{
  "schema": "rigor-public-proof-review-casebook-v1",
  "dataset_version": "2026-10-29-v1",
  "status": "public, project-authored, not independently adjudicated",
  "intended_use": "mathematical practice and unsolicited expert corrections only, NOT blinded evaluation",
  "disclosure": "The public main Arena still exposes author gold labels, which may leak. This separate file omits labels and original titles, but is NOT a blind study.",
  "source_family_count": 19,
  "fragment_count": 111,
  "items": [
    {
      "case_id": "11-A",
      "source_problem": "IMO 2011 / P3",
      "official_source": "https://www.imo-official.org/problems/2011/",
      "theorem_context": "A function f: ℝ → ℝ satisfies f(x+y) ≤ y·f(x)+f(f(x)) for every real x,y. Show that f(x)=0 for x≤0.",
      "topic": "Functional inequalities",
      "proof_steps": [
        "Assume f(x+y) ≤ y f(x)+f(f(x)) for all real x,y.",
        "Set a=f(0) and b=f(a).",
        "With x=0 and y=t, infer f(t) ≤ ta+b.",
        "Thus the universal affine upper bound f(t) ≤ at+b holds."
      ],
      "training_only": true
    },
    {
      "case_id": "11-B",
      "source_problem": "IMO 2011 / P3",
      "official_source": "https://www.imo-official.org/problems/2011/",
      "theorem_context": "A function f: ℝ → ℝ satisfies f(x+y) ≤ y·f(x)+f(f(x)) for every real x,y. Show that f(x)=0 for x≤0.",
      "topic": "Functional inequalities",
      "proof_steps": [
        "From the preceding lemma, f(u) ≤ au+b for every u∈ℝ.",
        "Using y=−a in the original inequality gives f(x−a) ≤ −a f(x)+f(f(x)).",
        "Apply the envelope at u=f(x): f(f(x)) ≤ a f(x)+b.",
        "Hence f(x−a) ≤ b, and therefore f(t) ≤ b for every real t."
      ],
      "training_only": true
    },
    {
      "case_id": "11-C",
      "source_problem": "IMO 2011 / P3",
      "official_source": "https://www.imo-official.org/problems/2011/",
      "theorem_context": "A function f: ℝ → ℝ satisfies f(x+y) ≤ y·f(x)+f(f(x)) for every real x,y. Show that f(x)=0 for x≤0.",
      "topic": "Functional inequalities",
      "proof_steps": [
        "Suppose the global upper bound f(t) ≤ b has been established.",
        "By definition b=f(a), where a=f(0).",
        "Every function with a finite upper bound has b ≥ 0.",
        "Therefore b is nonnegative without any other argument."
      ],
      "training_only": true
    },
    {
      "case_id": "11-D",
      "source_problem": "IMO 2011 / P3",
      "official_source": "https://www.imo-official.org/problems/2011/",
      "theorem_context": "A function f: ℝ → ℝ satisfies f(x+y) ≤ y·f(x)+f(f(x)) for every real x,y. Show that f(x)=0 for x≤0.",
      "topic": "Functional inequalities",
      "proof_steps": [
        "Assume f(t) ≤ b for all real t, where a=f(0) and b=f(a).",
        "Set x=a and y=0 in the original inequality: f(a) ≤ f(f(a)).",
        "Since f(a)=b, obtain b ≤ f(b).",
        "The global bound also gives f(b) ≤ b, hence f(b)=b."
      ],
      "training_only": true
    },
    {
      "case_id": "11-E",
      "source_problem": "IMO 2011 / P3",
      "official_source": "https://www.imo-official.org/problems/2011/",
      "theorem_context": "A function f: ℝ → ℝ satisfies f(x+y) ≤ y·f(x)+f(f(x)) for every real x,y. Show that f(x)=0 for x≤0.",
      "topic": "Functional inequalities",
      "proof_steps": [
        "Assume we have already established f(0)=0 and f(t)≤0 for every real t.",
        "Fix x<0. The original inequality at y=−x gives 0≤(−x)f(x)+f(f(x)).",
        "Because f(f(x))≤0, it follows that (−x)f(x)≥0.",
        "As −x>0, dividing shows f(x)<0."
      ],
      "training_only": true
    },
    {
      "case_id": "17-A",
      "source_problem": "IMO 2017 / P2",
      "official_source": "https://www.imo-official.org/problems/2017/",
      "theorem_context": "Determine f: ℝ→ℝ satisfying f(f(x)f(y))+f(x+y)=f(xy) for every x,y∈ℝ.",
      "topic": "Functional equations",
      "proof_steps": [
        "Suppose f(f(x)f(y))+f(x+y)=f(xy) for all x,y.",
        "Set c=f(0).",
        "Taking y=0 gives f(c f(x))+f(x)=c.",
        "Therefore f(c f(x))=c−f(x) for every real x."
      ],
      "training_only": true
    },
    {
      "case_id": "17-B",
      "source_problem": "IMO 2017 / P2",
      "official_source": "https://www.imo-official.org/problems/2017/",
      "theorem_context": "Determine f: ℝ→ℝ satisfying f(f(x)f(y))+f(x+y)=f(xy) for every x,y∈ℝ.",
      "topic": "Functional equations",
      "proof_steps": [
        "Let c=f(0), and assume f(c f(x))=c−f(x).",
        "Evaluate the identity at x=0.",
        "This gives f(c·f(0))=c−f(0).",
        "Since f(0)=c, conclude f(c²)=0."
      ],
      "training_only": true
    },
    {
      "case_id": "17-C",
      "source_problem": "IMO 2017 / P2",
      "official_source": "https://www.imo-official.org/problems/2017/",
      "theorem_context": "Determine f: ℝ→ℝ satisfying f(f(x)f(y))+f(x+y)=f(xy) for every x,y∈ℝ.",
      "topic": "Functional equations",
      "proof_steps": [
        "From the equation, derive f(c f(x))=c−f(x) with c=f(0).",
        "Suppose c≠0.",
        "This identity immediately proves that f is injective.",
        "Consequently f(x)=f(y) forces x=y."
      ],
      "training_only": true
    },
    {
      "case_id": "17-D",
      "source_problem": "IMO 2017 / P2",
      "official_source": "https://www.imo-official.org/problems/2017/",
      "theorem_context": "Determine f: ℝ→ℝ satisfying f(f(x)f(y))+f(x+y)=f(xy) for every x,y∈ℝ.",
      "topic": "Functional equations",
      "proof_steps": [
        "Assume c=f(0)≠0 and p≠0 satisfies f(t+p)=f(t) for all t.",
        "Periodicity gives f(p)=f(0)=c.",
        "Compare the original equation at (p,y) and at (0,y): the left sides are equal.",
        "Therefore f(py)=c for every y; as p≠0, f is constant, contradicting c≠0."
      ],
      "training_only": true
    },
    {
      "case_id": "17-E",
      "source_problem": "IMO 2017 / P2",
      "official_source": "https://www.imo-official.org/problems/2017/",
      "theorem_context": "Determine f: ℝ→ℝ satisfying f(f(x)f(y))+f(x+y)=f(xy) for every x,y∈ℝ.",
      "topic": "Functional equations",
      "proof_steps": [
        "Consider the candidate f(x)=x−1.",
        "Then f(f(x)f(y))=(x−1)(y−1)−1=xy−x−y.",
        "Also f(x+y)=x+y−1.",
        "Their sum is xy−1=f(xy), so this candidate satisfies the equation."
      ],
      "training_only": true
    },
    {
      "case_id": "17-F",
      "source_problem": "IMO 2017 / P2",
      "official_source": "https://www.imo-official.org/problems/2017/",
      "theorem_context": "Determine f: ℝ→ℝ satisfying f(f(x)f(y))+f(x+y)=f(xy) for every x,y∈ℝ.",
      "topic": "Functional equations",
      "proof_steps": [
        "Suppose f is constant: f(x)=k for all x.",
        "The functional equation becomes k+k=k.",
        "This equality holds for every real k.",
        "Thus every constant function solves the equation."
      ],
      "training_only": true
    },
    {
      "case_id": "16-A",
      "source_problem": "IMO 2016 / P4",
      "official_source": "https://www.imo-official.org/problems/2016/",
      "theorem_context": "Write P(n)=n²+n+1. Find the shortest run P(a+1),…,P(a+b) (a≥0) in which each member shares a prime factor with another member.",
      "topic": "Number theory",
      "proof_steps": [
        "Let P(n)=n²+n+1 and d≥1.",
        "Expand P(n+d)=(n+d)²+(n+d)+1.",
        "Subtract P(n) to obtain P(n+d)−P(n)=2nd+d²+d.",
        "Factor: P(n+d)−P(n)=d(2n+d+1)."
      ],
      "training_only": true
    },
    {
      "case_id": "16-B",
      "source_problem": "IMO 2016 / P4",
      "official_source": "https://www.imo-official.org/problems/2016/",
      "theorem_context": "Write P(n)=n²+n+1. Find the shortest run P(a+1),…,P(a+b) (a≥0) in which each member shares a prime factor with another member.",
      "topic": "Number theory",
      "proof_steps": [
        "Suppose an odd prime q divides both P(n) and P(n+1). Every P(n) is odd.",
        "The difference P(n+1)−P(n)=2(n+1), so q divides n+1.",
        "Modulo n+1, P(n)=n(n+1)+1 ≡ 1.",
        "This contradicts q|P(n), so gcd(P(n),P(n+1))=1."
      ],
      "training_only": true
    },
    {
      "case_id": "16-C",
      "source_problem": "IMO 2016 / P4",
      "official_source": "https://www.imo-official.org/problems/2016/",
      "theorem_context": "Write P(n)=n²+n+1. Find the shortest run P(a+1),…,P(a+b) (a≥0) in which each member shares a prime factor with another member.",
      "topic": "Number theory",
      "proof_steps": [
        "Suppose q is a prime dividing P(n+d)−P(n).",
        "Then q necessarily divides P(n) and P(n+d) separately.",
        "Consequently q is a common prime factor of these two values.",
        "Thus their greatest common divisor is always divisible by q."
