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`a,` Để `P\inZZ`
`=> 4\vdots 2n-1`
`=> 2n-1 \in {1;2;4;-1;-2;-4}`
`=> 2n \in {2;3;5;0;-1;-3}`
`=> n \in {1;3/2;5/2;0;-1/2;-3/2}`
Mà `n\inZZ`
`=> n \in {1;0}`
Vậy `n={1;0}`
`b, D = [4n-1]/[n+2] = [4n+4-5]/[n+2] = [2(n+2)-5]/[n+2] = 2 - 5/[n+2]`
Để `D\inZZ`
`=>5/[n+2] \in ZZ`
`=> 5 \vdots n + 2`
`=> n + 2 \in {1;5;-1;-5}`
`=> n \in {-1;3;-3;-7}`
Vậy `n = {-1;3;-3;-7}`