Ta có: n²-2n-1$\vdots$2n+3
⇒2(n²-2n-1)$\vdots$2n+3
⇒2n²-4n-2$\vdots$2n+3
⇒n(2n+3)-(7n+2)$\vdots$2n+3
⇒7n+2$\vdots$2n+3
⇒2(7n+2)$\vdots$2n+3
⇒7(2n+3)-17$\vdots$2n+3
⇒2n+3∈Ư(17)={±1;±17}
2n+3=1⇒n=-1 (tm)
2n+3=-1⇒n=-2 (tm)
2n+3=17⇒n=7 (tm)
2n+3=-17⇒n=-10 (tm)
Vậy n∈{-1;-2;7;-10}