      ],
      "training_only": true
    },
    {
      "case_id": "16-D",
      "source_problem": "IMO 2016 / P4",
      "official_source": "https://www.imo-official.org/problems/2016/",
      "theorem_context": "Write P(n)=n²+n+1. Find the shortest run P(a+1),…,P(a+b) (a≥0) in which each member shares a prime factor with another member.",
      "topic": "Number theory",
      "proof_steps": [
        "For n=2, calculate P(2)=2²+2+1=7.",
        "Calculate P(4)=4²+4+1=21.",
        "Since these arguments differ by 2, gcd(P(2),P(4))=1.",
        "Therefore every pair of P-values at distance 2 is coprime."
      ],
      "training_only": true
    },
    {
      "case_id": "16-E",
      "source_problem": "IMO 2016 / P4",
      "official_source": "https://www.imo-official.org/problems/2016/",
      "theorem_context": "Write P(n)=n²+n+1. Find the shortest run P(a+1),…,P(a+b) (a≥0) in which each member shares a prime factor with another member.",
      "topic": "Number theory",
      "proof_steps": [
        "A fragrant set contains at least two positive integers; each must share a prime factor with another member.",
        "For a run of length b=2, the two P-values are consecutive.",
        "The adjacent-coprimality lemma gives gcd(P(a+1),P(a+2))=1.",
        "Hence b=2 cannot be fragrant, so any successful run must have b≥3."
      ],
      "training_only": true
    },
    {
      "case_id": "24-A",
      "source_problem": "IMO 2024 / P3",
      "official_source": "https://www.imo-official.org/problems/2024/",
      "theorem_context": "For n>N, let aₙ count earlier occurrences of aₙ₋₁ among a₁,…,aₙ₋₁. Show that the odd- or even-indexed subsequence is eventually periodic.",
      "topic": "Combinatorics",
      "proof_steps": [
        "For each positive integer x define Cₓ(n)=#{j≤n : aⱼ=x}.",
        "The stated rule counts occurrences of aₙ among a₁,…,aₙ.",
        "Thus for every n≥N, aₙ₊₁=Cₐₙ(n).",
        "Equivalently, aₙ=Cₐₙ₋₁(n−1) whenever n>N."
      ],
      "training_only": true
    },
    {
      "case_id": "24-B",
      "source_problem": "IMO 2024 / P3",
      "official_source": "https://www.imo-official.org/problems/2024/",
      "theorem_context": "For n>N, let aₙ count earlier occurrences of aₙ₋₁ among a₁,…,aₙ₋₁. Show that the odd- or even-indexed subsequence is eventually periodic.",
      "topic": "Combinatorics",
      "proof_steps": [
        "For n≥N, aₙ₊₁ is the number of aₙ occurrences among the first n entries.",
        "Assume an ordered pair (x,y) occurs as (aₙ,aₙ₊₁) and again later as (aₘ,aₘ₊₁), m>n≥N.",
        "At the second occurrence of x, its cumulative count has strictly increased.",
        "Thus the second following value cannot again equal y, contradicting repetition."
      ],
      "training_only": true
    },
    {
      "case_id": "24-C",
      "source_problem": "IMO 2024 / P3",
      "official_source": "https://www.imo-official.org/problems/2024/",
      "theorem_context": "For n>N, let aₙ count earlier occurrences of aₙ₋₁ among a₁,…,aₙ₋₁. Show that the odd- or even-indexed subsequence is eventually periodic.",
      "topic": "Combinatorics",
      "proof_steps": [
        "The recurrence ensures that every repeated occurrence of x has a larger following count.",
        "Therefore no ordered pair (aₙ,aₙ₊₁) can repeat in the governed part of the sequence.",
        "Hence no individual term aₙ can ever repeat.",
        "In particular, all aₙ are distinct for n>N."
      ],
      "training_only": true
    },
    {
      "case_id": "24-D",
      "source_problem": "IMO 2024 / P3",
      "official_source": "https://www.imo-official.org/problems/2024/",
      "theorem_context": "For n>N, let aₙ count earlier occurrences of aₙ₋₁ among a₁,…,aₙ₋₁. Show that the odd- or even-indexed subsequence is eventually periodic.",
      "topic": "Combinatorics",
      "proof_steps": [
        "Assume a derived state Sₙ lies in a finite set for all sufficiently large n.",
        "Suppose the next state is determined uniquely by the current state: Sₙ₊₁=F(Sₙ).",
        "By the pigeonhole principle, two states Sᵢ=Sⱼ occur for i<j.",
        "Determinism gives Sᵢ₊ₖ=Sⱼ₊ₖ for all k≥0, so the state sequence is eventually periodic."
      ],
      "training_only": true
    },
    {
      "case_id": "24-E",
      "source_problem": "IMO 2024 / P3",
      "official_source": "https://www.imo-official.org/problems/2024/",
      "theorem_context": "For n>N, let aₙ count earlier occurrences of aₙ₋₁ among a₁,…,aₙ₋₁. Show that the odd- or even-indexed subsequence is eventually periodic.",
      "topic": "Combinatorics",
      "proof_steps": [
        "Suppose states Sₙ take values in a finite set.",
        "The pigeonhole principle guarantees Sᵢ=Sⱼ for some i<j.",
        "There is no assumption relating Sₙ to Sₙ₊₁.",
        "Therefore the entire sequence must be eventually periodic."
      ],
      "training_only": true
    },
    {
      "case_id": "F05-1",
      "source_problem": "IMO 2006 / P5",
      "official_source": "https://www.imo-official.org/problems/2006/",
      "theorem_context": "An integer polynomial P of degree d>1 is iterated k≥1 times. Bound the integer fixed points of its k-fold iterate by d.",
      "topic": "Number theory · Polynomials",
      "proof_steps": [
        "Take distinct integers u,v and an integer polynomial P.",
        "For each j≥1, uʲ−vʲ=(u−v)∑_{r=0}^{j−1}u^{j−1−r}vʳ.",
        "Every coefficient of P is an integer, hence u−v divides P(u)−P(v).",
        "This holds with u,v replaced by any pair of integer iterates."
      ],
      "training_only": true
    },
    {
      "case_id": "F05-2",
      "source_problem": "IMO 2006 / P5",
      "official_source": "https://www.imo-official.org/problems/2006/",
      "theorem_context": "An integer polynomial P of degree d>1 is iterated k≥1 times. Bound the integer fixed points of its k-fold iterate by d.",
      "topic": "Number theory · Polynomials",
      "proof_steps": [
        "Let S={t∈ℤ:Pᵏ(t)=t} and choose u≠v in S.",
        "Define dᵢ=Pⁱ(u)−Pⁱ(v), so dₖ=d₀≠0.",
        "Since P has degree greater than 1, its restriction to ℤ is injective.",
        "Thus every dᵢ≠0 and the divisibility chain may begin."
      ],
      "training_only": true
    },
    {
      "case_id": "F05-3",
      "source_problem": "IMO 2006 / P5",
      "official_source": "https://www.imo-official.org/problems/2006/",
      "theorem_context": "An integer polynomial P of degree d>1 is iterated k≥1 times. Bound the integer fixed points of its k-fold iterate by d.",
      "topic": "Number theory · Polynomials",
      "proof_steps": [
        "For u≠v in S, all dᵢ=Pⁱ(u)−Pⁱ(v) are nonzero because Pᵏ(u)=u and Pᵏ(v)=v.",
        "Integer polynomial divisibility gives dᵢ|dᵢ₊₁ for 0≤i<k.",
        "But dₖ=d₀; hence |d₀|≤···≤|dₖ|=|d₀|.",
        "Every inequality is equality and |P(u)−P(v)|=|u−v|."
      ],
      "training_only": true
    },
    {
      "case_id": "F05-4",
      "source_problem": "IMO 2006 / P5",
      "official_source": "https://www.imo-official.org/problems/2006/",
      "theorem_context": "An integer polynomial P of degree d>1 is iterated k≥1 times. Bound the integer fixed points of its k-fold iterate by d.",
      "topic": "Number theory · Polynomials",
      "proof_steps": [
        "Suppose |P(u)−P(v)|=|u−v| for all u,v in a finite set S.",
        "Let a=min S and b=max S. The image P(S) lies inside S.",
        "Then {P(a),P(b)}={a,b}.",
        "Therefore P(t)=t for every t∈S."
      ],
      "training_only": true
    },
    {
      "case_id": "F05-5",
      "source_problem": "IMO 2006 / P5",
      "official_source": "https://www.imo-official.org/problems/2006/",
      "theorem_context": "An integer polynomial P of degree d>1 is iterated k≥1 times. Bound the integer fixed points of its k-fold iterate by d.",
      "topic": "Number theory · Polynomials",
      "proof_steps": [
        "The k-fold iterate Q=Pᵏ has degree dᵏ.",
        "The integer fixed points solve Q(t)−t=0.",
        "Because P has degree d, this equation has at most d real roots.",
        "Hence there are at most d integer fixed points."
      ],
      "training_only": true
    },
    {
      "case_id": "F05-6",
      "source_problem": "IMO 2006 / P5",
      "official_source": "https://www.imo-official.org/problems/2006/",
      "theorem_context": "An integer polynomial P of degree d>1 is iterated k≥1 times. Bound the integer fixed points of its k-fold iterate by d.",
      "topic": "Number theory · Polynomials",
      "proof_steps": [
        "Suppose S is a finite set of integer fixed points and P preserves distances on S.",
        "Let a=min S, b=max S, with a<b, and note P(S)⊆S.",
        "If P(a)=a then P(t)=t for every t∈S; if P(a)=b then P(t)=a+b−t.",
        "Thus S lies among the roots of either P(x)−x or P(x)+x−(a+b), both degree d>1."
