$m$ is said to be related to $n$, if $m$ and $n$ are integers and $m-n$ is divisible by $13$. Does this define an equivalence relation?

Now this relation is reflexive as $(x,x)\in R$ as $x-x=0$ is always divisible by $13$ for all $x$.

Similarly the relation is also symmetric as if $(x,y)\in R$ this gives $(y,x)\in R$ as if $x-y$ is divisible by $13$ then $y-x$ is also divisible by $13$ for all $x,y$.

Again this relation $R$ is also transitive as if $(x,y)\in R,(y,z)\in R$ this gives $(x,z)\in R$ as $x-y=13k_1$....(1) [ Since $x-y$ is divisible by $13$] and $y-z=13k_2$....(2) [ Since $y-z$ is also divisible by $13$]

Now adding (1) and (2) we get, 

$x-z=13(k_1+k_2)$. [ For $k_1$ and $k_2$ integers]

This gives $x-z$ is also divisible by $13$.

So the relation is also transitive.

Hence the relation is reflexive, transitive and symmetric so it is an equivalance relation.