{
  "index": "1968-A-2",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "A-2. Given integers \\( a, b, e, c, d \\), and \\( f \\) with \\( a d \\neq b c \\), and given a real number \\( \\epsilon>0 \\), show that there exist rational numbers \\( r \\) and \\( s \\) for which\n\\[\n\\begin{array}{l}\n0<|r a+s b-e|<\\epsilon \\\\\n0<|r c+s d-f|<\\epsilon .\n\\end{array}\n\\]",
  "solution": "A-2 The easy solution is obtained by selecting a rational number \\( \\rho \\) with \\( 0<\\rho<\\epsilon \\) and solving the linear system\n\\[\n\\begin{array}{l}\na r+b s=e+\\rho \\\\\nc r+d s=f+\\rho\n\\end{array}\n\\]\n\nThe solution for \\( r \\) and \\( s \\) exist, since \\( a d \\neq b c \\), and are rational numbers which satisfy the given inequalities.",
  "vars": [
    "r",
    "s",
    "\\\\rho"
  ],
  "params": [
    "a",
    "b",
    "c",
    "d",
    "e",
    "f",
    "\\\\epsilon"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "r": "rationalr",
        "s": "rationals",
        "\\rho": "auxrho",
        "a": "coeffa",
        "b": "coeffb",
        "c": "coeffc",
        "d": "coeffd",
        "e": "coeffe",
        "f": "coefff",
        "\\epsilon": "tolerance"
      },
      "question": "A-2. Given integers \\( coeffa, coeffb, coeffe, coeffc, coeffd \\), and \\( coefff \\) with \\( coeffa coeffd \\neq coeffb coeffc \\), and given a real number \\( tolerance>0 \\), show that there exist rational numbers \\( rationalr \\) and \\( rationals \\) for which\n\\[\n\\begin{array}{l}\n0<| rationalr coeffa + rationals coeffb - coeffe |< tolerance \\\\\n0<| rationalr coeffc + rationals coeffd - coefff |< tolerance .\n\\end{array}\n\\]",
      "solution": "A-2 The easy solution is obtained by selecting a rational number \\( auxrho \\) with \\( 0<auxrho< tolerance \\) and solving the linear system\n\\[\n\\begin{array}{l}\ncoeffa rationalr+coeffb rationals=coeffe+auxrho \\\\\ncoeffc rationalr+coeffd rationals=coefff+auxrho\n\\end{array}\n\\]\n\nThe solution for \\( rationalr \\) and \\( rationals \\) exist, since \\( coeffa coeffd \\neq coeffb coeffc \\), and are rational numbers which satisfy the given inequalities."
    },
    "descriptive_long_confusing": {
      "map": {
        "r": "squirrel",
        "s": "blueprint",
        "\\rho": "gemstone",
        "a": "rectangle",
        "b": "envelope",
        "c": "lanterns",
        "d": "sunflower",
        "e": "horizons",
        "f": "avalanche",
        "\\epsilon": "seashore"
      },
      "question": "Given integers \\( rectangle, envelope, horizons, lanterns, sunflower \\), and \\( avalanche \\) with \\( rectangle sunflower \\neq envelope lanterns \\), and given a real number \\( seashore>0 \\), show that there exist rational numbers \\( squirrel \\) and \\( blueprint \\) for which\n\\[\n\\begin{array}{l}\n0<|squirrel rectangle+ blueprint envelope- horizons|<seashore \\\\\n0<|squirrel lanterns+ blueprint sunflower- avalanche|<seashore .\n\\end{array}\n\\]",
      "solution": "A-2 The easy solution is obtained by selecting a rational number \\( gemstone \\) with \\(0<gemstone<seashore\\) and solving the linear system\n\\[\n\\begin{array}{l}\nrectangle squirrel+envelope blueprint=horizons+gemstone \\\\\nlanterns squirrel+sunflower blueprint=avalanche+gemstone\n\\end{array}\n\\]\n\nThe solution for \\( squirrel \\) and \\( blueprint \\) exist, since \\( rectangle sunflower \\neq envelope lanterns \\), and are rational numbers which satisfy the given inequalities."
