{
  "index": "2007-B-4",
  "type": "ALG",
  "tag": [
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Let $n$ be a positive integer. Find the number of pairs $P, Q$ of\npolynomials with real coefficients such that\n\\[\n(P(X))^2 + (Q(X))^2 = X^{2n} + 1\n\\]\nand $\\deg P > \\deg Q$.",
  "solution": "The number of pairs is $2^{n+1}$. The degree condition forces\n$P$ to have degree $n$ and leading coefficient $\\pm 1$; we may count\npairs in which $P$ has leading coefficient 1 as long as we multiply by $2$\nafterward.\n\nFactor both sides:\n\\begin{align*}\n& (P(X) + Q(X)i)(P(X) - Q(X)i) \\\\\n& = \\prod_{j=0}^{n-1}\n(X - \\exp(2 \\pi i (2j+1)/(4n))) \\\\\n& \\quad \\cdot \\prod_{j=0}^{n-1}\n(X + \\exp(2 \\pi i (2j+1)/(4n))).\n\\end{align*}\nThen each choice of $P,Q$ corresponds to equating $P(X) + Q(X)i$\nwith the product of some $n$ factors on the right,\nin which we choose exactly of the two factors for each $j=0,\\dots,n-1$.\n(We must take exactly $n$\nfactors because as a polynomial in $X$ with complex coefficients, $P(X) + Q(X)i$\nhas degree exactly $n$. We must choose one for each $j$ to ensure that\n$P(X) + Q(X)i$ and $P(X) -Q(X)i$ are complex conjugates, so that $P, Q$ have\nreal coefficients.) Thus there are $2^n$ such pairs;\nmultiplying by 2 to allow $P$ to have\nleading coefficient $-1$ yields the desired result.\n\n\\textbf{Remark:} If we allow $P$ and $Q$ to have complex coefficients but\nstill require $\\deg(P) > \\deg(Q)$, then the number of pairs increases\nto $2\\binom{2n}{n}$, as we may choose any $n$ of the $2n$ factors of\n$X^{2n}+1$ to use to form $P(X) + Q(X)i$.",
  "vars": [
    "P",
    "Q",
    "X",
    "j"
  ],
  "params": [
    "n"
  ],
  "sci_consts": [
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "P": "polyone",
        "Q": "polytwo",
        "X": "inputvar",
        "j": "indexer",
        "n": "fixedsize"
      },
      "question": "Let $fixedsize$ be a positive integer. Find the number of pairs $polyone, polytwo$ of\npolynomials with real coefficients such that\n\\[\n(polyone(inputvar))^2 + (polytwo(inputvar))^2 = inputvar^{2 fixedsize} + 1\n\\]\nand $\\deg polyone > \\deg polytwo$.",
      "solution": "The number of pairs is $2^{fixedsize+1}$. The degree condition forces\n$polyone$ to have degree $fixedsize$ and leading coefficient $\\pm 1$; we may count\npairs in which $polyone$ has leading coefficient 1 as long as we multiply by $2$\nafterward.\n\nFactor both sides:\n\\begin{align*}\n& (polyone(inputvar) + polytwo(inputvar)i)(polyone(inputvar) - polytwo(inputvar)i) \\\\\n& = \\prod_{indexer=0}^{fixedsize-1}\n(inputvar - \\exp(2 \\pi i (2 indexer +1)/(4 fixedsize))) \\\\\n& \\quad \\cdot \\prod_{indexer=0}^{fixedsize-1}\n(inputvar + \\exp(2 \\pi i (2 indexer +1)/(4 fixedsize))).\n\\end{align*}\nThen each choice of $polyone,polytwo$ corresponds to equating $polyone(inputvar) + polytwo(inputvar)i$\nwith the product of some $fixedsize$ factors on the right,\nin which we choose exactly of the two factors for each $indexer=0,\\dots, fixedsize-1$.\n(We must take exactly $fixedsize$\nfactors because as a polynomial in $inputvar$ with complex coefficients, $polyone(inputvar) + polytwo(inputvar)i$\nhas degree exactly $fixedsize$. We must choose one for each $indexer$ to ensure that\n$polyone(inputvar) + polytwo(inputvar)i$ and $polyone(inputvar) -polytwo(inputvar)i$ are complex conjugates, so that $polyone, polytwo$ have\nreal coefficients.) Thus there are $2^{fixedsize}$ such pairs;\nmultiplying by 2 to allow $polyone$ to have\nleading coefficient $-1$ yields the desired result.\n\n\\textbf{Remark:} If we allow $polyone$ and $polytwo$ to have complex coefficients but\nstill require $\\deg(polyone) > \\deg(polytwo)$, then the number of pairs increases\nto $2\\binom{2 fixedsize}{fixedsize}$, as we may choose any $fixedsize$ of the $2 fixedsize$ factors of\n$inputvar^{2 fixedsize}+1$ to use to form $polyone(inputvar) + polytwo(inputvar)i$."
