For any integer $n$, we have $n / n \Rightarrow n R n$
So, $n R n$ for all $n \in Z \Rightarrow R$ is reflexive.
Now, $2 / 6$ but $6+2 \Rightarrow (2,6) \in R$ but $(6,2) \notin R$
So, $R$ is not symmetric.
Let $(m, n) \in R$ and $(n, p) \in R$
So, $R$ is transitive. Hence, $R$ is reflexive and transitive but it is not symmetric.