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  "target_id": "lem:kEllD",
  "proof_status": "locally_supported",
  "summary": "The supplied proof is valid under the usual convention that d is a positive integer modulus. The determinant divisibility justifies both completing the square and the otherwise abbreviated assertion that C is invertible in Case 2. The construction works also at p=2 and produces coefficients independent of m,n. No external literature is needed or imported to close this obligation.",
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    {
      "kind": "context",
      "source_quote": "If $(p,A)>1$, then by primitivity, $(p,C)=1$.",
      "explanation": "This uses both hypotheses, not primitivity alone: since p divides d and AC\u2212B\u00b2, p dividing A forces p to divide B. Then gcd(A,B,C)=1 forces p not to divide C. Thus the asserted inverse of C modulo p^e exists."
    },
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      "source_quote": "(m+\\bar A Bn)^{2}-\\bar A ^{2}B^{2}n^{2}+\\bar ACn^{2}",
      "explanation": "After multiplying the original congruence by the inverse of A, the remaining coefficient is A^{-2}(AC\u2212B\u00b2), which vanishes modulo p^e because p^e divides d. The same calculation with A and C interchanged proves Case 2. Neither calculation divides by 2, so there is no exceptional dyadic case."
    },
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      "source_quote": "By the Chinese Remainder Theorem, there are integers $k$ and $\\ell$ so that $k\\equiv k_{p}(\\mod p^{e})$, and similarly with $\\ell$.",
      "explanation": "The assembly preserves the implication for every m,n: the global linear form agrees with the chosen local form modulo each p^e. Explicitly, put D_p=d/p^e, choose u_p with D_p u_p\u22611 modulo p^e, and sum D_p u_p k_p and D_p u_p \u2113_p to obtain k and \u2113. Their squared linear form is divisible by every prime-power factor of d, hence by d. At each prime dividing d, one coefficient is congruent to 1, proving gcd(k,\u2113,d)=1. For d=1, taking k=1 and \u2113=0 suffices."
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