my soution was $\frac{n^2+n-27}{n-5}=\frac{n^2-10n+25+11n-55+3}{n-5}=n-5+11+\frac{3}{n-5}$ $n-5\in{-3,3,-1,1}$

Taking $n-5=x$, we obtain: $x|(x+5)^2+(x+5)-27\Leftrightarrow x|3 \Leftrightarrow n-5|3$

LOL! This is simple polynomial division theorem: $n^2 + n - 27 = d(n)(n-5) + r(n) \implies r(n)$ is constant $\implies $ sub in $n = 5 \implies r(5) = 3 \implies n-5 \mid 3$.

IDMasterz wrote: LOL! This is simple polynomial division theorem: $n^2 + n - 27 = d(n)(n-5) + r(n) \implies r(n)$ is constant $\implies $ sub in $n = 5 \implies r(5) = 3 \implies n-5 \mid 3$. So, 2,4,6,8.