`3n+5 \vdots 2n+1`
`=> 6n+10 \vdots 2n+1`
`=> 6n + 3 + 7 \vdots 2n+1`
`=> 3(2n+1) + 7 \vdots 2n+1`
`=> 7 \vdots 2n+1` ( vì `3(2n+1) \vdots 2n+1` )
`=> 2n +1 \in Ư(7)={±1 ; ±7}`
`=> 2n \in {0 ; -2 ; 6 ; -8}`
`=> n \in {0 ; -1 ; 3 ; -4}`
Mà `n \in N => n \in {0 ; 3}`
Vậy `n \in {0;3}`