Đáp án:
`n cancelvdots 3,n in NN^{**}`
`=>` \(\left[ \begin{array}{l}n=3k+1(k \ge 0,k \in N^*)\\n=3k+2(k \ge 0,k \in N^*)\end{array} \right.\)
`**n=3k+1`
`=>n^2=(3k+1)^2`
`=(3k+1)(3k+1)`
`=3k.3k+3k+3k+1:3` dư 1`
`=>n=3k+1` thì `n^2 cancelvdots 3(1)`
`**n=3k+2`
`=>n^2=(3k+2)^2`
`=(3k+2)(3k+2)`
`=3k.3k+3k.2+3k.2+4`
`=3k.3k+3k.2+3k.2+3+1:3` dư 1
`=>n=3k+1` thì `n^2 cancelvdots 3(2)`
`(1)(2)=>n in NN^{**},n cancel vdots 3` thì `n^2:3` dư 1.
