n$; hence $f_{n}$ is reducible.\n\n“$\\Rightarrow$’’  Conversely, suppose $f_{n}$ is reducible.  \nIf $n+1$ is composite we are done, so assume $n+1=p$ is prime.  \nThen $f_{n}(x)=\\varphi_{2,p}(x)$.  \nIf $f_{n}$ were irreducible, its degree $n$ would have to equal $\\operatorname{ord}_{p}(2)$ by (B), so $2$ would be primitive modulo $p$, contradicting the hypothesis.  \nThus the stated equivalence holds.\\;\\Box\n\n-------------------------------------------------\n(D)  The prime case\n-------------------------------------------------\nFor an odd prime $p$ one has $f_{p-1}(x)=\\varphi_{2,p}(x)$.  \nApplying (C) with $n=p-1$ gives\n\\[\nf_{p-1}\\text{ irreducible}\\;\\Longleftrightarrow\\;2\\text{ primitive modulo }p.\\;\\Box\n\\]\n\n-------------------------------------------------\n(E)  Explicit factorisations\n-------------------------------------------------\n(i)  $n=10$ ($n+1=11$ is prime and $\\operatorname{ord}_{11}(2)=10$).  \nBy (D) $f_{10}(x)$ is irreducible of degree $10$.\n\n(ii)  $n=30$ ($n+1=31$ is prime and $\\operatorname{ord}_{31}(2)=5$).  \nBy (B) every irreducible factor of $f_{30}$ has degree $5$; by degree count there are exactly $30/5=6$ such factors.\n\nThere are precisely six monic irreducible quintics over $\\mathbb F_{2}$.  Indeed, the standard counting formula gives  \n\\[\n\\frac{1}{5}\\sum_{d\\mid 5}\\mu(d)\\,2^{5/d}\n=\\frac{1}{5}\\bigl(\\mu(1)2^{5}+\\mu(5)2^{1}\\bigr)\n=\\frac{1}{5}(32-2)=6 .\n\\]\n\nA complete list is  \n\\[\n\\begin{aligned}\ng_{1}(x)&=x^{5}+x^{2}+1,\\\\\ng_{2}(x)&=x^{5}+x^{3}+1,\\\\\ng_{3}(x)&=x^{5}+x^{4}+1,\\\\\ng_{4}(x)&=x^{5}+x^{4}+x^{3}+x+1,\\\\\ng_{5}(x)&=x^{5}+x^{4}+x^{3}+x^{2}+1,\\\\\ng_{6}(x)&=x^{5}+x^{4}+x^{2}+x+1.\n\\end{aligned}\n\\]\n\nWhy does each $g_{i}$ divide $f_{30}(x)$?  \nBecause $\\deg g_{i}=5$, each $g_{i}$ splits completely in $\\mathbb F_{32}^{\\times}$, whose non-zero elements all have order $31$.  Thus every root $\\alpha$ of any $g_{i}$ satisfies $\\alpha^{31}=1$ but $\\alpha\\ne 1$, so $\\alpha^{31}+1=0$ and hence $\\alpha^{30}=1+\\alpha+\\dots+\\alpha^{29}=0$.  Consequently $g_{i}(x)\\mid x^{31}+1$ yet $g_{i}(x)\\nmid x+1$, and therefore $g_{i}(x)\\mid\\frac{x^{31}+1}{x+1}=f_{30}(x)$.\n\nThe polynomials $g_{1},\\dots,g_{6}$ are pairwise distinct and irreducible, and their product has degree $6\\times5=30=\\deg f_{30}$.  Hence\n\\[\nf_{30}(x)=g_{1}(x)g_{2}(x)g_{3}(x)g_{4}(x)g_{5}(x)g_{6}(x).\n\\]\n\nSummary of degrees  \n\\[\n\\begin{array}{ll}\n\\bullet\\ f_{10}: & \\text{one irreducible factor, degree }10,\\\\[2pt]\n\\bullet\\ f_{30}: & \\text{six irreducible factors, each of degree }5.\n\\end{array}\n\\quad\\Box\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.530813",
        "was_fixed": false,
        "difficulty_analysis": "• Extra theory The problem now demands familiarity with cyclotomic polynomials, the Frobenius automorphism, multiplicative orders, and the structure of finite-field extensions.  \n• Richer logical structure Instead of a single “composite ⇒ reducible’’ claim, the student must prove a bi-implication that splits into three non-trivial cases.  \n• Deep irreducibility criterion Irreducibility for prime indices is tied to the primitive-root property of 2, a classical number-theoretic concept far beyond the scope of the original exercise.  \n• Quantitative factorisation Part (E) obliges the solver not only to decide reducibility but to exhibit full factorizations, requiring concrete field-theoretic or computational techniques.  \n• Multiple interacting ideas Algebra (cyclotomic factorisation), field theory (Galois orbits under Frobenius), number theory (orders modulo a prime), and explicit computation all play essential roles, raising the overall conceptual and technical load well above the original and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let 𝔽₂={0,1} be the field with two elements and let 𝔽₂[x] be the polynomial ring in one indeterminate over 𝔽₂.  \nFor a positive integer n put  \n  fₙ(x)=1+x+x²+⋯+xⁿ ∈𝔽₂[x]   (deg fₙ=n).\n\nFor a positive integer m write Φₘ(x)∈ℤ[x] for the m-th cyclotomic polynomial and denote its reduction modulo 2 by  \n  φ₂,m(x):=Φₘ(x) (mod 2) ∈𝔽₂[x].\n\nThroughout Parts (B)–(E) we assume  \n  n is even ⇔ n+1 is an odd integer ≥3.                       (★)  \n(Consequently gcd(2,d)=1 for every divisor d>1 of n+1, so the multiplicative\norder ord_d(2) is well defined.)\n\nAnswer the following questions (all rings and polynomials are taken over 𝔽₂).\n\n(A) (Cyclotomic factorisation)  \n  Prove that\n   fₙ(x)=∏_{d∣(n+1), d>1} φ₂,d(x).\n\n(B) (Degrees of irreducible factors)  \n  Show that every irreducible factor of fₙ(x) has degree equal to ord_d(2) for some (necessarily odd) divisor d>1 of n+1.\n\n(C) (Reducibility criterion)  \n  Under the standing assumption (★) prove the equivalence\n   fₙ(x) reducible in 𝔽₂[x]  \n    ⇔ (n+1 is composite) or (2 is not a primitive root modulo n+1).\n\n(D) (Prime index)  \n  Let p≥3 be prime.  Show that  \n   f_{p−1}(x) is irreducible in 𝔽₂[x] ⇔ 2 is a primitive root modulo p.\n\n(E) (Explicit examples)  \n  Determine the complete factorisations in 𝔽₂[x] and the degrees of the irreducible factors for  \n\n  (i) f₁₀(x)=1+x+⋯+x¹⁰,   and  \n\n  (ii) f₃₀(x)=1+x+⋯+x³⁰.",
      "solution": "We fix an algebraic closure 𝔽̄ of 𝔽₂ and write ord_m(2) for the multiplicative order of 2 modulo an odd integer m.\n\n-------------------------------------------------\n(A) Cyclotomic factorisation of fₙ\n-------------------------------------------------\nIn characteristic 2 we have\n x^{n+1}+1=(x+1)·(1+x+⋯+xⁿ)=(x+1)·fₙ(x).                      (1)\n\nOver ℤ the classical cyclotomic factorisation is\n x^{n+1}−1=∏_{d∣(n+1)} Φ_d(x).                                (2)\n\nReplacing “−1’’ by “+1’’ makes no difference modulo 2, so reducing (2) gives\n x^{n+1}+1=∏_{d∣(n+1)} φ₂,d(x) in 𝔽₂[x].                     (3)\n\nBecause φ₂,1(x)=Φ₁(x)=x+1, dividing (3) by the linear factor x+1 and invoking (1) yields\n fₙ(x)=∏_{d∣(n+1), d>1} φ₂,d(x). □\n\n\n\n-------------------------------------------------\n(B) Degrees of the irreducible factors of fₙ\n-------------------------------------------------\nLet d>1 be an (odd) divisor of n+1 and let ζ∈𝔽̄ be a primitive d-th root of unity.  \nThe Frobenius automorphism F:α↦α² acts on the set μ_d of d-th roots of unity; its orbit through ζ has length\n k=ord_d(2).  \nHence the minimal polynomial of ζ over 𝔽₂ is the product of the conjugates\n M_ζ(x)=∏_{j=0}^{k-1}(x−F^{j}(ζ))  \nand therefore has degree k.  