      ],
      "training_only": true
    },
    {
      "case_id": "F06-1",
      "source_problem": "IMO 2009 / P6",
      "official_source": "https://www.imo-official.org/problems/2009/",
      "theorem_context": "Distinct positive jump lengths must be ordered so no partial sum falls in a forbidden set of at most n−1 positive integers, with the final total excluded.",
      "topic": "Combinatorics · Extremal",
      "proof_steps": [
        "Let a₁<···<aₙ be distinct positive jumps, s=∑aᵢ, t=s−aₙ.",
        "For 1≤j<n define uⱼ=t−aⱼ and vⱼ=s−aⱼ.",
        "Each uⱼ<t<vⱼ since aₙ>aⱼ>0.",
        "The 2(n−1) values form n−1 disjoint two-element pairs {uⱼ,vⱼ}."
      ],
      "training_only": true
    },
    {
      "case_id": "F06-2",
      "source_problem": "IMO 2009 / P6",
      "official_source": "https://www.imo-official.org/problems/2009/",
      "theorem_context": "Distinct positive jump lengths must be ordered so no partial sum falls in a forbidden set of at most n−1 positive integers, with the final total excluded.",
      "topic": "Combinatorics · Extremal",
      "proof_steps": [
        "Let t belong to a forbidden set M with |M|≤n−1.",
        "Consider n−1 pairwise disjoint pairs {uⱼ,vⱼ}, none containing t.",
        "M∖{t} has at most n−2 elements.",
        "Consequently every one of the n−1 pairs contains a forbidden point."
      ],
      "training_only": true
    },
    {
      "case_id": "F06-3",
      "source_problem": "IMO 2009 / P6",
      "official_source": "https://www.imo-official.org/problems/2009/",
      "theorem_context": "Distinct positive jump lengths must be ordered so no partial sum falls in a forbidden set of at most n−1 positive integers, with the final total excluded.",
      "topic": "Combinatorics · Extremal",
      "proof_steps": [
        "Suppose a partial jump sequence first lands on m=max M via a jump of length b.",
        "Insert the largest available jump A immediately before b, with A>b.",
        "The new landing is m−b+A>m.",
        "All subsequent landings increase further, so none belongs to M."
      ],
      "training_only": true
    },
    {
      "case_id": "F06-4",
      "source_problem": "IMO 2009 / P6",
      "official_source": "https://www.imo-official.org/problems/2009/",
      "theorem_context": "Distinct positive jump lengths must be ordered so no partial sum falls in a forbidden set of at most n−1 positive integers, with the final total excluded.",
      "topic": "Combinatorics · Extremal",
      "proof_steps": [
        "Inductively order a subset of jumps to avoid all members of M below its total t.",
        "Append another positive jump A.",
        "All new intermediate sums except the endpoint are unchanged.",
        "Therefore the extended sequence avoids M without checking its endpoint t+A."
      ],
      "training_only": true
    },
    {
      "case_id": "F06-5",
      "source_problem": "IMO 2009 / P6",
      "official_source": "https://www.imo-official.org/problems/2009/",
      "theorem_context": "Distinct positive jump lengths must be ordered so no partial sum falls in a forbidden set of at most n−1 positive integers, with the final total excluded.",
      "topic": "Combinatorics · Extremal",
      "proof_steps": [
        "Remove two jumps A and aⱼ leaving total uⱼ=s−A−aⱼ.",
        "Let M′={x∈M:x<uⱼ}; suppose |M′|≤(n−2)−1.",
        "Apply strong induction to the n−2 remaining distinct positive lengths, using M′.",
        "If uⱼ∉M, all resulting partial sums avoid M, because they do not exceed uⱼ."
      ],
      "training_only": true
    },
    {
      "case_id": "F06-6",
      "source_problem": "IMO 2009 / P6",
      "official_source": "https://www.imo-official.org/problems/2009/",
      "theorem_context": "Distinct positive jump lengths must be ordered so no partial sum falls in a forbidden set of at most n−1 positive integers, with the final total excluded.",
      "topic": "Combinatorics · Extremal",
      "proof_steps": [
        "Assume all partial sums for the remaining n−2 lengths avoid M′={x∈M:x<uⱼ}.",
        "Every such partial sum is at most uⱼ.",
        "Thus every such partial sum avoids M, regardless of whether uⱼ∈M.",
        "Appending the final two jumps produces a valid order."
      ],
      "training_only": true
    },
    {
      "case_id": "F07-1",
      "source_problem": "IMO 2010 / P6",
      "official_source": "https://www.imo-official.org/problems/2010/",
      "theorem_context": "A positive sequence satisfies aₙ=max{aₖ+aₙ₋ₖ : 1≤k<n} for n>s. Study eventual fixed-step decompositions.",
      "topic": "Sequences · Tropical algebra",
      "proof_steps": [
        "Set c=max_{1≤i≤s}(aᵢ/i) and bₙ=cn−aₙ.",
        "For each 1≤i≤s, bᵢ≥0, and for at least one ℓ≤s, bℓ=0.",
        "For n>s, aₙ=max_{1≤k<n}(aₖ+aₙ₋ₖ).",
        "Substituting aᵢ=ci−bᵢ yields bₙ=min_{1≤k<n}(bₖ+bₙ₋ₖ)."
      ],
      "training_only": true
    },
    {
      "case_id": "F07-2",
      "source_problem": "IMO 2010 / P6",
      "official_source": "https://www.imo-official.org/problems/2010/",
      "theorem_context": "A positive sequence satisfies aₙ=max{aₖ+aₙ₋ₖ : 1≤k<n} for n>s. Study eventual fixed-step decompositions.",
      "topic": "Sequences · Tropical algebra",
      "proof_steps": [
        "Normalize as bₙ=cn−aₙ and choose ℓ≤s with bℓ=0.",
        "For n+ℓ>s the minimum recurrence applies.",
        "Using k=ℓ as an admissible decomposition gives bₙ₊ℓ≤bℓ+bₙ=bₙ.",
        "Since bℓ=0, equality bₙ₊ℓ=bₙ follows for all n."
      ],
      "training_only": true
    },
    {
      "case_id": "F07-3",
      "source_problem": "IMO 2010 / P6",
      "official_source": "https://www.imo-official.org/problems/2010/",
      "theorem_context": "A positive sequence satisfies aₙ=max{aₖ+aₙ₋ₖ : 1≤k<n} for n>s. Study eventual fixed-step decompositions.",
      "topic": "Sequences · Tropical algebra",
      "proof_steps": [
        "Every bₙ for n>s is the minimum of finitely many sums of earlier b-values.",
        "By induction every bₙ is a nonnegative integer combination of b₁,…,bₛ.",
        "There are only finitely many initial generators b₁,…,bₛ.",
        "Therefore the range {bₙ:n≥1} is finite."
      ],
      "training_only": true
    },
    {
      "case_id": "F07-4",
      "source_problem": "IMO 2010 / P6",
      "official_source": "https://www.imo-official.org/problems/2010/",
      "theorem_context": "A positive sequence satisfies aₙ=max{aₖ+aₙ₋ₖ : 1≤k<n} for n>s. Study eventual fixed-step decompositions.",
      "topic": "Sequences · Tropical algebra",
      "proof_steps": [
        "Suppose bℓ=0 and bₙ₊ℓ≤bₙ for every n≥s+1.",
        "Every bₙ is nonnegative.",
        "Thus bₙ₊qℓ eventually equals zero as q increases.",
        "The min-plus recursion is eventually periodic with period ℓ."
      ],
      "training_only": true
    },
    {
      "case_id": "F07-5",
      "source_problem": "IMO 2010 / P6",
      "official_source": "https://www.imo-official.org/problems/2010/",
      "theorem_context": "A positive sequence satisfies aₙ=max{aₖ+aₙ₋ₖ : 1≤k<n} for n>s. Study eventual fixed-step decompositions.",
      "topic": "Sequences · Tropical algebra",
      "proof_steps": [
        "Assume bᵢ≥0 for 1≤i≤s.",
        "For n>s, write bₙ=min_{1≤k<n}(bₖ+bₙ₋ₖ).",
        "All arguments on the right are sums of previously nonnegative terms.",
        "By induction every bₙ≥0."
      ],
      "training_only": true
    },
    {
      "case_id": "F07-6",
      "source_problem": "IMO 2010 / P6",
      "official_source": "https://www.imo-official.org/problems/2010/",
      "theorem_context": "A positive sequence satisfies aₙ=max{aₖ+aₙ₋ₖ : 1≤k<n} for n>s. Study eventual fixed-step decompositions.",
      "topic": "Sequences · Tropical algebra",
      "proof_steps": [
        "Write aₙ=cn−bₙ and aₖ+aₙ₋ₖ=cn−(bₖ+bₙ₋ₖ).",
        "The recurrence says cn−bₙ=maxₖ[cn−(bₖ+bₙ₋ₖ)].",
        "Subtract cn to obtain −bₙ=maxₖ[−(bₖ+bₙ₋ₖ)].",
        "Therefore bₙ=maxₖ(bₖ+bₙ₋ₖ)."
      ],
      "training_only": true
    },
    {
      "case_id": "F08-1",
      "source_problem": "IMO 2014 / P6",
      "official_source": "https://www.imo-official.org/problems/2014/",
      "theorem_context": "In an arrangement of n lines in general position, choose at least √n blue lines without producing a bounded cell whose entire boundary is blue.",
      "topic": "Geometry · Combinatorics",
      "proof_steps": [
        "Let n lines have no parallels and no three concurrent.",
        "Insert them one at a time.",
        "The kth line has k−1 distinct intersection points with earlier lines, hence is divided into k segments.",
        "It adds k regions. The total number of plane regions is 1+n(n+1)/2."
      ],
      "training_only": true
    },
    {
      "case_id": "F08-2",
      "source_problem": "IMO 2014 / P6",
      "official_source": "https://www.imo-official.org/problems/2014/",
      "theorem_context": "In an arrangement of n lines in general position, choose at least √n blue lines without producing a bounded cell whose entire boundary is blue.",
      "topic": "Geometry · Combinatorics",
      "proof_steps": [
        "Take any finite set of lines in general position.",
        "Every region is a convex intersection of half-planes.",
        "A convex polygon cut out by lines necessarily has three sides.",
        "Thus a line arrangement's bounded faces are all triangles."
