28 On Non-Linear Coresolvents. 



and form the function F supposed to be such that 



F=£ 2 + 2(aj)+AP)j + 5| 2 + 2|jPj? + BP 2 ... (4) 

 and wherein the multipliers a, A, b, ft, and B are indeterminate. 



2. "We know that £>, q and P can be expressed as rational func- 

 tions of y, and by the theory of equations we also know that p, 

 q, and P, and, consequently, F can be expressed as rational and 

 entire functions of integral powers ofy, none of which transcends 

 the third power. "We are therefore at liberty to write 



p = X^ 3 + py 2 + vy + p (5) 



where X, p, v, and p, are linear functions of a, A, b, j3, and B. 

 By means of the four conditions 



X = o, p. = o, v = o, p = o (6, 7, 8, 9) 



eliminate four of the indeterminate multipliers, say A, b, £}, and 

 B from the dexter of (4). Then, these four conditions causing 

 P to vanish, we have a result which we may write, 



q* +2(ap+ CV)q + ~Dp* + 2EPp + HP 2 =o...(10) 

 and in which C, D, E, and H are now to be regarded as functions 

 of the coefficients of the quartic and of the uneliminated inde- 

 terminate a. It will be remarked that this indeterminate a only 

 enters linearly into C, D, E, and H. 



3. Put (10) under the form 



(j + «|) + CP)2 = (a 2 — D)^2 + 2 (a C— E) 



Vp + (C 2 — H)P 2 (11) 



then the dexter of (11) will be a perfect square if 



(«2_D) (C 2 — H) = (« C — E) 2 (12) 



that is to say, if 



D H — « 2 H — C 2 D = E 2 — 2 a C E...(13) 

 But (13) is a cubic in a. Determining a so as to satisfy it, and 

 then extracting the square root of either side of (11) we have 



q + a p + C?=^_ | y(« 2 _D) j p+ :7 === ) PJ (14) 



whence we may obtain an equation which I shall write 

 j + Ip + JP =o (15) 



4. Had the equation in y been a cubic instead of a quartic, we 

 might have dispensed with the term \y of F. Thus we should 

 have had for the solution of (13) two disposable indeterminates, 

 say a and b, in place of a only, and in such case (13) would have 

 been capable of taking the form 



N (a 3 + 3 L a 2 + 3 M a) = N Q (16) 



wherein N is free from b, and b enters L linearly and M and Q 

 to the second and third degrees respectively. Now if we use the 

 arbitrary quantity b for the purpose of satisfying the condition 



M = L 2 (17) 