    },
    "descriptive_long_misleading": {
      "map": {
        "r": "irrationalnumber",
        "s": "transcendentalvalue",
        "\\rho": "emptinessconst",
        "a": "effectvalue",
        "b": "terminalvalue",
        "c": "outcomevalue",
        "d": "dependentvalue",
        "e": "irrationalterm",
        "f": "unfixedterm",
        "\\epsilon": "gigantictolerance"
      },
      "question": "A-2. Given integers \\( effectvalue, terminalvalue, irrationalterm, outcomevalue, dependentvalue \\), and \\( unfixedterm \\) with \\( effectvalue dependentvalue \\neq terminalvalue outcomevalue \\), and given a real number \\( gigantictolerance>0 \\), show that there exist rational numbers \\( irrationalnumber \\) and \\( transcendentalvalue \\) for which\n\\[\n\\begin{array}{l}\n0<|irrationalnumber effectvalue+transcendentalvalue terminalvalue-irrationalterm|<gigantictolerance \\\\\n0<|irrationalnumber outcomevalue+transcendentalvalue dependentvalue-unfixedterm|<gigantictolerance .\n\\end{array}\n\\]\n",
      "solution": "A-2 The easy solution is obtained by selecting a rational number \\( emptinessconst \\) with \\( 0<emptinessconst<gigantictolerance \\) and solving the linear system\n\\[\n\\begin{array}{l}\n effectvalue irrationalnumber+terminalvalue transcendentalvalue=irrationalterm+emptinessconst \\\\\n outcomevalue irrationalnumber+dependentvalue transcendentalvalue=unfixedterm+emptinessconst\n\\end{array}\n\\]\n\nThe solution for \\( irrationalnumber \\) and \\( transcendentalvalue \\) exist, since \\( effectvalue dependentvalue \\neq terminalvalue outcomevalue \\), and are rational numbers which satisfy the given inequalities.\n"
    },
    "garbled_string": {
      "map": {
        "r": "lkjhgfdsa",
        "s": "asdfghjkl",
        "\\rho": "zxcvbnmas",
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "mnbvtrew",
        "d": "plokijuh",
        "e": "qazwsxed",
        "f": "wsxcvbnm",
        "\\epsilon": "poiulkjh"
      },
      "question": "A-2. Given integers \\( qzxwvtnp, hjgrksla, qazwsxed, mnbvtrew, plokijuh \\), and \\( wsxcvbnm \\) with \\( qzxwvtnp plokijuh \\neq hjgrksla mnbvtrew \\), and given a real number \\( poiulkjh>0 \\), show that there exist rational numbers \\( lkjhgfdsa \\) and \\( asdfghjkl \\) for which\n\\[\n\\begin{array}{l}\n0<|lkjhgfdsa qzxwvtnp+asdfghjkl hjgrksla-qazwsxed|<poiulkjh \\\\\n0<|lkjhgfdsa mnbvtrew+asdfghjkl plokijuh-wsxcvbnm|<poiulkjh .\n\\end{array}\n\\]",
      "solution": "A-2 The easy solution is obtained by selecting a rational number \\( zxcvbnmas \\) with \\( 0<zxcvbnmas<poiulkjh \\) and solving the linear system\n\\[\n\\begin{array}{l}\nqzxwvtnp lkjhgfdsa+hjgrksla asdfghjkl=qazwsxed+zxcvbnmas \\\\\nmnbvtrew lkjhgfdsa+plokijuh asdfghjkl=wsxcvbnm+zxcvbnmas\n\\end{array}\n\\]\n\nThe solution for \\( lkjhgfdsa \\) and \\( asdfghjkl \\) exist, since \\( qzxwvtnp plokijuh \\neq hjgrksla mnbvtrew \\), and are rational numbers which satisfy the given inequalities."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\na,b,c,d,e,f\\in\\mathbb Z ,\\qquad   \n\\Delta:=ad-bc\\neq 0 .\n\\]\n\nLet $m>1$ be a square–free \\emph{integer} and put  \n\\[\nK:=\\mathbb Q(\\sqrt m), \\qquad {\\rm Gal}(K/\\mathbb Q)=\\{1,\\ \\iota\\},\n\\]\nso that $K$ is a totally real quadratic field.  \n\nFix a non–empty finite set of rational primes  \n\\[\nS=\\{p_{1},\\dots ,p_{t}\\}\\subset\\mathbb P\n\\]\nsubject to  \n\n(i) $p\\nmid 2m\\Delta\\quad(\\forall\\,p\\in S)$,  \n\n(ii) $p$ is \\emph{inert} in $K/\\mathbb Q$, i.e.  \n\\[\n\\bigl(\\tfrac{m}{p}\\bigr)=-1\\qquad(\\forall\\,p\\in S),\n\\]\nso that each $p\\in S$ gives rise to a unique prime ideal\n\\[\n\\mathfrak p_{p}:=p\\,\\mathcal O_{K}\\subset\\mathcal O_{K}.\n\\]\n\nFor every $p\\in S$ fix a positive integer $k_{p}$, and fix one positive real number  \n\\[\n\\varepsilon>0\\qquad(\\text{the same }\\varepsilon\\text{ for all }p\\in S).\n\\]\n\n(Local non–degeneracy hypothesis)  \n\\[\n(\\star)\\qquad  \np\\nmid(de-bf)\\ \\text{ and }\\ p\\nmid(af-ce)\\qquad(\\forall\\,p\\in S).\n\\]\n\nProve that there exist $r,s\\in K$ such that  \n\n(A) Archimedean control  \n\\[\n0<\\lvert\\sigma(a)\\sigma(r)+\\sigma(b)\\sigma(s)-e\\rvert\n=\\lvert\\sigma(c)\\sigma(r)+\\sigma(d)\\sigma(s)-f\\rvert<\\varepsilon\n\\quad(\\forall\\,\\sigma:K\\hookrightarrow\\mathbb R);\n\\]\n\n(B) Prescribed $p$–adic size  \n\\[\nv_{p}\\!\\bigl(N_{K/\\mathbb Q}(ar+bs-e)\\bigr)=\nv_{p}\\!\\bigl(N_{K/\\mathbb Q}(cr+ds-f)\\bigr)=2k_{p}\n\\qquad(\\forall\\,p\\in S);\n\\]\n\n(C) Local coprimality  \n\\[\nv_{\\mathfrak p_{p}}(r)=v_{\\mathfrak p_{p}}(s)=0\n\\qquad(\\forall\\,p\\in S).\n\\]\n(Thus $r$ and $s$ are $\\mathfrak p_{p}$–units for every $p\\in S$.)\n\nShow, moreover, that $(\\star)$ is \\emph{necessary}: if $(\\star)$ fails for some $p\\in S$, then no pair $(r,s)\\in K^{2}$ can satisfy simultaneously (B) and (C).\n\n--------------------------------------------------------------------",
      "solution": "Throughout write  \n\\[\nM:=\\prod_{p\\in S}p^{k_{p}},\\qquad  \n\\lambda=\\frac{u}{v}\\in\\mathbb Q,\\qquad \\beta:=\\Delta\\lambda .\n\\]\n\n\\textbf{Step 1.  A rational parameter with prescribed valuations.}  \nChoose coprime integers $u,v$ with  \n\\[\n\\begin{aligned}\n&u=M\\,w,\\quad w\\in\\mathbb Z,\\quad w\\not\\equiv 0\\bmod p\\;(\\forall\\,p\\in S),\\\\\n&v\\not\\equiv 0\\bmod p\\;(\\forall\\,p\\mid 2m\\Delta M),\\\\\n&0<\\lvert\\lambda\\rvert<\n\\frac{\\varepsilon}{\\bigl(\\lvert d-b\\rvert+\\lvert a-c\\rvert+1\\bigr)\\lvert\\Delta\\rvert}.\n\\end{aligned}\n\\]\nBecause $\\mathbb Q$ is dense in $\\mathbb R$ we can choose $w,v$ to satisfy the\nlast inequality.  Then\n\\[\nv_{p}(\\beta)=k_{p}\\quad(\\forall\\,p\\in S),\\qquad\n0<\\lvert\\beta\\rvert<\\varepsilon .\n\\]\n\n\\textbf{Step 2.  A perturbed linear system.}  \nImpose the \\emph{same} rational error $\\beta$ on both linear forms:\n\\[\n\\begin{cases}\nar+bs=e+\\beta,\\\\[2pt]\ncr+ds=f+\\beta .\n\\end{cases}\\tag{1}\n\\]\nSince $\\Delta\\neq 0$, (1) has the unique solution\n\\[\n\\begin{aligned}\nr&=\\dfrac{d(e+\\beta)-b(f+\\beta)}{\\Delta}=C+\\alpha\\beta,\\\\\ns&=\\dfrac{-c(e+\\beta)+a(f+\\beta)}{\\Delta}=D+\\gamma\\beta,\n\\end{aligned}\n\\]\nwhere\n\\[\nC=\\dfrac{de-bf}{\\Delta},\\quad D=\\dfrac{af-ce}{\\Delta},\\quad\n\\alpha=\\dfrac{d-b}{\\Delta},\\quad \\gamma=\\dfrac{a-c}{\\Delta}.\n\\]\nThus $r,s\\in\\mathbb Q\\subset K$.\n\n\\textbf{Step 3.  Local coprimality (C).}  \nFix $p\\in S$ and write $\\mathfrak p:=\\mathfrak p_{p}$.  \nBecause $p\\nmid\\Delta$, the denominators of $C,D$ are $\\mathfrak p$–units.\nHypothesis $(\\star)$ gives $v_{\\mathfrak p}(C)=v_{\\mathfrak p}(D)=0$.  \nMoreover $v_{\\mathfrak p}(\\beta)=k_{p}\\ge 1$, so\n\\[\nv_{\\mathfrak p}(\\alpha\\beta)\\ge 1,\\qquad v_{\\mathfrak p}(\\gamma\\beta)\\ge 1.\n\\]\nConsequently\n\\[\nv_{\\mathfrak p}(r)=v_{\\mathfrak p}(C+\\alpha\\beta),\\qquad\nv_{\\mathfrak p}(s)=v_{\\mathfrak p}(D+\\gamma\\beta).\n\\]\nExact cancellation $C\\equiv-\\alpha\\beta\\bmod\\mathfrak p$ is impossible because\n$C$ is a $\\mathfrak p$–unit whereas $\\alpha\\beta$ is divisible by $\\mathfrak p$.\n(The same argument applies to $D+\\gamma\\beta$.)  Hence\n\\[\nv_{\\mathfrak p}(r)=v_{\\mathfrak p}(s)=0,\n\\]\nproving (C).\n\n\\textbf{Step 4.  Archimedean control (A).}  \nFor every real embedding $\\sigma:K\\hookrightarrow\\mathbb R$, applying $\\sigma$\nto (1) yields\n\\[\n\\sigma(a)\\sigma(r)+\\sigma(b)\\sigma(s)-e\n=\\sigma(c)\\sigma(r)+\\sigma(d)\\sigma(s)-f=\\beta ,\n\\]\nwhence $0<\\lvert\\beta\\rvert<\\varepsilon$, establishing (A).\n\n\\textbf{Step 5.  Prescribed $p$–adic size (B).}  \nBecause both error terms equal the \\emph{rational} number $\\beta$,\n\\[\nN_{K/\\mathbb Q}(ar+bs-e)=N_{K/\\mathbb Q}(cr+ds-f)=\\beta^{2},\n\\]\nso for every $p\\in S$\n\\[\nv_{p}\\!\\bigl(N_{K/\\mathbb Q}(ar+bs-e)\\bigr)=\nv_{p}(\\beta^{2})=2k_{p},\n\\]\nand likewise for the second norm.  Thus (B) holds.\n\n\\textbf{Step 6.  Necessity of $(\\star)$.}  \n\nAssume, for contradiction, that $p\\in S$ divides $de-bf$ and that a pair\n$(r,s)\\in K^{2}$ satisfies (B) and (C).  Put\n\\[\nx:=ar+bs-e,\\qquad y:=cr+ds-f .\n\\]\n\n\\emph{First degeneracy $p\\mid(de-bf)$.}  \nCompute in $K$:\n\\[\nd\\,x-b\\,y\n=d(ar+bs-e)-b(cr+ds-f)=\\Delta r-(de-bf).   \\tag{2}\n\\]\nBecause $p\\nmid\\Delta$ and $v_{\\mathfrak p}(r)=0$ by (C),\nthe term $\\Delta r$ is a $\\mathfrak p$–unit, while $v_{\\mathfrak p}}(de-bf)\\ge 1$.\nHence\n\\[\nv_{\\mathfrak p}(d\\,x-b\\,y)=0. \\tag{3}\n\\]\nThe coefficients $b,d$ are $\\mathfrak p$–units, so (3) forces\n\\[\n\\min\\bigl\\{v_{\\mathfrak p}(x),\\,v_{\\mathfrak p}(y)\\bigr\\}=0 .\\tag{4}\n\\]\nYet (B) and the inertness of $p$ in $K$ imply\n\\[\nv_{\\mathfrak p}(x)=v_{\\mathfrak p}\\bigl((x)\\bigr)\n=\\frac12\\,v_{p}\\!\\bigl(N_{K/\\mathbb Q}(x)\\bigr)=k_{p}\\ge 1,\n\\]\nand similarly $v_{\\mathfrak p}(y)=k_{p}\\ge 1$, contradicting (4).\nTherefore $p\\nmid(de-bf)$.\n\n\\emph{Second degeneracy $p\\mid(af-ce)$.}  \nAssume instead $p\\mid(af-ce)$.  A symmetric computation gives\n\\[\na\\,y-c\\,x\n=a(cr+ds-f)-c(ar+bs-e)=\\Delta s-(af-ce).   \\tag{5}\n\\]\nAgain $\\Delta s$ is a $\\mathfrak p$–unit by $v_{\\mathfrak p}}(s)=0$, so\n$v_{\\mathfrak p}}(a\\,y-c\\,x)=0$ and, because $a,c$ are $\\mathfrak p$–units,\n\\[\n\\min\\bigl\\{v_{\\mathfrak p}(x),\\,v_{\\mathfrak p}(y)\\bigr\\}=0 ,\n\\]\ncontradicting $v_{\\mathfrak p}(x)=v_{\\mathfrak p}(y)=k_{p}\\ge 1$.\nThus $p\\nmid(af-ce)$.\n\nHence $(\\star)$ is necessary.\n\n\\hfill$\\square$\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.575311",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher mathematical setting  \n • The unknowns r and s now live in the ring of integers of a quadratic number field, not merely in ℚ or in the quadratic field itself.  \n • Both Archimedean (real) and non-Archimedean (p-adic) valuations are forced to obey simultaneous, highly specific constraints.\n\n2. Multiple interacting concepts  \n • Algebraic number theory (rings of integers, embeddings, norms, prime ideals).  \n • p-adic valuation control for several primes at prescribed exact exponents.  \n • Simultaneous real approximation with equality of absolute values.  \n • Copimality of ideals, demanding an additional arithmetic restriction on (r) and (s).\n\n3. Technical obstacles  \n • Ensuring integrality of r and s despite the presence of δ forced by the p-adic requirements.  \n • Keeping δ small in the real metric while guaranteeing the exact p-adic divisibilities.  \n • Maintaining coprimality with all primes above S, which requires delicate denominator management.\n\n4. Depth of the solution  \n The proof needs careful selection of δ using Chinese-remainder–type reasoning across valuations, explicit manipulation inside O_K, and an understanding of how valuations behave under field norms—tools absent from the original elementary linear-algebraic argument.\n\nHence the enhanced variant is substantially harder than both the original problem and the current kernel variant, combining advanced number-theoretic structures with simultaneous metric and arithmetic constraints."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\n• a, b, c, d, e, f ∈ ℤ with  \n  Δ := ad − bc ≠ 0,  \n\n• m > 1 square-free with gcd(m, 2Δ) = 1 and K := ℚ(√m),  \n\nand let S = {p₁,…,p_t} be a non–empty finite set of rational primes satisfying  \n p ∤ 2mΔ for every p ∈ S.  \nWith every p ∈ S fix a positive integer k_p and a real number ε > 0.\n\n(⋆)  (Local non–degeneracy hypothesis)  \n For every p ∈ S we have simultaneously  \n  p ∤ (d e − b f) and p ∤ (a f − c e).\n\nProve that there exist elements r, s ∈ K such that  \n\n(A) (archimedean condition) For every real embedding σ : K ↪ ℝ  \n  0 < |σ(a)σ(r)+σ(b)σ(s) − e|  \n   = |σ(c)σ(r)+σ(d)σ(s) − f| < ε;  \n\n(B) (p–adic condition) For every p ∈ S  \n  v_p( N_{K/ℚ}(a r + b s − e) )  \n   = v_p( N_{K/ℚ}(c r + d s − f) ) = 2k_p;  \n\n(C) (local coprimality) For every prime ideal 𝔭 of 𝔒_K lying above a prime p ∈ S we have  \n  v_𝔭(r) = v_𝔭(s) = 0.  \n(Thus r and s are 𝔭-units for every 𝔭 | p with p ∈ S.)\n\nShow moreover that hypothesis (⋆) is necessary: in its absence conditions (B) and (C) can not be fulfilled simultaneously.\n\n——————————————————————————————————————",
      "solution": "Throughout write  \n M := ∏_{p∈S} p^{k_p}.    (1)\n\nStep 1.  A rational parameter with prescribed p-adic valuations.  \nChoose a rational number  \n λ = u/v with gcd(u,v)=1  \nthat satisfies  \n\n(i) u is divisible by M but by no higher power of any p ∈ S (so v_p(λ)=k_p);  \n\n(ii) v is coprime to 2mΔM;  \n\n(iii) 0 < |λ| < ε / ( (|d−b| + |a−c| + 1)|Δ| ).  (2)\n\nSuch a λ exists because the admissible numerators form an arithmetic progression and ℚ is dense in ℝ.\n\nPut  \n β := Δ λ ∈ ℚ ⊂ K.        (3)\n\nThen  \n\n v_p(β)=k_p for every p∈S,    (4a)  \n 0<|β|<ε.            (4b)\n\nStep 2.  A linear system with common “error’’ β.  \nConsider  \n\n a r + b s = e + β,      (5a)  \n c r + d s = f + β.      (5b)\n\nBecause Δ≠0 this system has the unique solution  \n\n r = [d(e+β) − b(f+β)]/Δ = C + αβ,  (6a)  \n s = [−c(e+β)+a(f+β)]/Δ = D + γβ,  (6b)\n\nwhere  \n\n C := (d e − b f)/Δ,   α := (d−b)/Δ,  \n D := (a f − c e)/Δ,   γ := (a−c)/Δ.         (7)\n\nAll five constants C, D, α, γ, β are rational, hence r,s∈ℚ⊂K.\n\nStep 3.  Local coprimality at the primes in S.  \nFix p ∈ S and a prime ideal 𝔭 | p of 𝔒_K.\n\n• Because p ∤ Δ, the denominators of C and D are 𝔭-units.  \n• Hypothesis (⋆) gives v_𝔭(C)=v_𝔭(D)=0.  \n• By (4a) we have v_𝔭(β)=k_p ≥1, hence v_𝔭(αβ) ≥1 and v_𝔭(γβ) ≥1.\n\nTherefore  \n\n v_𝔭(r)=v_𝔭(C + αβ)=0, v_𝔭(s)=v_𝔭(D + γβ)=0. (8)\n\nThus condition (C) holds.\n\nStep 4.  The archimedean requirement.  \nFor every real embedding σ of K we have, using (5a),(5b),\n\n σ(a)σ(r)+σ(b)σ(s)−e = σ(c)σ(r)+σ(d)σ(s)−f = β. (9)\n\nInequality (4b) gives 0<|β|<ε, establishing (A).\n\nStep 5.  The p-adic sizes of the error terms.  \nEquation (9) shows that the two error expressions equal the same rational number β, so\n\n N_{K/ℚ}(a r + b s − e) = N_{K/ℚ}(c r + d s − f) = β². (10)\n\nBecause β is rational, N_{K/ℚ}(β)=β²; hence for every p ∈ S\n\n v_p( N_{K/ℚ}(a r + b s − e) )  \n  = v_p(β²) = 2 v_p(β) = 2k_p, (11)\n\nand likewise for the second norm, yielding (B).\n\nStep 6.  Necessity of hypothesis (⋆).  \nSuppose p ∈ S divides, say, d e − b f.  \nThen v_𝔭(C) > 0 for every 𝔭 | p.  \nSince v_𝔭(β)=k_p ≥ 1, formula (6a) shows v_𝔭(r) > 0, contradicting (C).  \nThe same argument with (6b) shows that p ∤ a f − c e is also necessary. ∎\n\n——————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.466434",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher mathematical setting  \n • The unknowns r and s now live in the ring of integers of a quadratic number field, not merely in ℚ or in the quadratic field itself.  \n • Both Archimedean (real) and non-Archimedean (p-adic) valuations are forced to obey simultaneous, highly specific constraints.\n\n2. Multiple interacting concepts  \n • Algebraic number theory (rings of integers, embeddings, norms, prime ideals).  \n • p-adic valuation control for several primes at prescribed exact exponents.  \n • Simultaneous real approximation with equality of absolute values.  \n • Copimality of ideals, demanding an additional arithmetic restriction on (r) and (s).\n\n3. Technical obstacles  \n • Ensuring integrality of r and s despite the presence of δ forced by the p-adic requirements.  \n • Keeping δ small in the real metric while guaranteeing the exact p-adic divisibilities.  \n • Maintaining coprimality with all primes above S, which requires delicate denominator management.\n\n4. Depth of the solution  \n The proof needs careful selection of δ using Chinese-remainder–type reasoning across valuations, explicit manipulation inside O_K, and an understanding of how valuations behave under field norms—tools absent from the original elementary linear-algebraic argument.\n\nHence the enhanced variant is substantially harder than both the original problem and the current kernel variant, combining advanced number-theoretic structures with simultaneous metric and arithmetic constraints."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}