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feat(ErdosProblems): prove erdos_885 variants k_eq_2 and k_eq_3 - #5055

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feat(ErdosProblems): prove erdos_885 variants k_eq_2 and k_eq_3#5055
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@Alexispap Alexispap commented Aug 19, 2026

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Closes #5050

Replaces the sorry in the 885 variants erdos_885.variants.k_eq_2 and k_eq_3 with explicit witness proofs and adds one shared API lemma (factorDifferenceSet_finite: for n ≥ 1, D(n) is finite since every element is at most n).

For k=2: Ns = {12, 42}. 1 = |3−4| = |6−7| and 11 = |1−12| = |3−14|.
And for k=3: Ns = {1936, 4900, 32400} (= 44², 70², 180²). Common differences {0, 105, 480}.
The main statement and the k=4 variant remain untouched.

The mathematical results are due to [ErRo97] and [Ji99]. This PR only contributes the first machine-checked proofs with witnesses found independently by my automated search system.
AI disclosure: the proofs are generated by MathPaperAI, which is my automated Lean proof-search pipeline (LLM-driven).
Everything is kernel-checked and replayed against this repository's pinned toolchain (lake --wfail build 'FormalConjectures.ErdosProblems.«885»' passes). #print axioms for all three declarations: [propext, Classical.choice, Quot.sound].
I read and verified the proofs myself. Please tell me, if you prefer a formal_proof link!

@github-actions github-actions Bot added the erdos-problems Erdős Problems label Aug 19, 2026
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Formal proofs for erdos_885 variants k_eq_2 and k_eq_3

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