      ],
      "training_only": true
    },
    {
      "case_id": "F08-3",
      "source_problem": "IMO 2014 / P6",
      "official_source": "https://www.imo-official.org/problems/2014/",
      "theorem_context": "In an arrangement of n lines in general position, choose at least √n blue lines without producing a bounded cell whose entire boundary is blue.",
      "topic": "Geometry · Combinatorics",
      "proof_steps": [
        "Let B be an inclusion-maximal set of blue lines such that no bounded face has entirely blue boundary.",
        "Fix any line ℓ not in B.",
        "If adding ℓ created no all-blue bounded face, B∪{ℓ} would remain permitted.",
        "Therefore adding each outside line produces some bounded face whose boundary lies in B∪{ℓ}."
      ],
      "training_only": true
    },
    {
      "case_id": "F08-4",
      "source_problem": "IMO 2014 / P6",
      "official_source": "https://www.imo-official.org/problems/2014/",
      "theorem_context": "In an arrangement of n lines in general position, choose at least √n blue lines without producing a bounded cell whose entire boundary is blue.",
      "topic": "Geometry · Combinatorics",
      "proof_steps": [
        "Let B be an inclusion-maximal valid blue set.",
        "Every line ℓ outside B is part of a forbidden face if recoloured blue.",
        "Hence there is one bounded face F whose boundary contains every line outside B.",
        "Counting that face gives the desired lower bound."
      ],
      "training_only": true
    },
    {
      "case_id": "F08-5",
      "source_problem": "IMO 2014 / P6",
      "official_source": "https://www.imo-official.org/problems/2014/",
      "theorem_context": "In an arrangement of n lines in general position, choose at least √n blue lines without producing a bounded cell whose entire boundary is blue.",
      "topic": "Geometry · Combinatorics",
      "proof_steps": [
        "Take an arrangement of n≥4 lines in general position.",
        "Any selected pair of lines meets in a unique point.",
        "The intersection is a vertex of exactly one bounded region.",
        "Consequently each pair determines a distinct bounded face."
      ],
      "training_only": true
    },
    {
      "case_id": "F08-6",
      "source_problem": "IMO 2014 / P6",
      "official_source": "https://www.imo-official.org/problems/2014/",
      "theorem_context": "In an arrangement of n lines in general position, choose at least √n blue lines without producing a bounded cell whose entire boundary is blue.",
      "topic": "Geometry · Combinatorics",
      "proof_steps": [
        "A bounded face is defined by its supporting boundary lines.",
        "Colour some subset B of lines blue.",
        "If at least one boundary line of each bounded face lies outside B, no bounded face is all blue.",
        "This condition remains true after uncolouring any number of blue lines."
      ],
      "training_only": true
    },
    {
      "case_id": "F09-1",
      "source_problem": "IMO 2015 / P5",
      "official_source": "https://www.imo-official.org/problems/2015/",
      "theorem_context": "Find f:ℝ→ℝ with f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x) for all real x,y.",
      "topic": "Functional equations",
      "proof_steps": [
        "Let c=f(0) in the IMO 2015 P5 equation.",
        "Put x=0 and y=t: f(f(t))+c=f(t)+tc.",
        "Rearrange to f(f(t))=f(t)+c(t−1).",
        "The identity holds for every real t."
      ],
      "training_only": true
    },
    {
      "case_id": "F09-2",
      "source_problem": "IMO 2015 / P5",
      "official_source": "https://www.imo-official.org/problems/2015/",
      "theorem_context": "Find f:ℝ→ℝ with f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x) for all real x,y.",
      "topic": "Functional equations",
      "proof_steps": [
        "Assume c=f(0)≠0 and f(r)=r.",
        "Use f(f(r))=f(r)+c(r−1).",
        "The left side is f(r)=r, so c(r−1)=0.",
        "Since c≠0, r=1."
      ],
      "training_only": true
    },
    {
      "case_id": "F09-3",
      "source_problem": "IMO 2015 / P5",
      "official_source": "https://www.imo-official.org/problems/2015/",
      "theorem_context": "Find f:ℝ→ℝ with f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x) for all real x,y.",
      "topic": "Functional equations",
      "proof_steps": [
        "From the original equation at y=1, obtain f(x+f(x+1))=x+f(x+1).",
        "Thus every real number x+f(x+1) is a fixed point of f.",
        "Assume c≠0; the only possible fixed point is 1.",
        "Therefore x+f(x+1)=1 implies f(x)=1−x."
      ],
      "training_only": true
    },
    {
      "case_id": "F09-4",
      "source_problem": "IMO 2015 / P5",
      "official_source": "https://www.imo-official.org/problems/2015/",
      "theorem_context": "Find f:ℝ→ℝ with f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x) for all real x,y.",
      "topic": "Functional equations",
      "proof_steps": [
        "Assume c=0; then f(f(t))=f(t) for all real t.",
        "The function f is idempotent under composition.",
        "Every idempotent self-map of ℝ is the identity map.",
        "Thus f(t)=t without needing any other substitutions."
      ],
      "training_only": true
    },
    {
      "case_id": "F09-5",
      "source_problem": "IMO 2015 / P5",
      "official_source": "https://www.imo-official.org/problems/2015/",
      "theorem_context": "Find f:ℝ→ℝ with f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x) for all real x,y.",
      "topic": "Functional equations",
      "proof_steps": [
        "Set f(t)=2−t.",
        "Then f(x+f(x+y))=2−(2−y)=y.",
        "Also f(xy)=2−xy.",
        "The right side x+f(x+y)+yf(x)=x+(2−x−y)+y(2−x)=2+y−xy, agreeing with the left."
      ],
      "training_only": true
    },
    {
      "case_id": "F09-6",
      "source_problem": "IMO 2015 / P5",
      "official_source": "https://www.imo-official.org/problems/2015/",
      "theorem_context": "Find f:ℝ→ℝ with f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x) for all real x,y.",
      "topic": "Functional equations",
      "proof_steps": [
        "Suppose c=0, so f(f(t))=f(t).",
        "Every r in im(f) is a fixed point of f.",
        "Since f maps all reals into its own image, each output f(t) is fixed.",
        "Therefore each input t is fixed by f."
      ],
      "training_only": true
    },
    {
      "case_id": "F10-1",
      "source_problem": "IMO 2018 / P3",
      "official_source": "https://www.imo-official.org/problems/2018/",
      "theorem_context": "An anti-Pascal triangle records absolute differences above neighboring entries. Can a 2018-row triangle contain every integer 1,…,2018·2019/2 exactly once?",
      "topic": "Combinatorics · Extremal",
      "proof_steps": [
        "In an anti-Pascal triangle let Mᵣ and mᵣ be the row maxima and minima.",
        "Every entry immediately above row r is |u−v| for two entries u,v in row r.",
        "Thus Mᵣ₋₁≤Mᵣ−mᵣ.",
        "Summing over rows yields ∑ᵣ mᵣ≤Mₙ."
      ],
      "training_only": true
    },
    {
      "case_id": "F10-2",
      "source_problem": "IMO 2018 / P3",
      "official_source": "https://www.imo-official.org/problems/2018/",
      "theorem_context": "An anti-Pascal triangle records absolute differences above neighboring entries. Can a 2018-row triangle contain every integer 1,…,2018·2019/2 exactly once?",
      "topic": "Combinatorics · Extremal",
      "proof_steps": [
        "Assume all entries 1,…,N occur exactly once and there are n rows.",
        "Row minima are n distinct positive integers, hence ∑mᵣ≥n(n+1)/2=N.",
        "The row-maximum inequality gives ∑mᵣ≤Mₙ≤N.",
        "Equality holds and the minima form precisely {1,2,…,n}."
      ],
      "training_only": true
    },
    {
      "case_id": "F10-3",
      "source_problem": "IMO 2018 / P3",
      "official_source": "https://www.imo-official.org/problems/2018/",
      "theorem_context": "An anti-Pascal triangle records absolute differences above neighboring entries. Can a 2018-row triangle contain every integer 1,…,2018·2019/2 exactly once?",
      "topic": "Combinatorics · Extremal",
      "proof_steps": [
        "A large entry v>N−n is |x−y| between adjacent children in the next row.",
        "With x>y and x≤N, one gets y=x−v<n.",
        "Therefore every large parent has at least one small child.",
        "Hence every large parent has two small children."
      ],
      "training_only": true
    },
    {
      "case_id": "F10-4",
      "source_problem": "IMO 2018 / P3",
      "official_source": "https://www.imo-official.org/problems/2018/",
      "theorem_context": "An anti-Pascal triangle records absolute differences above neighboring entries. Can a 2018-row triangle contain every integer 1,…,2018·2019/2 exactly once?",
      "topic": "Combinatorics · Extremal",
      "proof_steps": [
        "There is exactly one small entry in each row.",
        "Any large entry in row r−1 must have the small entry of row r as a child.",
        "Therefore there is at most one large entry in row r−1.",
        "Hence the entire triangle has at most n large entries."
      ],
      "training_only": true
    },
    {
      "case_id": "F10-5",
      "source_problem": "IMO 2018 / P3",
      "official_source": "https://www.imo-official.org/problems/2018/",
      "theorem_context": "An anti-Pascal triangle records absolute differences above neighboring entries. Can a 2018-row triangle contain every integer 1,…,2018·2019/2 exactly once?",
      "topic": "Combinatorics · Extremal",
      "proof_steps": [
        "For n=2018, the sum of any 64 distinct positive minima is at least 1+···+64=2080.",
        "If row r has 64 rows strictly beneath it, Mᵣ≤N−∑_{j>r}mⱼ.",
        "Therefore Mᵣ≤N−2080<N−2018.",
        "No large entry (value>N−2018) occurs in row r or any higher row."
      ],
      "training_only": true
    },
    {
      "case_id": "F10-6",
      "source_problem": "IMO 2018 / P3",
      "official_source": "https://www.imo-official.org/problems/2018/",
      "theorem_context": "An anti-Pascal triangle records absolute differences above neighboring entries. Can a 2018-row triangle contain every integer 1,…,2018·2019/2 exactly once?",
      "topic": "Combinatorics · Extremal",
      "proof_steps": [
        "Partition the bottom row of length 2018 into 1009 adjacent disjoint pairs.",
        "If both entries of a pair are large, the entry immediately above that pair is small.",
        "The row above contains exactly one small entry, so at most one pair has two large entries.",
        "Therefore the bottom row contains at most 1009 large entries."
