Problem
erdos:204False ↔ ∃ n a, have D := {d | d ∣ n ∧ d > 1}; (∀ (x : ℤ), ∃ d ∈ D, x ≡ a d [ZMOD ↑d]) ∧ ∀ d ∈ D, ∀ d' ∈ D, d ≠ d' → (∃ x, x ≡ a d [ZMOD ↑d] ∧ x ≡ a d' [ZMOD ↑d']) → d.gcd d' = 1
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