The given relation is $R = {(x, y) : x, y ∈ Z$ and $x + y$ is even $}$.

Reflexive: For any $x ∈ Z$, $x + x = 2x$, which is even. So, $(x, x) ∈ R$. Thus, $R$ is reflexive.

Symmetric: If $(x, y) ∈ R$, then $x + y$ is even. Since $x + y = y + x$, $y + x$ is also even. So, $(y, x) ∈ R$. Thus, $R$ is symmetric.

Transitive: If $(x, y) ∈ R$ and $(y, z) ∈ R$, then $x + y$ is even and $y + z$ is even. Then $(x + y) + (y + z) = x + 2y + z$ is even. Since $2y$ is even, it follows that $x + z$ is even. So, $(x, z) ∈ R$. Thus, $R$ is transitive.

Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.