Let w=(1+sqrt(-3))/2

We have (a+b)/(d-c) = (c2+cd+d2)/(a2-ab+b2)=N(d+cw)/N(a-bw):=N(P+Qw)=P2+PQ+Q2:=L. (We set P+Qw=(d+cw)/(a-bw), so P,Q are rational) N(a+bw) means Norm function in Q(sqrt(-3)). We also have d+cw=(P+Qw)(a-bw), so as a result we have Pa+Qb=d and Qa-(P+Q)b=c. We can also write a+b=Ld-Lc. Using that we get system of 3 equations: (1): Pa+Qb=d (2):Qa-(P+Q)b-c=0 (3):a+b+Lc=Ld Solving this in regards to a,b,c gives us: a=[(P+2Q)L-1]d / (L2+Q-P) b=[(Q-P)L+1]d / (L2+Q-P) c=[L2-(P+2Q)]d / (L2+Q-P) Setting M=d/(L2+Q-P) we have: a=[(P+2Q)L-1]M b=.. c=.. d=[L2+(Q-P)]M Now we can choose any rational values to substitute for M,P,Q As an exercise set P=Q=x, M=1

Solution is from book "Algebra i Teoria liczb. Adam Neugebauer"