Inverse Problem of Criticoids. 193 



and, introducing for convenience two new quantities defined 



by 



A + E, A-E = X2, o>, 

 we get 



b 2 =b — w(o) + 3). 



19. Again, we have 

 P 2 =6(A-1)L + 2A 3 -2A, 

 Q 2 =6(A 2 E+AM + EL) + 3M-24L-4A--2E, 



R 2 = 6(AE 2 + AN + EM)-15M + 12N-20A + 14E, 

 S 2 =6(E-1)N + 2E 3 -2E. 



20. Hence we find 



B 2 = (a> + 3){a>(<w--3) + 3Z>} ? 

 C 2 =&(fl-2)-(2e» + 3)(N-L) + ©(a>n + 3) 

 and, consequently, 



C 2 + & 2 = (5 + 0) 2 )(O-l)-(2ft) + 3)(N-L). 



21. The equations corresponding to (4) and (5) are 



(2A-1){A(A-1) + 3L}=0, . . . (4) 2 

 (2E-1){E(E-1) + 3N}=0, . . . (5) 2 



which give rise to four sets of relations. 



22. First, take the set 



A(A-1) + 3L=0, E(E-1) + 3N = 0; 

 we deduce from it 



3(N-L) = ©(fl-1), (6) 



3(N + L)=-K^ 2 + ^ 2 ) + n ; .... (7) 



and the penultimate combined with the last equation of art. 20 

 gives 



C 2 + 6 2 =(3^ + <»-3)(N-L). 



23. Again, 



N 2 -L 2 =N-L4-0)(O-l) = 4(N-L); 



so that the equation (2) 2 becomes, after reduction and substi- 

 tution, 



&>(a) + 3) + 3Z> + 2a>v/9-4/> + 4a)(a) + 3) = 0. 



24. Write (1) 2 in the form 



B 2 -%b 2 = ±b 2 s/9^4b 2 , 