      ],
      "training_only": true
    },
    {
      "case_id": "F11-1",
      "source_problem": "IMO 2020 / P3",
      "official_source": "https://www.imo-official.org/problems/2020/",
      "theorem_context": "Four pebbles of each of n colours bear weights 1,…,4n. Divide them equally by weight and colour: two of each colour per pile.",
      "topic": "Combinatorics · Graph theory",
      "proof_steps": [
        "Pair weights i and 4n+1−i for i=1,…,2n.",
        "Every such pair has total weight 4n+1.",
        "If each pile receives exactly n complete pairs, both weight sums are n(4n+1).",
        "Thus total-weight equality follows from the pair count."
      ],
      "training_only": true
    },
    {
      "case_id": "F11-2",
      "source_problem": "IMO 2020 / P3",
      "official_source": "https://www.imo-official.org/problems/2020/",
      "theorem_context": "Four pebbles of each of n colours bear weights 1,…,4n. Divide them equally by weight and colour: two of each colour per pile.",
      "topic": "Combinatorics · Graph theory",
      "proof_steps": [
        "Make a multigraph with one vertex per colour and one edge per complementary-weight pair.",
        "An edge joins the colours of its two pebbles.",
        "A same-colour pair becomes a loop, contributing two incidences.",
        "Every vertex has degree exactly four, including loops."
      ],
      "training_only": true
    },
    {
      "case_id": "F11-3",
      "source_problem": "IMO 2020 / P3",
      "official_source": "https://www.imo-official.org/problems/2020/",
      "theorem_context": "Four pebbles of each of n colours bear weights 1,…,4n. Divide them equally by weight and colour: two of each colour per pile.",
      "topic": "Combinatorics · Graph theory",
      "proof_steps": [
        "Every colour vertex has degree four.",
        "A graph in which all vertex degrees are even contains no odd cycle.",
        "Hence the graph is bipartite.",
        "Colour its two vertex classes to produce the piles."
      ],
      "training_only": true
    },
    {
      "case_id": "F11-4",
      "source_problem": "IMO 2020 / P3",
      "official_source": "https://www.imo-official.org/problems/2020/",
      "theorem_context": "Four pebbles of each of n colours bear weights 1,…,4n. Divide them equally by weight and colour: two of each colour per pile.",
      "topic": "Combinatorics · Graph theory",
      "proof_steps": [
        "In a connected component of a 4-regular multigraph with v vertices, sum degrees=4v.",
        "The handshaking lemma gives 2e=4v, so e=2v is even.",
        "Every degree is even, so the connected component admits an Euler circuit.",
        "Its edges can be coloured alternately A,B along the circuit, including across its endpoint."
      ],
      "training_only": true
    },
    {
      "case_id": "F11-5",
      "source_problem": "IMO 2020 / P3",
      "official_source": "https://www.imo-official.org/problems/2020/",
      "theorem_context": "Four pebbles of each of n colours bear weights 1,…,4n. Divide them equally by weight and colour: two of each colour per pile.",
      "topic": "Combinatorics · Graph theory",
      "proof_steps": [
        "Follow an alternatingly coloured Euler circuit in a 4-regular multigraph.",
        "At each vertex, incoming and outgoing incidences pair cyclically.",
        "Every such pair has one A and one B incidence.",
        "Therefore each vertex has four A incidences and four B incidences."
      ],
      "training_only": true
    },
    {
      "case_id": "F11-6",
      "source_problem": "IMO 2020 / P3",
      "official_source": "https://www.imo-official.org/problems/2020/",
      "theorem_context": "Four pebbles of each of n colours bear weights 1,…,4n. Divide them equally by weight and colour: two of each colour per pile.",
      "topic": "Combinatorics · Graph theory",
      "proof_steps": [
        "Suppose the alternating Euler labelling gives each colour two pebbles in each pile.",
        "There are n colours.",
        "Therefore each pile contains n pebbles, hence n/2 complementary pairs.",
        "Its weight must equal (n/2)(4n+1)."
      ],
      "training_only": true
    },
    {
      "case_id": "F12-1",
      "source_problem": "IMO 2021 / P6",
      "official_source": "https://www.imo-official.org/problems/2021/",
      "theorem_context": "A finite integer set A has subsets with sums m,m²,…,mᵐ. Show that |A|≥m/2, for m≥2.",
      "topic": "Number theory · Information",
      "proof_steps": [
        "For c₁,…,cₘ∈{0,…,m−1}, define S=∑_{k=1}^m cₖmᵏ.",
        "Suppose two coefficient vectors give the same S.",
        "Take the least index r where the vectors differ; division by mʳ leaves a nonzero residue cᵣ−dᵣ modulo m.",
        "But 0<|cᵣ−dᵣ|<m, contradiction. All mᵐ sums are distinct."
      ],
      "training_only": true
    },
    {
      "case_id": "F12-2",
      "source_problem": "IMO 2021 / P6",
      "official_source": "https://www.imo-official.org/problems/2021/",
      "theorem_context": "A finite integer set A has subsets with sums m,m²,…,mᵐ. Show that |A|≥m/2, for m≥2.",
      "topic": "Number theory · Information",
      "proof_steps": [
        "Let A={a₁,…,aₙ} and write the sums of Bₖ as mᵏ.",
        "For cₖ∈{0,…,m−1}, collect coefficients tᵢ=∑_{k:aᵢ∈Bₖ} cₖ.",
        "Because each cₖ≤m−1, conclude tᵢ≤m−1.",
        "Thus there are at most mⁿ resulting coefficient vectors."
      ],
      "training_only": true
    },
    {
      "case_id": "F12-3",
      "source_problem": "IMO 2021 / P6",
      "official_source": "https://www.imo-official.org/problems/2021/",
      "theorem_context": "A finite integer set A has subsets with sums m,m²,…,mᵐ. Show that |A|≥m/2, for m≥2.",
      "topic": "Number theory · Information",
      "proof_steps": [
        "Write every weighted subset sum as ∑ᵢ tᵢaᵢ.",
        "Each coefficient tᵢ is an integer between 0 and m(m−1).",
        "The number of coefficient vectors is at most (m(m−1)+1)ⁿ≤m²ⁿ.",
        "This count remains valid even if some aᵢ are negative."
      ],
      "training_only": true
    },
    {
      "case_id": "F12-4",
      "source_problem": "IMO 2021 / P6",
      "official_source": "https://www.imo-official.org/problems/2021/",
      "theorem_context": "A finite integer set A has subsets with sums m,m²,…,mᵐ. Show that |A|≥m/2, for m≥2.",
      "topic": "Number theory · Information",
      "proof_steps": [
        "There are mᵐ distinct weighted sums S.",
        "There are at most m²ⁿ coefficient vectors (t₁,…,tₙ).",
        "Distinct coefficient vectors always yield distinct sums ∑tᵢaᵢ, even for arbitrary signed aᵢ.",
        "Therefore mᵐ≤m²ⁿ."
      ],
      "training_only": true
    },
    {
      "case_id": "F12-5",
      "source_problem": "IMO 2021 / P6",
      "official_source": "https://www.imo-official.org/problems/2021/",
      "theorem_context": "A finite integer set A has subsets with sums m,m²,…,mᵐ. Show that |A|≥m/2, for m≥2.",
      "topic": "Number theory · Information",
      "proof_steps": [
        "There are mᵐ distinct base-m sums.",
        "Every one has a coefficient representation from a pool of at most m²ⁿ vectors.",
        "Thus mᵐ≤m²ⁿ.",
        "As m≥2, compare exponents to get m≤2n, hence |A|=n≥m/2."
      ],
      "training_only": true
    },
    {
      "case_id": "F12-6",
      "source_problem": "IMO 2021 / P6",
      "official_source": "https://www.imo-official.org/problems/2021/",
      "theorem_context": "A finite integer set A has subsets with sums m,m²,…,mᵐ. Show that |A|≥m/2, for m≥2.",
      "topic": "Number theory · Information",
      "proof_steps": [
        "Because B₁,…,Bₘ are sets, any aᵢ occurs in each Bₖ at most once.",
        "In ∑ₖ cₖ ∑_{aᵢ∈Bₖ} aᵢ, the total coefficient of aᵢ is therefore 0 or 1.",
        "At most 2ⁿ distinct sums can occur.",
        "This is a stronger bound than the published m²ⁿ."
      ],
      "training_only": true
    },
    {
      "case_id": "F13-1",
      "source_problem": "IMO 2022 / P5",
      "official_source": "https://www.imo-official.org/problems/2022/",
      "theorem_context": "Determine positive integers a,b and prime p satisfying aᵖ=b!+p.",
      "topic": "Number theory · p-adic",
      "proof_steps": [
        "Assume aᵖ=b!+p with p prime and b≥p.",
        "Since p≤b, p divides b!.",
        "Reducing modulo p gives aᵖ≡0 mod p.",
        "Therefore p divides a."
      ],
      "training_only": true
    },
    {
      "case_id": "F13-2",
      "source_problem": "IMO 2022 / P5",
      "official_source": "https://www.imo-official.org/problems/2022/",
      "theorem_context": "Determine positive integers a,b and prime p satisfying aᵖ=b!+p.",
      "topic": "Number theory · p-adic",
      "proof_steps": [
        "Suppose b≥2p and p|a.",
        "The factorial b! contains both p and 2p, so p²|b!.",
        "Since p≥2, p²|aᵖ.",
        "Thus aᵖ−b! is divisible by p², contradicting aᵖ−b!=p."
      ],
      "training_only": true
    },
    {
      "case_id": "F13-3",
      "source_problem": "IMO 2022 / P5",
      "official_source": "https://www.imo-official.org/problems/2022/",
      "theorem_context": "Determine positive integers a,b and prime p satisfying aᵖ=b!+p.",
      "topic": "Number theory · p-adic",
      "proof_steps": [
        "From b≥p the equation gives p|a.",
        "If b≥2p, then p²|b! and p²|aᵖ, contradiction.",
        "Thus p≤b<2p.",
        "Consequently the same argument also proves b<p+1."
