8o ON THE ALGEBRAICAL SOLUTION OF 

 3. QUESTION 2. Solve the equation 



Po^P^+P^+Pg^P^+Ps^Pe 4 . (1) 



In a former paper 1 there occurs the identity 



[6c 3 (c 8 -6 8 + 22z,, 4 )] 4 + [2c*2x, 4 -& 4 (c 8 -6 8 )] 4 

 =[6c 3 (c 8 -6 8 -2Sa: n 4 )] 4 +[2c 4 Sa; B 4 +6 4 (c 8 -6 8 )] 4 +[26c(c 8 -6 8 )]* 



From this we see that if 



6 8 -c 8 =2Sx B 4 (2) 



then [2c 4 Sx ll 4 -6 4 (c 8 -6 8 )] 4 =[6c 3 (c 8 -& 8 -2Sa: n 4 )] 4 + 



[2c 4 Sa; n 4 +6 4 (c 8 -6 8 )] 4 +[26c(6 8 -c 8 )] 4 (a; 1 4 +a; 2 4 + .... +,*) 

 i.e. on making use of (2) and dividing each root by c 8 & 8 , 



a result which is immediately obvious from the fact that we 

 have identically 



(&4+ C 4)4= (&4_ C 4)4 + 8&4 C 4( 6 8 + C 8) 



= (& 4 -c 4 ) 4 + (26c 3 ) 4 + 86 4 c 4 (6 8 -c 8 ). 



Now the equation (2) is soluble algebraically when n is of 

 the form 2 r+2 . For, if in the identity 



we replace x by ic 4 and y by 4/ 4 we derive 

 (z 4 +4t/ 4 ) 4 -(z 4 -4y 4 ) 4 =2(2zi/) 4 (:e 

 so that multiplying each side by (a; 4 +42/ 4 ) 4 +(a; 4 4?/ 4 ) 4 we have 



+ (* 4 -42/ 4 ) 4 ] (4) 



Hence as a solution of equation (1) we have, putting 

 6=x 4 +4?/ 4 , c=z 4 -4t/ 4 in (3) 



P 6 = 8xy 3 (x*+ 4?/ 4 ) (a; 4 - 



1 Viz., Part II. 



