62 ON THE ALGEBRAICAL SOLUTION OF 



2. The equation Pi*+Pf=P\*+P'+k 

 is always soluble whatever be the value of k, provided a 

 particular solution be known. For by putting P^ 

 P 2 =uv, P\=xy 9 and P' 2 =u+v it becomes 



which is an indeterminate cubic in x and u. We proceed to 

 solve the case where &=P' 3 4 P 3 4 . 



QUESTION 2. Solve the equation 



PS+Pf+Pjf^P'S+P'S+P'if (1) 



First method. We may assume 



(Xjr+a)^ (x z r+ 6) 4 + (x 3 r+ c) 4 = (x^r+ d) 4 + (z 2 r+ e) 4 + (z 3 r+/) 4 (2) 

 where by hypothesis 



4 +& 4 +c 4 =d 4 +e 4 +/ 4 . (3) 



On expansion and rearrangement (2) becomes 



a . 3 ) r== 0. (4) 



To make the coefficient of r vanish we must take 



* 3 H(a 3 -d 3 )*i+ (& 3 -e 3 W/(/ 3 -c 3 ). (5) 



Equation (4) is then satisfied by taking 



_ 



2[(a-d)x 1 *+(b-e)x 2 *+(c-f)x 3 *] 



a - <P) V + (ft* - e2)o: 2 2}(/3 - c 3 )^/ 2 +/g + c 2 ) - (c 



on substituting for a: 3 its value given by (5). These values of 

 x s and r when substituted in (2) render it an identity and 

 constitute a solution which is clearly algebraical. 



We may satisfy equation (3) in several ways. Thus we 

 may put 



d=a, /=&, e=c. 



Hence # 3 =[2a 3 # 1 + (& 3 -c 3 )z 2 ]/(6 3 -c 3 ), 



and 