      ],
      "training_only": true
    },
    {
      "case_id": "F13-4",
      "source_problem": "IMO 2022 / P5",
      "official_source": "https://www.imo-official.org/problems/2022/",
      "theorem_context": "Determine positive integers a,b and prime p satisfying aᵖ=b!+p.",
      "topic": "Number theory · p-adic",
      "proof_steps": [
        "Suppose p≤b<2p and a=pc.",
        "Assume an independent size estimate has already established a<p², so 1≤c<p.",
        "Therefore c=1 immediately, since p is prime.",
        "Thus a=p."
      ],
      "training_only": true
    },
    {
      "case_id": "F13-5",
      "source_problem": "IMO 2022 / P5",
      "official_source": "https://www.imo-official.org/problems/2022/",
      "theorem_context": "Determine positive integers a,b and prime p satisfying aᵖ=b!+p.",
      "topic": "Number theory · p-adic",
      "proof_steps": [
        "Assume earlier arguments have established a=p.",
        "Substitute into aᵖ=b!+p.",
        "Then b!=pᵖ−p=p(p^{p−1}−1).",
        "This identity sharply restricts b and p."
      ],
      "training_only": true
    },
    {
      "case_id": "F13-6",
      "source_problem": "IMO 2022 / P5",
      "official_source": "https://www.imo-official.org/problems/2022/",
      "theorem_context": "Determine positive integers a,b and prime p satisfying aᵖ=b!+p.",
      "topic": "Number theory · p-adic",
      "proof_steps": [
        "Suppose p is prime and b≥p.",
        "Then b!≥p!.",
        "From aᵖ=b!+p conclude aᵖ≤p!+p.",
        "This bounds a from above by data depending only on p."
      ],
      "training_only": true
    },
    {
      "case_id": "F14-1",
      "source_problem": "IMO 2023 / P6",
      "official_source": "https://www.imo-official.org/problems/2023/",
      "theorem_context": "A difficult equilateral-triangle configuration produces three circumcircles. Prove that all three share two common points under the scalene hypothesis.",
      "topic": "Geometry · Circle systems",
      "proof_steps": [
        "Consider two circles with different centres, written F(P)=|P|²+u·P+c=0 and G(P)=|P|²+v·P+d=0, where u≠v.",
        "At any common point P, both vanish.",
        "Subtract: (u−v)·P+(c−d)=0.",
        "Because u≠v, the common-point condition is a genuine line: the radical axis."
      ],
      "training_only": true
    },
    {
      "case_id": "F14-2",
      "source_problem": "IMO 2023 / P6",
      "official_source": "https://www.imo-official.org/problems/2023/",
      "theorem_context": "A difficult equilateral-triangle configuration produces three circumcircles. Prove that all three share two common points under the scalene hypothesis.",
      "topic": "Geometry · Circle systems",
      "proof_steps": [
        "Suppose the three normalized circle equations satisfy αF_A+βF_B+γF_C≡0.",
        "Assume αβγ≠0.",
        "Any common point of the first two circles lies on the third.",
        "Therefore the three circles necessarily have two common real points."
      ],
      "training_only": true
    },
    {
      "case_id": "F14-3",
      "source_problem": "IMO 2023 / P6",
      "official_source": "https://www.imo-official.org/problems/2023/",
      "theorem_context": "A difficult equilateral-triangle configuration produces three circumcircles. Prove that all three share two common points under the scalene hypothesis.",
      "topic": "Geometry · Circle systems",
      "proof_steps": [
        "Assume a closed disk D contains the vertices A,B,C of a triangle.",
        "A Euclidean disk is convex.",
        "The entire triangle conv{A,B,C} lies in D.",
        "Every point strictly inside ABC lies strictly inside D if the triangle vertices are in D and D is a genuine disk."
      ],
      "training_only": true
    },
    {
      "case_id": "F14-4",
      "source_problem": "IMO 2023 / P6",
      "official_source": "https://www.imo-official.org/problems/2023/",
      "theorem_context": "A difficult equilateral-triangle configuration produces three circumcircles. Prove that all three share two common points under the scalene hypothesis.",
      "topic": "Geometry · Circle systems",
      "proof_steps": [
        "Let three circles bound disks all containing O in their interiors.",
        "Each pair of disks therefore has overlapping interiors.",
        "Any two disks with overlapping interiors have two intersecting boundary points.",
        "Hence every pair of circles crosses twice."
      ],
      "training_only": true
    },
    {
      "case_id": "F14-5",
      "source_problem": "IMO 2023 / P6",
      "official_source": "https://www.imo-official.org/problems/2023/",
      "theorem_context": "A difficult equilateral-triangle configuration produces three circumcircles. Prove that all three share two common points under the scalene hypothesis.",
      "topic": "Geometry · Circle systems",
      "proof_steps": [
        "Take two distinct circle equations F=0 and G=0.",
        "Their radical axis is the line F−G=0.",
        "A line and a circle always meet in two distinct real points.",
        "Consequently their two-circle intersection has exactly two points."
      ],
      "training_only": true
    },
    {
      "case_id": "F14-6",
      "source_problem": "IMO 2023 / P6",
      "official_source": "https://www.imo-official.org/problems/2023/",
      "theorem_context": "A difficult equilateral-triangle configuration produces three circumcircles. Prove that all three share two common points under the scalene hypothesis.",
      "topic": "Geometry · Circle systems",
      "proof_steps": [
        "Suppose αF_A+βF_B+γF_C≡0 with γ≠0.",
        "Let P satisfy F_A(P)=F_B(P)=0.",
        "Then γF_C(P)=0.",
        "Therefore F_C(P)=0, so every common point of circles A,B lies on circle C."
      ],
      "training_only": true
    },
    {
      "case_id": "F15-1",
      "source_problem": "IMO 2025 / P6",
      "official_source": "https://www.imo-official.org/problems/2025/",
      "theorem_context": "Tile a 2025×2025 grid by nonoverlapping rectangles, leaving one uncovered square in every row and every column, while minimizing the number of tiles.",
      "topic": "Combinatorics · Tiling",
      "proof_steps": [
        "Each row of the n×n board has exactly one uncovered cell.",
        "Each column also has exactly one uncovered cell.",
        "The coordinates of holes are (i,π(i)) for some permutation π of {1,…,n}.",
        "There are exactly n holes and no two share a row or column."
      ],
      "training_only": true
    },
    {
      "case_id": "F15-2",
      "source_problem": "IMO 2025 / P6",
      "official_source": "https://www.imo-official.org/problems/2025/",
      "theorem_context": "Tile a 2025×2025 grid by nonoverlapping rectangles, leaving one uncovered square in every row and every column, while minimizing the number of tiles.",
      "topic": "Combinatorics · Tiling",
      "proof_steps": [
        "Assume t nonoverlapping rectangular tiles cover all cells except the n holes.",
        "Treat every hole as an additional 1×1 rectangular tile.",
        "The resulting t+n rectangles partition all n² cells.",
        "Any rectangle-boundary counting lemma can now be applied to this partition."
      ],
      "training_only": true
    },
    {
      "case_id": "F15-3",
      "source_problem": "IMO 2025 / P6",
      "official_source": "https://www.imo-official.org/problems/2025/",
      "theorem_context": "Tile a 2025×2025 grid by nonoverlapping rectangles, leaving one uncovered square in every row and every column, while minimizing the number of tiles.",
      "topic": "Combinatorics · Tiling",
      "proof_steps": [
        "Suppose a geometric counting argument proves t≥n+2√n−3 for any arrangement.",
        "Take n=2025=45².",
        "Substituting yields t≥2025+90−3=2112.",
        "Therefore the minimum is exactly 2112 without exhibiting a tiling."
      ],
      "training_only": true
    },
    {
      "case_id": "F15-4",
      "source_problem": "IMO 2025 / P6",
      "official_source": "https://www.imo-official.org/problems/2025/",
      "theorem_context": "Tile a 2025×2025 grid by nonoverlapping rectangles, leaving one uncovered square in every row and every column, while minimizing the number of tiles.",
      "topic": "Combinatorics · Tiling",
      "proof_steps": [
        "Suppose every row of a square grid contains exactly one hole.",
        "There are n holes in total.",
        "Therefore the n columns each contain exactly one hole.",
        "The holes form a permutation matrix."
      ],
      "training_only": true
    },
    {
      "case_id": "F15-5",
      "source_problem": "IMO 2025 / P6",
      "official_source": "https://www.imo-official.org/problems/2025/",
      "theorem_context": "Tile a 2025×2025 grid by nonoverlapping rectangles, leaving one uncovered square in every row and every column, while minimizing the number of tiles.",
      "topic": "Combinatorics · Tiling",
      "proof_steps": [
        "In the established lower bound t≥n+2√n−3, set n=2025.",
        "Since 2025=45², we have √n=45.",
        "Thus t≥2025+90−3=2112.",
        "To prove this is optimal one must provide a valid arrangement with 2112 tiles."
      ],
      "training_only": true
    },
    {
      "case_id": "F15-6",
      "source_problem": "IMO 2025 / P6",
      "official_source": "https://www.imo-official.org/problems/2025/",
      "theorem_context": "Tile a 2025×2025 grid by nonoverlapping rectangles, leaving one uncovered square in every row and every column, while minimizing the number of tiles.",
      "topic": "Combinatorics · Tiling",
      "proof_steps": [
        "Suppose a real lower bound gives t≥n+2√n−3 and t is an integer.",
        "Since n and 3 are integers, t≥n+⌈2√n⌉−3.",
        "Replace ⌈2√n⌉ by 2⌈√n⌉ for every positive n.",
        "Conclude t≥n+2⌈√n⌉−3."
