I solved this both for N,Z
For N,let 1066|a,b,c,d for simplicity.Then we need a3+b3+c3+d3=102,using the identity 13+23+....+n3=n2(n+1)242,we get (a1,b1,c1,d1)=(1,2,3,4) and all its permutation, since then a1+b1+c1+d1=10.Then the numbers are 1066,2.1066,3.1066,4.1066
Now for Z,we take a=5.1066,b=c=1066,d=−3.1066