`3n^3+10n^2-5\vdots 3n+1`
`⇒(3n^3+n^2)+(9n^2+3n)-(3n+1)-4\vdots 3n+1`
`⇒n^2(3n+1)+3n(3n+1)-(3n+1)-4\vdots 3n+1`
`⇒(3n+1)(n^2+3n-1)-4\vdots 3n+1`
`⇒-4\vdots 3n+1`
`⇒3n+1∈Ư(-4)={±1; ±2; ±4}`
`⇒n=0; -2/3; 1/3; -1; 1; -5/3`
Vậy `n=0; 1/3; -2/3; -5/3; -1; 1`