      ],
      "training_only": true
    },
    {
      "case_id": "F16-1",
      "source_problem": "IMO 2007 / P6",
      "official_source": "https://www.imo-official.org/problems/2007/",
      "theorem_context": "For n≥1, let S={0,…,n}³∖{(0,0,0)}. Determine the minimum number of planes needed to cover S without any plane containing the origin.",
      "topic": "Polynomial method · Discrete geometry",
      "proof_steps": [
        "Let S={0,…,n}³∖{(0,0,0)} and n≥1.",
        "For each k∈{1,…,3n}, consider the plane x+y+z=k.",
        "Every point of S has integer coordinate sum between 1 and 3n, hence lies on one of these planes.",
        "No such plane passes through (0,0,0); therefore 3n planes suffice."
      ],
      "training_only": true
    },
    {
      "case_id": "F16-2",
      "source_problem": "IMO 2007 / P6",
      "official_source": "https://www.imo-official.org/problems/2007/",
      "theorem_context": "For n≥1, let S={0,…,n}³∖{(0,0,0)}. Determine the minimum number of planes needed to cover S without any plane containing the origin.",
      "topic": "Polynomial method · Discrete geometry",
      "proof_steps": [
        "Suppose m affine planes cover S and none contains the origin.",
        "Write their equations L_r(x,y,z)=0, with L_r(0,0,0)≠0 for every r.",
        "Define the polynomial P(x,y,z)=∏_{r=1}^m L_r(x,y,z), which has total degree at most m.",
        "Then P vanishes on all S but P(0,0,0)≠0."
      ],
      "training_only": true
    },
    {
      "case_id": "F16-3",
      "source_problem": "IMO 2007 / P6",
      "official_source": "https://www.imo-official.org/problems/2007/",
      "theorem_context": "For n≥1, let S={0,…,n}³∖{(0,0,0)}. Determine the minimum number of planes needed to cover S without any plane containing the origin.",
      "topic": "Polynomial method · Discrete geometry",
      "proof_steps": [
        "Suppose a real polynomial P vanishes at every point of S.",
        "The set S contains (n+1)³−1 different integer triples.",
        "Since these are many distinct zeros, P must be the zero polynomial.",
        "In particular P(0,0,0)=0, without any degree hypothesis."
      ],
      "training_only": true
    },
    {
      "case_id": "F16-4",
      "source_problem": "IMO 2007 / P6",
      "official_source": "https://www.imo-official.org/problems/2007/",
      "theorem_context": "For n≥1, let S={0,…,n}³∖{(0,0,0)}. Determine the minimum number of planes needed to cover S without any plane containing the origin.",
      "topic": "Polynomial method · Discrete geometry",
      "proof_steps": [
        "Fix n≥1 and define w_i=1/∏_{0≤j≤n, j≠i}(i−j) for i=0,…,n.",
        "Lagrange interpolation reconstructs every polynomial q of degree at most n from q(0),…,q(n).",
        "Comparing coefficients of xⁿ yields ∑_{i=0}^n w_i i^a=0 for every integer 0≤a<n.",
        "For a=n, the same coefficient comparison yields ∑_{i=0}^n w_i iⁿ=1."
      ],
      "training_only": true
    },
    {
      "case_id": "F16-5",
      "source_problem": "IMO 2007 / P6",
      "official_source": "https://www.imo-official.org/problems/2007/",
      "theorem_context": "For n≥1, let S={0,…,n}³∖{(0,0,0)}. Determine the minimum number of planes needed to cover S without any plane containing the origin.",
      "topic": "Polynomial method · Discrete geometry",
      "proof_steps": [
        "Let P(x,y,z) have total degree strictly below 3n, and let w_i be the Lagrange weights for {0,…,n}.",
        "Define T(P)=∑_{i,j,k=0}^n w_i w_j w_k P(i,j,k).",
        "Each monomial of total degree below 3n has at least one exponent below n, so T(P)=0.",
        "If P vanishes on S, only the origin contributes: T(P)=w₀³P(0,0,0).",
        "Since w₀≠0, P(0,0,0)=0."
      ],
      "training_only": true
    },
    {
      "case_id": "F16-6",
      "source_problem": "IMO 2007 / P6",
      "official_source": "https://www.imo-official.org/problems/2007/",
      "theorem_context": "For n≥1, let S={0,…,n}³∖{(0,0,0)}. Determine the minimum number of planes needed to cover S without any plane containing the origin.",
      "topic": "Polynomial method · Discrete geometry",
      "proof_steps": [
        "Let P have total degree at most 3n.",
        "For a monomial xᵃyᵇzᶜ with a+b+c≤3n, one of a,b,c is strictly less than n.",
        "Therefore the tensor Lagrange functional T annihilates every such monomial.",
        "Consequently any polynomial P of degree at most 3n that vanishes on S must vanish at the origin."
      ],
      "training_only": true
    },
    {
      "case_id": "F17-1",
      "source_problem": "IMO 2013 / P6",
      "official_source": "https://www.imo-official.org/problems/2013/",
      "theorem_context": "Place the labels 0,…,n on n+1 equally spaced circle points, up to rotation. A labelling is beautiful if chords (a,d) and (b,c) never cross when a<b<c<d and a+d=b+c. Show the number of beautiful labellings equals 1 plus the number of coprime ordered pairs x,y≥1 with x+y≤n (n≥3).",
      "topic": "Combinatorics · Farey geometry",
      "proof_steps": [
        "Take four distinct points A,B,C,D on a circle.",
        "Their chords AC and BD cross in the open disk precisely when their endpoints alternate around the circle.",
        "The criterion depends only on cyclic order, not on equal arc lengths.",
        "Thus the crossing restriction can be analysed through circular permutations of labels."
      ],
      "training_only": true
    },
    {
      "case_id": "F17-2",
      "source_problem": "IMO 2013 / P6",
      "official_source": "https://www.imo-official.org/problems/2013/",
      "theorem_context": "Place the labels 0,…,n on n+1 equally spaced circle points, up to rotation. A labelling is beautiful if chords (a,d) and (b,c) never cross when a<b<c<d and a+d=b+c. Show the number of beautiful labellings equals 1 plus the number of coprime ordered pairs x,y≥1 with x+y≤n (n≥3).",
      "topic": "Combinatorics · Farey geometry",
      "proof_steps": [
        "Fix integers a<b<c<d satisfying a+d=b+c.",
        "Label four distinct points of a circle with these four numbers.",
        "The equality of sums forces the endpoints of chords (a,d) and (b,c) to alternate around the circle.",
        "Therefore the two chords cross, regardless of circular placement."
      ],
      "training_only": true
    },
    {
      "case_id": "F17-3",
      "source_problem": "IMO 2013 / P6",
      "official_source": "https://www.imo-official.org/problems/2013/",
      "theorem_context": "Place the labels 0,…,n on n+1 equally spaced circle points, up to rotation. A labelling is beautiful if chords (a,d) and (b,c) never cross when a<b<c<d and a+d=b+c. Show the number of beautiful labellings equals 1 plus the number of coprime ordered pairs x,y≥1 with x+y≤n (n≥3).",
      "topic": "Combinatorics · Farey geometry",
      "proof_steps": [
        "Put n+1 distinct labels 0,…,n on n+1 equally spaced points.",
        "Count labellings up to rotations of the circle.",
        "Rotate each labelling uniquely until the label 0 occupies a chosen reference point.",
        "The remaining n labels may be arranged in n! orders, so there are exactly n! rotation classes."
      ],
      "training_only": true
    },
    {
      "case_id": "F17-4",
      "source_problem": "IMO 2013 / P6",
      "official_source": "https://www.imo-official.org/problems/2013/",
      "theorem_context": "Place the labels 0,…,n on n+1 equally spaced circle points, up to rotation. A labelling is beautiful if chords (a,d) and (b,c) never cross when a<b<c<d and a+d=b+c. Show the number of beautiful labellings equals 1 plus the number of coprime ordered pairs x,y≥1 with x+y≤n (n≥3).",
      "topic": "Combinatorics · Farey geometry",
      "proof_steps": [
        "Choose α∈(0,1) that is not rational with denominator at most n.",
        "The residues {α},{2α},…,{nα} are pairwise distinct and none is 0.",
        "Their circular order can change continuously only when {iα}={jα} for some 0≤i<j≤n.",
        "Such a collision requires (j−i)α∈ℤ, a rational point of denominator at most n.",
        "Therefore the circular order is constant inside any interval avoiding these rational boundary points."
      ],
      "training_only": true
    },
    {
      "case_id": "F17-5",
      "source_problem": "IMO 2013 / P6",
      "official_source": "https://www.imo-official.org/problems/2013/",
      "theorem_context": "Place the labels 0,…,n on n+1 equally spaced circle points, up to rotation. A labelling is beautiful if chords (a,d) and (b,c) never cross when a<b<c<d and a+d=b+c. Show the number of beautiful labellings equals 1 plus the number of coprime ordered pairs x,y≥1 with x+y≤n (n≥3).",
      "topic": "Combinatorics · Farey geometry",
      "proof_steps": [
        "There are N reduced rationals strictly between 0 and 1 with denominator at most n, where N counts coprime pairs x,y≥1 with x+y≤n.",
        "These N boundary points partition (0,1) into N+1 open intervals.",
        "Suppose each interval produces a beautiful circular labelling and distinct intervals produce distinct labellings.",
        "Therefore there are exactly N+1 beautiful labellings, with no need to exclude other beautiful labellings."
      ],
      "training_only": true
    },
    {
      "case_id": "F17-6",
      "source_problem": "IMO 2013 / P6",
      "official_source": "https://www.imo-official.org/problems/2013/",
      "theorem_context": "Place the labels 0,…,n on n+1 equally spaced circle points, up to rotation. A labelling is beautiful if chords (a,d) and (b,c) never cross when a<b<c<d and a+d=b+c. Show the number of beautiful labellings equals 1 plus the number of coprime ordered pairs x,y≥1 with x+y≤n (n≥3).",
      "topic": "Combinatorics · Farey geometry",
      "proof_steps": [
        "Let α,β∈(0,1) avoid rationals with denominator at most n.",
        "The order of 0,{α},…,{nα} determines ε_j(α)=1 if {jα}<{(j−1)α}, and 0 otherwise, for 1≤j≤n.",
        "Because 0<α<1, the recurrence of fractional parts gives ⌊qα⌋=∑_{j=1}^q ε_j(α) for every q≤n.",
        "Thus identical orders for α and β force ⌊qα⌋=⌊qβ⌋ for all q≤n.",
        "If a reduced p/q, q≤n, lay strictly between α and β, then ⌊qα⌋≤p−1<⌊qβ⌋, contradiction."
