72 ON THE ALGEBRAICAL SOLUTION OF 



Again, putting d0, we have that if Xa 4 + [i& 4 =Xc 4 , then also 

 X[2a(a 2 +ac+c 2 )(a 3 -c 3 ) 2 X 2 +36 4 (a 2 +ac+c 2 )(a 3 -c 3 )X(i 



+ (3c+a)&V] 4 



+ (x[6(a 2 +ac+c 2 )(a 3 -c 3 ) 2 X 2 +3(c+a)6 5 (a 3 -c 2 )X|i+26V 2 ] 4 

 =X[2c(a 2 +ac+c 2 )(a 3 -c 3 ) 2 X 2 +36 4 (a 2 +ac+c 2 )(a 3 -c 3 )6 4 X(i 



+ (c+3a)&V] 4 

 + ^[36(a 2 +ac+c 2 )(a 3 -c 3 ) 2 X 2 +3(c+a)(a 3 -c 3 )& 5 X!A] 4 . 



Thus to solve P 1 4 +5P 2 4 =P' 1 4 +5P' 2 4 , since ! 4 +5-2 4 =3 4 +5-0 4 , 

 we derive 



2. The equation XPj 4 + [iP 2 4 = 

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

 particular solution be known. For by putting P-^ 

 P z =uv, P' l =xy, and P' z =u+v, it becomes 



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

 solve the case where Z=v(P' 3 4 P 3 4 ). 



QUESTION 2. Solve the equation 



xP 1 4 + ! *P 2 4 + I ,p 3 4 =xPY+ I ,p' 2 4 + J 'P / 3 4 > (i) 



knowing a particular solution, say 



Xa 4 +[/.& 4 +j/c 4 =Xd 4 +(z.e 4 +i'/ 4 (2) 



Here we may assume for equation (1) 

 X(Xir+ a) 4 + \.(x 2 r+ 6) 4 + v(x a r+ c) 4 = X(x 1 r+ d)*+ i>.(x 2 r+ e)* 



+"(*V+/) 4 (3) 

 which on expansion and rearrangement becomes 



To make the coefficient of r vanish we must take 



=0 (4) 

 (5) 



