Inverse Problem of Criticoids. 191 



9. This last is obtained from 



C = (a + ey-4;ae(a + e)+(g-f+n)7) 

 = (a 2 — e 2 +g — /+ n)rj 



by eliminating g— /and introducing 6. 



10. These relations give rise to the system 



B^itfnWn^&y, ..... (1) 



C + b=-i(N-L)(n + ^rf^b), . . (2) 



*,= -i(nW?7=S), (3) 



wherein the radical is to be taken with the same sign 

 throughout. 



11. Thus far the process is direct. Now invert the problem 

 and suppose that we have given us the equation in z, and that 

 we are asked whether it can be made to take a binomial form. 

 The answer is this. When (1) and (2) are identically satisfied 

 it can be made to take such a form. And since (3) only de- 

 termines a— e, we may give any value we please to one of the 

 quantities a, e, and we shall nevertheless have four disposable 

 quantities, /, g, h, and k, wherewith to satisfy the four out- 

 standing equations which, apart from (1) 7 (2), and (3), have 

 still to be satisfied. 



12. Proceeding thus we shall be led to a symbolical teror- 

 dinal which I write 



/(D> + .^F(D>=0 ; 

 wherein 



/(D) = [D] 3 + 3a[D] 2 + 3/T> + A 3 



F(D) = [D] 3 + Se[DJ + SgJ) + *, 



and F(D) is obtained from /(D) by the change of a,f, Ii } into 

 e y g, k respectively. 



13. Now let 



h = S(a-l)f-a(a-l)(2a-l), . . . (iv) 



*=3(«~l)>-<«-l)(2i-l); . . . (v) 

 then 



/(D) = (D + a-l){D 2 + 2(a-l)D + 3/-a(2a-l)^ 



F(D) = (D + g-l){D 2 -f2(6~l)D + 3a-K26-l)}, 



and the factors of both /(D) and F(D) are in arithmetical 

 progression ; and the transfer of a? 1 from the left to the right 

 does not affect this property. 



14. But (iv) and (v) are respectively equivalent to 



P + 3L = 0, (4) 



S + 3N=0 (5) 



P2 