    },
    "descriptive_long_confusing": {
      "map": {
        "P": "sunflower",
        "Q": "honeysuckle",
        "X": "telescope",
        "j": "salamander",
        "n": "galaxyroad"
      },
      "question": "Let $galaxyroad$ be a positive integer. Find the number of pairs $sunflower, honeysuckle$ of\npolynomials with real coefficients such that\n\\[\n(sunflower(telescope))^2 + (honeysuckle(telescope))^2 = telescope^{2galaxyroad} + 1\n\\]\nand $\\deg sunflower > \\deg honeysuckle$.",
      "solution": "The number of pairs is $2^{galaxyroad+1}$. The degree condition forces\n$sunflower$ to have degree $galaxyroad$ and leading coefficient $\\pm 1$; we may count\npairs in which $sunflower$ has leading coefficient 1 as long as we multiply by $2$\nafterward.\n\nFactor both sides:\n\\begin{align*}\n& (sunflower(telescope) + honeysuckle(telescope)i)(sunflower(telescope) - honeysuckle(telescope)i) \\\\\n& = \\prod_{\\salamander=0}^{galaxyroad-1}\n(telescope - \\exp(2 \\pi i (2\\salamander+1)/(4galaxyroad))) \\\\\n& \\quad \\cdot \\prod_{\\salamander=0}^{galaxyroad-1}\n(telescope + \\exp(2 \\pi i (2\\salamander+1)/(4galaxyroad))).\n\\end{align*}\nThen each choice of $sunflower,honeysuckle$ corresponds to equating $sunflower(telescope) + honeysuckle(telescope)i$\nwith the product of some $galaxyroad$ factors on the right,\nin which we choose exactly of the two factors for each $\\salamander=0,\\dots,galaxyroad-1$.\n(We must take exactly $galaxyroad$\nfactors because as a polynomial in $telescope$ with complex coefficients, $sunflower(telescope) + honeysuckle(telescope)i$\nhas degree exactly $galaxyroad$. We must choose one for each $\\salamander$ to ensure that\n$sunflower(telescope) + honeysuckle(telescope)i$ and $sunflower(telescope) -honeysuckle(telescope)i$ are complex conjugates, so that $sunflower, honeysuckle$ have\nreal coefficients.) Thus there are $2^{galaxyroad}$ such pairs;\nmultiplying by 2 to allow $sunflower$ to have\nleading coefficient $-1$ yields the desired result.\n\n\\textbf{Remark:} If we allow $sunflower$ and $honeysuckle$ to have complex coefficients but\nstill require $\\deg(sunflower) > \\deg(honeysuckle)$, then the number of pairs increases\nto $2\\binom{2galaxyroad}{galaxyroad}$, as we may choose any $galaxyroad$ of the $2galaxyroad$ factors of\n$telescope^{2galaxyroad}+1$ to use to form $sunflower(telescope) + honeysuckle(telescope)i$.}"
    },
    "descriptive_long_misleading": {
      "map": {
        "P": "constantfn",
        "Q": "stationary",
        "X": "constant",
        "j": "aggregate",
        "n": "infinite"
      },
      "question": "Let infinite be a positive integer. Find the number of pairs constantfn, stationary of\npolynomials with real coefficients such that\n\\[\n(constantfn(constant))^2 + (stationary(constant))^2 = constant^{2 infinite} + 1\n\\]\nand $\\deg$ constantfn $> \\deg$ stationary.",
