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 e. Prime /\ M e. ZZ ) -> ( P || M <-> P || ( M ^ 2 ) ) )

Proof

Step Hyp Ref Expression
1 2nn
 |-  2 e. NN
2 prmdvdsexp
 |-  ( ( P e. Prime /\ M e. ZZ /\ 2 e. NN ) -> ( P || ( M ^ 2 ) <-> P || M ) )
3 1 2 mp3an3
 |-  ( ( P e. Prime /\ M e. ZZ ) -> ( P || ( M ^ 2 ) <-> P || M ) )
4 3 bicomd
 |-  ( ( P e. Prime /\ M e. ZZ ) -> ( P || M <-> P || ( M ^ 2 ) ) )