Let $x,y\ge 1$ be integers such that, for all $n\ge 1$, the set of primes dividing $x^n-1$ is equal to set of primes dividing $y^n-1$. Must $x=y$?

AMS 11 · solved (literature)

theorem erdos_1214 :
    answer(True) ↔ ∀ x y : ℕ, x ≥ 1 → y ≥ 1 →
      (∀ n : ℕ, n ≥ 1 → { p : ℕ | p.Prime ∧ p ∣ x ^ n - 1 } = { p : ℕ | p.Prime ∧ p ∣ y ^ n - 1 }) →
      x = y