[tex]\dfrac{n^2+n+1}{n-1}=\dfrac{(n-1)^2+3(n-1)+3}{n-1}=n-2+\dfrac3{n-1}[/tex]
The remainder term [tex]\dfrac3{n-1}[/tex] will never vanish, and will always be rational as long as the denominator is not 1, 3, or -3.
[tex]n-1=1\implies n=2[/tex]
[tex]n-1=3\implies n=4[/tex]
[tex]n-1=-3\implies n=-2[/tex]