Since ζ is a root of φ₂,d(x), every irreducible factor of φ₂,d(x), hence of fₙ(x), has degree ord_d(2). □\n\n\n\n-------------------------------------------------\n(C) Reducibility criterion (n even, n+1 odd ≥3)\n-------------------------------------------------\n“⇐’’  If n+1 is composite pick a proper divisor d>1.  \nBy (A) φ₂,d(x) is a non-constant factor of fₙ, so fₙ is reducible.\n\nAssume now that n+1=p is prime but 2 is not primitive modulo p.  \nSet k=ord_p(2)<p−1=n.  \nBy (A) and (B) all irreducible factors of fₙ(x)=φ₂,p(x) have degree k<n, so the polynomial itself is reducible.\n\n“⇒’’  Conversely, suppose fₙ is reducible.  \nIf n+1 is composite we are finished, so assume n+1=p is prime.\nThen fₙ(x)=φ₂,p(x).  \nIf fₙ were irreducible its degree n would have to equal ord_p(2) by (B), i.e. 2 would be primitive modulo p, contradicting the hypothesis.  \nHence the equivalence is proved. □\n\n\n\n-------------------------------------------------\n(D) The prime case\n-------------------------------------------------\nFor an odd prime p we have f_{p−1}(x)=φ₂,p(x).  \nApplying (C) with n=p−1 shows\n f_{p−1} irreducible ⇔ 2 primitive modulo p. □\n\n\n\n-------------------------------------------------\n(E) Explicit factorisations\n-------------------------------------------------\n(i) n=10 (so n+1=11 is prime and 2 has order ord_{11}(2)=10).  \nBy (D) f₁₀(x) is irreducible of degree 10.\n\n(ii) n=30 (so n+1=31 is prime and ord_{31}(2)=5).  \nBy (B) every irreducible factor of f₃₀ has degree 5; by degree count there are exactly 30/5=6 such factors.\n\nThere are precisely six monic irreducible quintics over 𝔽₂, namely  \n g₁(x)=x⁵+x²+1,  \n g₂(x)=x⁵+x³+1,  \n g₃(x)=x⁵+x⁴+1,  \n g₄(x)=x⁵+x⁴+x³+x+1,  \n g₅(x)=x⁵+x⁴+x³+x²+1,  \n g₆(x)=x⁵+x⁴+x²+x+1.\n\nAll six divide x³¹+1 but not x+1, hence each divides f₃₀(x); they are pairwise distinct and irreducible, so\n\n f₃₀(x)=g₁(x)·g₂(x)·g₃(x)·g₄(x)·g₅(x)·g₆(x).\n\nSummary of degrees  \n • f₁₀ : one irreducible factor, degree 10.  \n • f₃₀ : six irreducible factors, each of degree 5. □",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.443016",
        "was_fixed": false,
        "difficulty_analysis": "• Extra theory The problem now demands familiarity with cyclotomic polynomials, the Frobenius automorphism, multiplicative orders, and the structure of finite-field extensions.  \n• Richer logical structure Instead of a single “composite ⇒ reducible’’ claim, the student must prove a bi-implication that splits into three non-trivial cases.  \n• Deep irreducibility criterion Irreducibility for prime indices is tied to the primitive-root property of 2, a classical number-theoretic concept far beyond the scope of the original exercise.  \n• Quantitative factorisation Part (E) obliges the solver not only to decide reducibility but to exhibit full factorizations, requiring concrete field-theoretic or computational techniques.  \n• Multiple interacting ideas Algebra (cyclotomic factorisation), field theory (Galois orbits under Frobenius), number theory (orders modulo a prime), and explicit computation all play essential roles, raising the overall conceptual and technical load well above the original and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}