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 ) → ( 𝑃 ∥ 𝑀 ↔ 𝑃 ∥ ( 𝑀 ↑ 2 ) ) )

Proof

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