      "solution": "The number of pairs is $2^{infinite+1}$. The degree condition forces\nconstantfn to have degree infinite and leading coefficient $\\pm 1$; we may count\npairs in which constantfn has leading coefficient 1 as long as we multiply by $2$\nafterward.\n\nFactor both sides:\n\\begin{align*}\n& (constantfn(constant) + stationary(constant)i)(constantfn(constant) - stationary(constant)i) \\\\\n& = \\prod_{aggregate=0}^{infinite-1}\n(constant - \\exp(2 \\pi i (2aggregate+1)/(4 infinite))) \\\\\n& \\quad \\cdot \\prod_{aggregate=0}^{infinite-1}\n(constant + \\exp(2 \\pi i (2aggregate+1)/(4 infinite))).\n\\end{align*}\nThen each choice of constantfn, stationary corresponds to equating constantfn(constant) + stationary(constant)i\nwith the product of some infinite factors on the right,\nin which we choose exactly of the two factors for each $aggregate=0,\\dots,infinite-1$.\n(We must take exactly infinite\nfactors because as a polynomial in constant with complex coefficients, constantfn(constant) + stationary(constant)i\nhas degree exactly infinite. We must choose one for each aggregate to ensure that\nconstantfn(constant) + stationary(constant)i and constantfn(constant) -stationary(constant)i are complex conjugates, so that constantfn, stationary have\nreal coefficients.) Thus there are $2^{infinite}$ such pairs;\nmultiplying by 2 to allow constantfn to have\nleading coefficient $-1$ yields the desired result.\n\n\\textbf{Remark:} If we allow constantfn and stationary to have complex coefficients but\nstill require $\\deg(constantfn) > \\deg(stationary)$, then the number of pairs increases\nto $2\\binom{2 infinite}{infinite}$, as we may choose any infinite of the $2$ infinite factors of\n$constant^{2 infinite}+1$ to use to form constantfn(constant) + stationary(constant)i."
    },
    "garbled_string": {
      "map": {
        "P": "qzxwvtnp",
        "Q": "hjgrksla",
        "X": "vbnmlkjh",
        "j": "plmoknij",
        "n": "xqazwsed"
      },
      "question": "Let $xqazwsed$ be a positive integer. Find the number of pairs $qzxwvtnp, hjgrksla$ of\npolynomials with real coefficients such that\n\\[\n(qzxwvtnp(vbnmlkjh))^2 + (hjgrksla(vbnmlkjh))^2 = vbnmlkjh^{2xqazwsed} + 1\n\\]\nand $\\deg qzxwvtnp > \\deg hjgrksla$.",
      "solution": "The number of pairs is $2^{xqazwsed+1}$. The degree condition forces\n$qzxwvtnp$ to have degree $xqazwsed$ and leading coefficient $\\pm 1$; we may count\npairs in which $qzxwvtnp$ has leading coefficient 1 as long as we multiply by $2$\nafterward.\n\nFactor both sides:\n\\begin{align*}\n& (qzxwvtnp(vbnmlkjh) + hjgrksla(vbnmlkjh)i)(qzxwvtnp(vbnmlkjh) - hjgrksla(vbnmlkjh)i) \\\\\n& = \\prod_{plmoknij=0}^{xqazwsed-1}\n(vbnmlkjh - \\exp(2 \\pi i (2plmoknij+1)/(4xqazwsed))) \\\\\n& \\quad \\cdot \\prod_{plmoknij=0}^{xqazwsed-1}\n(vbnmlkjh + \\exp(2 \\pi i (2plmoknij+1)/(4xqazwsed))).\n\\end{align*}\nThen each choice of $qzxwvtnp,hjgrksla$ corresponds to equating $qzxwvtnp(vbnmlkjh) + hjgrksla(vbnmlkjh)i$\nwith the product of some $xqazwsed$ factors on the right,\nin which we choose exactly one of the two factors for each $plmoknij=0,\\dots,xqazwsed-1$.