Metamath Proof Explorer


Theorem prmdvdssq

Description: Condition for a prime dividing a square. (Contributed by Scott Fenton, 8-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014) (Proof shortened by SN, 21-Aug-2024)

Ref Expression
Assertion prmdvdssq ( ( 𝑃 ∈ ℙ ∧ 𝑀 ∈ ℤ ) → ( 𝑃𝑀𝑃 ∥ ( 𝑀 ↑ 2 ) ) )

Proof

Step Hyp Ref Expression
1 2nn 2 ∈ ℕ
2 prmdvdsexp ( ( 𝑃 ∈ ℙ ∧ 𝑀 ∈ ℤ ∧ 2 ∈ ℕ ) → ( 𝑃 ∥ ( 𝑀 ↑ 2 ) ↔ 𝑃𝑀 ) )
3 1 2 mp3an3 ( ( 𝑃 ∈ ℙ ∧ 𝑀 ∈ ℤ ) → ( 𝑃 ∥ ( 𝑀 ↑ 2 ) ↔ 𝑃𝑀 ) )
4 3 bicomd ( ( 𝑃 ∈ ℙ ∧ 𝑀 ∈ ℤ ) → ( 𝑃𝑀𝑃 ∥ ( 𝑀 ↑ 2 ) ) )