1. Given relation

   $R=\{(x,y): x,y\in \mathbb{Z} \text{ and } x+y \text{ is even}\}$

   We must check whether this relation on $\mathbb{Z}$ is:

   • reflexive
   • symmetric
   • transitive

2. Reflexive check

   A relation is reflexive if for every $x\in \mathbb{Z}$, $(x,x)\in R.$

   Here, $x+x=2x,$ and $2x$ is always even for every integer $x$.

   So $(x,x)\in R$ for all $x\in \mathbb{Z}$.

   Hence, $R$ is reflexive.

3. Symmetric check

   A relation is symmetric if whenever $(x,y)\in R$, then $(y,x)\in R$.

   If $(x,y)\in R$, then $x+y \text{ is even.}$

   But $y+x=x+y,$ which is also even.

   Therefore, $(y,x)\in R$.

   Hence, $R$ is symmetric.

4. Transitive check

   A relation is transitive if whenever $(x,y)\in R$ and $(y,z)\in R$, then $(x,z)\in R$.

   Suppose $x+y \text{ is even} \quad \text{and} \quad y+z \text{ is even.}$

   This means $x$ and $y$ have the same parity, and $y$ and $z$ have the same parity. Therefore, $x$ and $z$ also have the same parity.

   So, $x+z \text{ is even.}$

   Hence, $(x,z)\in R$.

   Therefore, $R$ is transitive.

5. Conclusion

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

6. Option check

   • A: reflexive and transitive but not symmetric — false
   • B: reflexive and symmetric but not transitive — false
   • C: an equivalence relation — true
   • D: symmetric and transitive but not reflexive — false

Therefore, the correct answer is: $\boxed{\text{C}}$

7. Comparison with stored answer

   Stored correct answer: C

   This matches our derived answer.