Yes. It is equivalence relation.
Check Reflexive: Consider a function f on Z to Z. f(x)-f(x)=0 always for all x in Z. Thus f(x)-f(x) = C where C ( =0 ) ∈ Z. Thus (f, f) ∈ Z
Check Symmetric: Let (f1, f2) ∈ R => f1(x)-f2(x) = C where C ∈ Z . Now f2(x)-f1(x)=-C and since C ∈ Z => -C ∈ Z. Thus (f2, f1) ∈ Z
Check Transitive: Let (f1, f2) ∈ R and (f2, f3) ∈ Z. Thus f1(x)-f2(x)=C1 and f2(x)-f3(x)=C2 where C1, C2 ∈ Z. Thus f1(x)-f3(x)=f1(x)-f2(x) + f2(x)-f3(x) = C1+C2 = K (say) . Since C1, C2 ∈ Z => K ∈ Z. Hence, (f1, f3) ∈ Z