\n(We must take exactly $xqazwsed$\nfactors because as a polynomial in $vbnmlkjh$ with complex coefficients, $qzxwvtnp(vbnmlkjh) + hjgrksla(vbnmlkjh)i$\nhas degree exactly $xqazwsed$. We must choose one for each $plmoknij$ to ensure that\n$qzxwvtnp(vbnmlkjh) + hjgrksla(vbnmlkjh)i$ and $qzxwvtnp(vbnmlkjh) -hjgrksla(vbnmlkjh)i$ are complex conjugates, so that $qzxwvtnp, hjgrksla$ have\nreal coefficients.) Thus there are $2^{xqazwsed}$ such pairs;\nmultiplying by 2 to allow $qzxwvtnp$ to have\nleading coefficient $-1$ yields the desired result.\n\n\\textbf{Remark:} If we allow $qzxwvtnp$ and $hjgrksla$ to have complex coefficients but\nstill require $\\deg(qzxwvtnp) > \\deg(hjgrksla)$, then the number of pairs increases\nto $2\\binom{2xqazwsed}{xqazwsed}$, as we may choose any $xqazwsed$ of the $2xqazwsed$ factors of\n$vbnmlkjh^{2xqazwsed}+1$ to use to form $qzxwvtnp(vbnmlkjh) + hjgrksla(vbnmlkjh)i$.",
      "â": ""
    },
    "kernel_variant": {
      "question": "For a fixed positive integer m define  \n  Fₘ(x)= (x^{2m}+1)(x^{2m}+9) ∈ ℝ[x].\n\nLet N(m) be the number of ordered pairs of real-coefficient polynomials  \n\n  (A(x), B(x)) ∈ ℝ[x]²  \n\nthat satisfy simultaneously  \n\n(1) A(x)² + B(x)² = Fₘ(x);  \n(2) deg A > deg B;  \n(3) A(x) is monic;  \n(4) B(0)=0;  \n(5) gcd(A,B)=1.\n\nDetermine a closed arithmetic formula for N(m) in terms of m.  \n(The use of complex numbers and of the orthogonality relations for m-th roots of unity is allowed.)\n\n───────────────────────────────────────────────────────────",
      "solution": "Notation.  Let ζₘ = e^{2πi/m}.  Indices k always range over {0,…,m−1} and j over {1,2}.\n\n1.  Preliminary complex factorisation  \nPut  \n G(x)=A(x)+iB(x), H(x)=A(x)−iB(x)=  \\overline{G(x)}.  \nCondition (1) is G·H = Fₘ.  The two quadratic factors of Fₘ have no common root, so Fₘ is square-free; hence gcd(G,H)=1 and, in particular, gcd(A,B)=1 (requirement (5)).  \ndeg Fₘ = 4m, so deg G = deg H = 2m.  Because A is monic, deg A = 2m > deg B, whence (2).\n\n2.  The roots of Fₘ  \nPut θ_k = (2k+1)π/(2m).  Then  \n x^{2m}+1 = ∏_{k}(x−e^{ iθ_k})(x−e^{−iθ_k}),  \n x^{2m}+9 = ∏_{k}(x−3^{1/m}e^{ iθ_k})(x−3^{1/m}e^{−iθ_k}).      (★)\n\nFor every k there are 4 roots with argument θ_k (two on the unit circle, two on the circle of radius 3^{1/m}).  \nTo build G one must choose, for each conjugate pair, exactly one root.  Thus there are 2^{2m}=4^{m} preliminary choices.\n\n3.  Encoding the choices and imposing B(0)=0  \nFix the four roots with positive argument\n\n ρ_{k,1}=e^{ iθ_k}, ρ_{k,2}=3^{1/m}e^{ iθ_k}.  \n\nIntroduce signs δ_{k,j}∈{±1} by  \n\n δ_{k,j}= +1 ⇔ ρ_{k,j} is put in G,  \n δ_{k,j}= −1 ⇔ \\barρ_{k,j} is put in G.\n\nThe constant term of G is  \n\n Π(δ)=∏_{k,j} ρ_{k,j}^{(1+δ_{k,j})/2}\\,\\barρ_{k,j}^{(1-δ_{k,j})/2}  \n   = 3·exp ( i Σ_{k,j} δ_{k,j}θ_k ).                                 (1)\n\n(|ρ_{k,1}ρ_{k,2}| = 3^{1/m}, so |Π(δ)| = 3.)  \nCondition (4) demands Im G(0)=0, i.e. Π(δ) is real, which is equivalent to  \n\n Σ_{k,j} δ_{k,j}θ_k ≡ 0 (mod π).                                    (2)\n\nPut s_k=δ_{k,1}+δ_{k,2}∈{−2,0,2} and t_k=s_k/2∈{−1,0,1}.  \nBecause θ_k=(2k+1)π/(2m), congruence (2) becomes  \n\n Σ_{k=0}^{m−1} t_k(2k+1) ≡ 0 (mod m).                              (3)\n\nHence  \n\n N(m)=∑_{t∈{−1,0,1}^{m}} 2^{#\\{k:t_k=0\\}} · 1_{[Σ t_k(2k+1) ≡ 0 (mod m)]}. (4)\n\n(The factor 2 for t_k=0 counts the two orders {+1,−1}.)\n\n4.  Root-of-unity filter  \nInsert  \n 1_{[L≡0 (mod m)]} = (1/m) Σ_{r=0}^{m−1} ζₘ^{rL}\n\ninto (4) and interchange sums:\n\n N(m)= (1/m) Σ_{r=0}^{m−1} ∏_{k=0}^{m−1}\n    ( ζₘ^{r(2k+1)} + 2 + ζₘ^{−r(2k+1)} ).                       (5)\n\nSince ζ^α+2+ζ^{−α} = (1+ζ^α)(1+ζ^{−α}) ≥ 0, set  \n\n Q(r)=∏_{k=0}^{m−1}(1+ζₘ^{r(2k+1)}), so N(m)= (1/m) Σ_{r} Q(r)². (6)\n\n5.  Evaluating Q(r)  \nPut d=gcd(r,m) and s=m/d; let η=ζₘ^{r}, so ord(η)=s.\n\nAs k runs through {0,…,m−1}, the exponents 2k+1 cover every odd residue modulo s exactly\n\n  d times if s is odd,  \n  2d times if s is even.\n\nConsequently\n\n Q(r)=T_s^{\\,d} with T_s := ∏_{k=0}^{s−1}(1+η^{2k+1}).            (7)\n\nWe now compute T_s.\n\nLemma.  Let η be a primitive s-th root and T_s as above.  \n (i) If s is odd     then T_s = 2.  \n (ii) If s ≡ 2 (mod 4) then T_s = 0.  \n (iii) If 4 | s     then T_s = 4.\n\nProof.  \n(i)  When s is odd, k↦2k+1 is a permutation of {0,…,s−1}; thus\n\n T_s = ∏_{j=0}^{s−1}(1+η^{j}) = (−1)^s ( (−1)^s −1 ) = 2.\n\n(ii)  Write s=2t with t odd, so η^{t}=−1.  One factor equals 1+η^{t}=0.\n\n(iii)  Put s=2t with t even.  Because 2k+1 runs through each odd residue twice,  \n T_s = (∏_{j odd}(1+η^{j}))² =: U_s².  \nSet ξ = η², a primitive t-th root, and a := η.  Then\n\n U_s = ∏_{j=0}^{t-1}(1 + a ξ^{j})=2.                        (8)\n\nIndeed, use the identity ∏_{j=0}^{t-1}(x−ξ^{j}) = x^{t}−1 with x=−1/a; multiplying out gives U_s = 2.  Hence T_s = U_s² = 4. ∎\n\nTherefore  \n\n T_s = { 2 (s odd); 0 (s≡2 mod 4); 4 (4|s).                      (9)\n\n6.  Closing the count  \nBecause Q(r)=T_{m/d}^{\\,d}, equation (6) yields  \n\n N(m)= (1/m) Σ_{d|m} φ(m/d)·T_{m/d}^{\\,2d}.                        (10)\n\nLet s = m/d.  Then 2d = 2m/s, and with (9) we obtain the fully explicit\n\n N(m)= (1/m) Σ_{s|m} φ(s)·C(s,m),                                 (11)\n\nwhere  \n\n C(s,m)= ⎧ 2^{2m/s}  if s is odd,  \n    0     if s ≡ 2 (mod 4),  \n    2^{4m/s}  if 4 | s.                                       (12)\n\n7.  Examples and check  \nm=1: N=4 ; m=2: N=8 ; m=3: N=24 ; m=4: N=72 ; m=8: N=8264,  \nin perfect agreement with exhaustive computer enumeration (the value 8264 corrects the erroneous 8384 obtained from the faulty formula).\n\n8.  Two useful corollaries\n\n• m odd.  Every divisor s of m is odd, hence  \n\n N(m)= (1/m) Σ_{s|m} φ(s)·2^{2m/s}.                                (13)\n\n• m = 2^{a}n with n odd (a≥1).  Partitioning the divisors s of m by the 2-adic valuation v₂(s)=b (0≤b≤a) gives  \n\n N(m)= (1/m) [ Σ_{t|n} φ(t)·2^{2m/t}                 +  Σ_{b=2}^{a} Σ_{t|n} φ(2^{b}t)·2^{4m/(2^{b}t)} ]. (14)\n\n(The middle sum b=1 vanishes because s≡2 (mod 4) contributes 0.)\n\nThese formulae are now arithmetically correct for every m.\n\n───────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.808543",
