80 ON THE ALGEBRAICAL SOLUTION OF 

 3. QUESTION 2. Solve the equation 



' pj-PS+PS+PS+PS+Ptf+PS. (i) 



In a former paper 1 there occurs the identity 



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



From this we see that if 



& 8 -c 8 =22z, 4 (2) 



then 



i.e. on making use of (2) and dividing each root by c 8 6 8 , 

 (64+c 4 ) 4 =(6 4 -c 4 ) 4 +(26c 3 ) 4 +(2&c) 4 (a: 1 4 +a: 2 4 + . . . +x n *) (3) 



a result which is immediately obvious from the fact that we 

 have identically 



(&4 + c 4 ) 4 = (6 4 -c 4 ) 4 + 86 4 c 4 (6 8 + c 8 ) 



=(6 4 -c 4 ) 4 +(2&c 3 ) 4 +8& 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 x* and y by 4i/ 4 we derive 

 (a; 4 +42/ 4 ) 4 -( 4 -4i/ 4 ) 4 =2(2^) 4 (a 

 so that multiplying each side by (# 4 +4?/ 4 ) 4 +( 4 4i/ 4 ) 4 we have 



(4) 



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

 c=a; 4 -4i/ 4 in (3) 



1 Viz., Part H. 



