62 ON THE ALGEBRAICAL SOLUTION OF 



2. The equation P 1 4 +P 2 4 =P' 1 4 +P' 2 4 +* 

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

 particular solution be known. For by putting P^x+y, 

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



8x 3 y+ 8xy 3 =8u 3 v+ Sui^+ k 



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

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



QUESTION 2. Solve the equation 



P 1 4 +P 2 4 +P 3 4 =P / 1 4 +PV+^ / 3 4 (1) 



First method. We may assume 



(X 1 r+a) i +(x z r+b)*+(x 3 r+c)*=(x 1 r+d^+(x z r+e^+(x a r+f)* (2) 

 where by hypothesis 



a 4+tf +c *=d*+e*+f*. (3) 



On expansion and rearrangement (2) becomes 



ayt- i?^f*x a )r=Q. (4) 

 To make the coefficient of r vanish we must take 



* 3 H(a 3 -rf 3 )*i+ (& 3 -e3)o; 2 !/(/ 3 -c 3 ). (5) 



Equation (4) is then satisfied by taking 





* - 



2[{(a - d>, + (6 - ifrfK/* - c 3 )'^ +/c + e) - {(a 3 - 



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

 x 3 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, f=b, e=c. 



Hence a; 3 =[2a 3 a; 1 + (6 3 -c 3 )x 2 ]/(6 3 -c 3 ), 



and 



