`ĐK: n in ZZ,n \ne -1`
`[n+4]/[n+1]=[n+1+3]/[n+1]=1+3/[n+1]`
Để `[n+4]/[n+1] in ZZ` thì `1+3/[n+1] in ZZ`
Hay `3/[n+1] in ZZ`
`=>n+1 in Ư_3`
Mà `U_3={+-1;+-3}`
`@n+1=1=>n=0`(t/m)
`@n+1=-1=>n=-2` (t/m)
`@n+1=3=>n=2` (t/m)
`@n+1=-3=>n=-4`(t/m)
Vậy `n = {-4;-2;0;2}`