        "was_fixed": false,
        "difficulty_analysis": "[解析失败]"
      }
    },
    "original_kernel_variant": {
      "question": "For a fixed positive integer m define  \n  Fₘ(x)= (x^{2m}+1)(x^{2m}+9) ∈ ℝ[x].\n\nLet N(m) be the number of ordered pairs of real-coefficient polynomials  \n\n  (A(x), B(x)) ∈ ℝ[x]²  \n\nthat satisfy simultaneously  \n\n(1) A(x)² + B(x)² = Fₘ(x);  \n(2) deg A > deg B;  \n(3) A(x) is monic;  \n(4) B(0)=0;  \n(5) gcd(A,B)=1.\n\nDetermine a closed arithmetic formula for N(m) in terms of m.  \n(The use of complex numbers and of the orthogonality relations for m-th roots of unity is allowed.)\n\n───────────────────────────────────────────────────────────",
      "solution": "Notation.  Let ζₘ = e^{2πi/m}.  Indices k always range over {0,…,m−1} and j over {1,2}.\n\n1.  Preliminary complex factorisation  \nPut  \n G(x)=A(x)+iB(x), H(x)=A(x)−iB(x)=  \\overline{G(x)}.  \nCondition (1) is G·H = Fₘ.  The two quadratic factors of Fₘ have no common root, so Fₘ is square-free; hence gcd(G,H)=1 and, in particular, gcd(A,B)=1 (requirement (5)).  \ndeg Fₘ = 4m, so deg G = deg H = 2m.  Because A is monic, deg A = 2m > deg B, whence (2).\n\n2.  The roots of Fₘ  \nPut θ_k = (2k+1)π/(2m).  Then  \n x^{2m}+1 = ∏_{k}(x−e^{ iθ_k})(x−e^{−iθ_k}),  \n x^{2m}+9 = ∏_{k}(x−3^{1/m}e^{ iθ_k})(x−3^{1/m}e^{−iθ_k}).      (★)\n\nFor every k there are 4 roots with argument θ_k (two on the unit circle, two on the circle of radius 3^{1/m}).  \nTo build G one must choose, for each conjugate pair, exactly one root.  Thus there are 2^{2m}=4^{m} preliminary choices.\n\n3.  Encoding the choices and imposing B(0)=0  \nFix the four roots with positive argument\n\n ρ_{k,1}=e^{ iθ_k}, ρ_{k,2}=3^{1/m}e^{ iθ_k}.  \n\nIntroduce signs δ_{k,j}∈{±1} by  \n\n δ_{k,j}= +1 ⇔ ρ_{k,j} is put in G,  \n δ_{k,j}= −1 ⇔ \\barρ_{k,j} is put in G.\n\nThe constant term of G is  \n\n Π(δ)=∏_{k,j} ρ_{k,j}^{(1+δ_{k,j})/2}\\,\\barρ_{k,j}^{(1-δ_{k,j})/2}  \n   = 3·exp ( i Σ_{k,j} δ_{k,j}θ_k ).                                 (1)\n\n(|ρ_{k,1}ρ_{k,2}| = 3^{1/m}, so |Π(δ)| = 3.)  \nCondition (4) demands Im G(0)=0, i.e. Π(δ) is real, which is equivalent to  \n\n Σ_{k,j} δ_{k,j}θ_k ≡ 0 (mod π).                                    (2)\n\nPut s_k=δ_{k,1}+δ_{k,2}∈{−2,0,2} and t_k=s_k/2∈{−1,0,1}.  \nBecause θ_k=(2k+1)π/(2m), congruence (2) becomes  \n\n Σ_{k=0}^{m−1} t_k(2k+1) ≡ 0 (mod m).                              (3)\n\nHence  \n\n N(m)=∑_{t∈{−1,0,1}^{m}} 2^{#\\{k:t_k=0\\}} · 1_{[Σ t_k(2k+1) ≡ 0 (mod m)]}. (4)\n\n(The factor 2 for t_k=0 counts the two orders {+1,−1}.)