Is $\rho$ equivalence relation and, if not, specify all the ways to expand it to the equivalence relation. On the set $\Bbb Z \times\Bbb Z$, the relation $\rho$ is given as it follows:
$(a,b) \rho (c,d) \iff (2 | (a-c) $ or $ 3|(b-d))$
Is $\rho$ equivalence relation and, if not, specify all the ways to expand it to the equivalence relation.

First, I checked if it's equivalence relation.

reflexivity

$(\forall a,b \in \mathbb Z )(a,b) \rho(a,b)$
This is true because $(2|(a-a)$ or $3|(b-b)) \iff(2|0$ or $3|0)$ is true.

symmetry

$(\forall a,b,c,d \in \mathbb Z )(a,b) \rho(c,d) \Rightarrow (c,d)\rho(a,b)$
Also true, $(a,b) \rho(c,d) \Rightarrow (2|(a-c)$ or $3|(b-d))\Rightarrow (2|(c-a)$ or $3|(d-b)) \Rightarrow (c,d)\rho(a,b)$.

transitivity

$(\forall a,b,c,d,e,f \in \mathbb Z )((a,b) \rho(c,d)$ and $(c,d) \rho(e,f)) \Rightarrow (a,b)\rho(e,f)$
Not true, for example while$(5,4) \rho (3,2)$ and $(3,2)\rho(2,-1)$ is true, $(5,4)\rho(2,-1)$ isn't.
Ok so now I have to expand this relation so that it can also be transitive. How do I do that?
I'm not sure what the question means by "all the ways" to make it an equivalence relation, however there is in general a way to make a symmetric, reflexive but not transitive $\rho$ into an equivalence relation.

Regardless of what $\rho$ is, we can define a new equivalence relation $\tilde{\rho}$ such that $x\tilde{\rho} y$ if and only if $x\rho y$ or there exists $z$ such that $x\rho z$ and $y\rho z$.

Note that symmetry and reflexivity of $\tilde{\rho}$ follow from symmetry and reflexivity of $\rho$.

In your case I would recommend writing explicitly what this $\tilde{\rho}$ is.
Hint: for any $a, b, c, d \in \mathbb{Z}$, we have $(a, b) \mathrel{\rho} (c, b)$ and $(c, b) \mathrel{\rho} (c, d)$.