Metamath Proof Explorer


Theorem odzid

Description: Any element raised to the power of its order is 1 . (Contributed by Mario Carneiro, 28-Feb-2014)

Ref Expression
Assertion odzid ⊢ N ∈ ℕ ∧ A ∈ ℤ ∧ A gcd N = 1 → N ∥ A odℤ ⁡ N ⁡ A − 1

Proof

Step Hyp Ref Expression
1 odzcllem ⊢ N ∈ ℕ ∧ A ∈ ℤ ∧ A gcd N = 1 → odℤ ⁡ N ⁡ A ∈ ℕ ∧ N ∥ A odℤ ⁡ N ⁡ A − 1
2 1 simprd ⊢ N ∈ ℕ ∧ A ∈ ℤ ∧ A gcd N = 1 → N ∥ A odℤ ⁡ N ⁡ A − 1