84 ON THE ALGEBRAICAL SOLUTION OF 

 4. QUESTION 3. Solve the equation 



Pf-Pf+Pf+Pf+Pf+Pt+ff+Pi* (1) 



In the identity 



(H- 1 )=(-!)+ 8*+ 8* 



put f=(o; 4 +2/ 4 +z 4 )/8 and we obtain 



4 +y 4 +2 4 +8 \4/a4 +y 4 +2 4_ 8 y 4 4 4 (x 4 +y 4 +s 4 ) 3 (2) 



8 / \ 8 / 64 



Let us now choose x, y, and z so that 



To do this, since we have identically 



it will be sufficient to take 



x=2ab+b 2 , y=a?b z , z=a z +2ab, 

 for since this makes x z +xy+y z =(a z +ab+b z ) z , we shall have 1 



(2a6+6 2 ) 4 +(a 2 -6 2 ) 4 +(a 2 +2a6) 4 =2(a 2 +a6+6 2 ) 4 . 

 Making these substitutions (2) becomes 



J L 4 



+ (a 2 +2a6) 4 +2(a 2 +a6+6 2 ) 12 /16. (3) 



Now since 2 - 



(3) may be written 



[(o 2 +a6+ 6 2 ) 4 +4] 4 =[(a 2 +a6+6 2 ) 4 -4] 4 + (8a6+46 2 ) 4 + (4a 2 -46 2 ) 4 



This is an algebraical solution of equation (1) and it may 

 be integrated by multiplying each root by (c z +cd+d 2 ), or 



1 This solution of the equation 2P 4 =P! 4 + P 2 4 + P 3 4 involves the assumption 

 P 1= P 2 + P 3 , a restriction which the identity 



2-484813 4 = 575528' + 155873* + 1 16745 4 

 shows to be unnecessary. 