      ],
      "training_only": true
    },
    {
      "case_id": "F18-1",
      "source_problem": "IMO 2019 / P5",
      "official_source": "https://www.imo-official.org/problems/2019/",
      "theorem_context": "For n coins in a line, if k>0 coins show H, flip the kth coin from the left. Prove the process terminates, and calculate the average number of flips over all 2ⁿ starting configurations.",
      "topic": "Combinatorics · Monovariants",
      "proof_steps": [
        "For a coin configuration, let k be the number of heads and let S be the sum of their positions.",
        "If heads occupy a₁<···<a_k, then a_j≥j for all j.",
        "Set V=2S−k². Then V≥2∑_{j=1}^k j−k²=k≥0.",
        "Equality V=0 is possible only if k=0, i.e. all coins show tails."
      ],
      "training_only": true
    },
    {
      "case_id": "F18-2",
      "source_problem": "IMO 2019 / P5",
      "official_source": "https://www.imo-official.org/problems/2019/",
      "theorem_context": "For n coins in a line, if k>0 coins show H, flip the kth coin from the left. Prove the process terminates, and calculate the average number of flips over all 2ⁿ starting configurations.",
      "topic": "Combinatorics · Monovariants",
      "proof_steps": [
        "If k>0, the operation flips coin k.",
        "Put ε=+1 if that coin changes T→H and ε=−1 if it changes H→T.",
        "Then k'=k+ε and S'=S+εk.",
        "Therefore V'=2(S+εk)−(k+ε)²=2S−k²−ε²=V−1."
      ],
      "training_only": true
    },
    {
      "case_id": "F18-3",
      "source_problem": "IMO 2019 / P5",
      "official_source": "https://www.imo-official.org/problems/2019/",
      "theorem_context": "For n coins in a line, if k>0 coins show H, flip the kth coin from the left. Prove the process terminates, and calculate the average number of flips over all 2ⁿ starting configurations.",
      "topic": "Combinatorics · Monovariants",
      "proof_steps": [
        "A configuration with k heads and head-position sum S has V=2S−k².",
        "Every move reduces V by one, and V≥k≥0.",
        "Since V≥k, the numerical values V and k must be equal.",
        "Thus every legal move reduces the head count k by exactly one."
      ],
      "training_only": true
    },
    {
      "case_id": "F18-4",
      "source_problem": "IMO 2019 / P5",
      "official_source": "https://www.imo-official.org/problems/2019/",
      "theorem_context": "For n coins in a line, if k>0 coins show H, flip the kth coin from the left. Prove the process terminates, and calculate the average number of flips over all 2ⁿ starting configurations.",
      "topic": "Combinatorics · Monovariants",
      "proof_steps": [
        "The integer-valued potential V is always nonnegative.",
        "Every legal move decreases V by exactly one.",
        "Consequently the process cannot continue indefinitely; it stops at all tails where V=0.",
        "Telescoping the changes shows that the number L(C) of moves equals V(C) for the initial configuration."
      ],
      "training_only": true
    },
    {
      "case_id": "F18-5",
      "source_problem": "IMO 2019 / P5",
      "official_source": "https://www.imo-official.org/problems/2019/",
      "theorem_context": "For n coins in a line, if k>0 coins show H, flip the kth coin from the left. Prove the process terminates, and calculate the average number of flips over all 2ⁿ starting configurations.",
      "topic": "Combinatorics · Monovariants",
      "proof_steps": [
        "Choose a starting configuration uniformly from all 2ⁿ possibilities.",
        "Let k be the number of heads; then E[k]=n/2.",
        "Since k is random, E[k²]=(E[k])²=n²/4.",
        "Substitute into E[L]=2E[S]−E[k²] to obtain the exact mean."
      ],
      "training_only": true
    },
    {
      "case_id": "F18-6",
      "source_problem": "IMO 2019 / P5",
      "official_source": "https://www.imo-official.org/problems/2019/",
      "theorem_context": "For n coins in a line, if k>0 coins show H, flip the kth coin from the left. Prove the process terminates, and calculate the average number of flips over all 2ⁿ starting configurations.",
      "topic": "Combinatorics · Monovariants",
      "proof_steps": [
        "Let X_i indicate that coin i starts heads. The X_i are independent Bernoulli(1/2) variables.",
        "Then k=∑X_i and S=∑iX_i, so E[2S]=n(n+1)/2.",
        "Using X_i²=X_i and E[X_iX_j]=1/4 for i≠j gives E[k²]=n/2+n(n−1)/4=n(n+1)/4.",
        "Since L=2S−k², the mean is E[L]=n(n+1)/4."
      ],
      "training_only": true
    },
    {
      "case_id": "F19-1",
      "source_problem": "IMO 2024 / P6",
      "official_source": "https://www.imo-official.org/problems/2024/",
      "theorem_context": "For f:ℚ→ℚ suppose for all rational x,y at least one equation holds: f(x+f(y))=f(x)+y OR f(f(x)+y)=x+f(y). Determine the smallest universal upper bound on |{f(r)+f(−r):r∈ℚ}|.",
      "topic": "Functional equations · Rational numbers",
      "proof_steps": [
        "Let f:ℚ→ℚ satisfy the aquaesulian disjunction for every rational x,y.",
        "Set y=x.",
        "Both permitted alternatives become the identical equality f(x+f(x))=x+f(x).",
        "Thus the diagonal fixed-point identity holds for all x∈ℚ."
      ],
      "training_only": true
    },
    {
      "case_id": "F19-2",
      "source_problem": "IMO 2024 / P6",
      "official_source": "https://www.imo-official.org/problems/2024/",
      "theorem_context": "For f:ℚ→ℚ suppose for all rational x,y at least one equation holds: f(x+f(y))=f(x)+y OR f(f(x)+y)=x+f(y). Determine the smallest universal upper bound on |{f(r)+f(−r):r∈ℚ}|.",
      "topic": "Functional equations · Rational numbers",
      "proof_steps": [
        "Suppose f(a)=f(b)=t.",
        "The diagonal identity gives f(a+t)=a+t and f(b+t)=b+t.",
        "Apply the original disjunction at (x,y)=(a,b). In the first branch f(a+t)=t+b, so a=b.",
        "In the second branch f(b+t)=a+t, so again a=b.",
        "Hence f is injective."
      ],
      "training_only": true
    },
    {
      "case_id": "F19-3",
      "source_problem": "IMO 2024 / P6",
      "official_source": "https://www.imo-official.org/problems/2024/",
      "theorem_context": "For f:ℚ→ℚ suppose for all rational x,y at least one equation holds: f(x+f(y))=f(x)+y OR f(f(x)+y)=x+f(y). Determine the smallest universal upper bound on |{f(r)+f(−r):r∈ℚ}|.",
      "topic": "Functional equations · Rational numbers",
      "proof_steps": [
        "From the diagonal case, f(x+f(x))=x+f(x) for every rational x.",
        "Therefore every number of the form x+f(x) is fixed under f.",
        "Every rational is of the form x+f(x) for some rational x.",
        "Thus f(t)=t for all rationals t."
      ],
      "training_only": true
    },
    {
      "case_id": "F19-4",
      "source_problem": "IMO 2024 / P6",
      "official_source": "https://www.imo-official.org/problems/2024/",
      "theorem_context": "For f:ℚ→ℚ suppose for all rational x,y at least one equation holds: f(x+f(y))=f(x)+y OR f(f(x)+y)=x+f(y). Determine the smallest universal upper bound on |{f(r)+f(−r):r∈ℚ}|.",
      "topic": "Functional equations · Rational numbers",
      "proof_steps": [
        "The diagonal identity yields f(f(0))=f(0).",
        "Also f(0) is the value of f at input 0.",
        "Since f has been shown injective, equality f(f(0))=f(0) forces f(0)=0.",
        "Hence e(0)=f(0)+f(−0)=0."
      ],
      "training_only": true
    },
    {
      "case_id": "F19-5",
      "source_problem": "IMO 2024 / P6",
      "official_source": "https://www.imo-official.org/problems/2024/",
      "theorem_context": "For f:ℚ→ℚ suppose for all rational x,y at least one equation holds: f(x+f(y))=f(x)+y OR f(f(x)+y)=x+f(y). Determine the smallest universal upper bound on |{f(r)+f(−r):r∈ℚ}|.",
      "topic": "Functional equations · Rational numbers",
      "proof_steps": [
        "Assume a derived formula gives D(u,v)=f(u+v)−f(u)−f(v)∈{0,−e(u)}∩{0,−e(v)}, where e(t)=f(t)+f(−t).",
        "If e(u)=e(v), the two option sets are identical.",
        "Therefore their intersection contains only 0.",
        "Hence f(u+v)=f(u)+f(v) whenever e(u)=e(v)."
      ],
      "training_only": true
    },
    {
      "case_id": "F19-6",
      "source_problem": "IMO 2024 / P6",
      "official_source": "https://www.imo-official.org/problems/2024/",
      "theorem_context": "For f:ℚ→ℚ suppose for all rational x,y at least one equation holds: f(x+f(y))=f(x)+y OR f(f(x)+y)=x+f(y). Determine the smallest universal upper bound on |{f(r)+f(−r):r∈ℚ}|.",
      "topic": "Functional equations · Rational numbers",
      "proof_steps": [
        "Define f(x)=2⌊x⌋−x for rational x, and write x=m+u,y=n+v with integers m,n and 0≤u,v<1.",
        "Then f(x)=m−u and f(y)=n−v.",
        "If u≥v, f(x+f(y))=m+n−u+v=f(x)+y; if u≤v, f(f(x)+y)=m+n+u−v=x+f(y).",
        "Thus the function satisfies at least one aquaesulian branch for every x,y.",
        "For rational r, f(r)+f(−r) equals 0 when r∈ℤ and −2 otherwise; both values occur."
      ],
      "training_only": true
    }
  ]
}