Số tự nhiên `n` khi chia cho `3` luôn có `3` số dư: `0;1;2`
`⇒n∈{3k;3k+1;3k+2}` `(k∈N)`
`TH1:n=3k`
`⇒n^2=(3k)^2`
`⇒n^2=9k^2`
`⇒n^2+n+6=9k^2+3k+6⋮3` `(KTM)`
`TH2:n=3k+1`
`⇒n^2=(3k+1)^2`
`⇒n^2=(3k+1).(3k+1)`
`⇒n^2=3k.(3k+1)+(3k+1)`
`⇒n^2=9k^2+3k+3k+1`
`⇒n^2=9k^2+6k+1`
`⇒n^2+n+6=9k^2+6k+1+3k+1+6`
`⇒n^2+n+6=9k^2+9k+8⋮̸3` `(TM)`
`TH3:n=3k+2`
`⇒n^2=(3k+2)^2`
`⇒n^2=(3k+2).(3k+2)`
`⇒n^2=3k.(3k+2)+2.(3k+2)`
`⇒n^2=9k^2+6k+6k+4`
`⇒n^2=9k^2+12k+4`
`⇒n^2+n+6=9k^2+12k+4+3k+2+6` `⇒n^2+n+6=9k^2+15k+12⋮3` `(KTM)`
`⇒n=3k+1`
`⇒n^2=9k^2+6k+1`
`⇒2n^2=18k^2+12k+2`
`⇒2n^2+n+8=18k^2+12k+2+3k+1+8`
`⇒2n^2+n+8=18k^2+15k+11`
`18k^2+15k+11` chia `3` dư `2`
`⇒2n^2+n+8` không phải là số chính phương
