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TheoremFord's theorem on totient multiplicities#●ready to prove

For every natural number \(k \ge 2\) there is a natural number \(m\) such that exactly \(k\) natural numbers \(x\) satisfy \(\varphi(x) = m\), where \(\varphi\) is Euler's totient function.

theorem_wanted ford_totient_multiplicity :
    ∀ k : ℕ, 1 < k → ∃ m : ℕ, Set.ncard {x : ℕ | Nat.totient x = m} = k
Sources
  • Kevin Ford, The Number of Solutions of phi(x) = m, Annals of Mathematics 150 (1999), DOI 10.2307/121103. Also: https://en.wikipedia.org/wiki/Euler%27s_totient_function

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