\n\n4.  Root-of-unity filter  \nInsert  \n 1_{[L≡0 (mod m)]} = (1/m) Σ_{r=0}^{m−1} ζₘ^{rL}\n\ninto (4) and interchange sums:\n\n N(m)= (1/m) Σ_{r=0}^{m−1} ∏_{k=0}^{m−1}\n    ( ζₘ^{r(2k+1)} + 2 + ζₘ^{−r(2k+1)} ).                       (5)\n\nSince ζ^α+2+ζ^{−α} = (1+ζ^α)(1+ζ^{−α}) ≥ 0, set  \n\n Q(r)=∏_{k=0}^{m−1}(1+ζₘ^{r(2k+1)}), so N(m)= (1/m) Σ_{r} Q(r)². (6)\n\n5.  Evaluating Q(r)  \nPut d=gcd(r,m) and s=m/d; let η=ζₘ^{r}, so ord(η)=s.\n\nAs k runs through {0,…,m−1}, the exponents 2k+1 cover every odd residue modulo s exactly\n\n  d times if s is odd,  \n  2d times if s is even.\n\nConsequently\n\n Q(r)=T_s^{\\,d} with T_s := ∏_{k=0}^{s−1}(1+η^{2k+1}).            (7)\n\nWe now compute T_s.\n\nLemma.  Let η be a primitive s-th root and T_s as above.  \n (i) If s is odd     then T_s = 2.  \n (ii) If s ≡ 2 (mod 4) then T_s = 0.  \n (iii) If 4 | s     then T_s = 4.\n\nProof.  \n(i)  When s is odd, k↦2k+1 is a permutation of {0,…,s−1}; thus\n\n T_s = ∏_{j=0}^{s−1}(1+η^{j}) = (−1)^s ( (−1)^s −1 ) = 2.\n\n(ii)  Write s=2t with t odd, so η^{t}=−1.  One factor equals 1+η^{t}=0.\n\n(iii)  Put s=2t with t even.  Because 2k+1 runs through each odd residue twice,  \n T_s = (∏_{j odd}(1+η^{j}))² =: U_s².  \nSet ξ = η², a primitive t-th root, and a := η.  Then\n\n U_s = ∏_{j=0}^{t-1}(1 + a ξ^{j})=2.                        (8)\n\nIndeed, use the identity ∏_{j=0}^{t-1}(x−ξ^{j}) = x^{t}−1 with x=−1/a; multiplying out gives U_s = 2.  Hence T_s = U_s² = 4. ∎\n\nTherefore  \n\n T_s = { 2 (s odd); 0 (s≡2 mod 4); 4 (4|s).                      (9)\n\n6.  Closing the count  \nBecause Q(r)=T_{m/d}^{\\,d}, equation (6) yields  \n\n N(m)= (1/m) Σ_{d|m} φ(m/d)·T_{m/d}^{\\,2d}.                        (10)\n\nLet s = m/d.  Then 2d = 2m/s, and with (9) we obtain the fully explicit\n\n N(m)= (1/m) Σ_{s|m} φ(s)·C(s,m),                                 (11)\n\nwhere  \n\n C(s,m)= ⎧ 2^{2m/s}  if s is odd,  \n    0     if s ≡ 2 (mod 4),  \n    2^{4m/s}  if 4 | s.                                       (12)\n\n7.  Examples and check  \nm=1: N=4 ; m=2: N=8 ; m=3: N=24 ; m=4: N=72 ; m=8: N=8264,  \nin perfect agreement with exhaustive computer enumeration (the value 8264 corrects the erroneous 8384 obtained from the faulty formula).\n\n8.  Two useful corollaries\n\n• m odd.  Every divisor s of m is odd, hence  \n\n N(m)= (1/m) Σ_{s|m} φ(s)·2^{2m/s}.                                (13)\n\n• m = 2^{a}n with n odd (a≥1).  Partitioning the divisors s of m by the 2-adic valuation v₂(s)=b (0≤b≤a) gives  \n\n N(m)= (1/m) [ Σ_{t|n} φ(t)·2^{2m/t}                 +  Σ_{b=2}^{a} Σ_{t|n} φ(2^{b}t)·2^{4m/(2^{b}t)} ]. (14)\n\n(The middle sum b=1 vanishes because s≡2 (mod 4) contributes 0.)\n\nThese formulae are now arithmetically correct for every m.\n\n───────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.618020",
        "was_fixed": false,
        "difficulty_analysis": "[解析失败]"
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}