Vì `2n+1` là `Ư(3n+2)`
`=> 3n+2 \vdots 2n + 1`
`=> 2(3n + 2) \vdots 2n+1`
`=> 6n +4 \vdots 2n+1`
`=>( 6n+3 )+1 \vdots 2n+1`
`=> 3(2n+1) +1 \vdots 2n+1`
`=> 1 \vdots 2n+1`
`=> 2n+1 \in Ư(1)={1 ; -1}`
`=> 2n \in {0 ; -2}`
`=> n \in {0 ; -1}`
Vậy `n \in {0; -1}`