88 ON THE ALGEBRAICAL SOLUTION OF 



sum and the difference of two squares has led to solutions in 

 all cases where n is greater than 4, it is natural to try this 

 assumption here. Now if 



then P 4 -P 1 *=(w 2 +t) 4 -(*-0 8 ) 4 =8M 8 v a (***+v*)' (2) 



Now every bi quadrate is of the form 5n or 5n+l ; hence 

 three of the roots on the right-hand side of (1) are multiples 

 of 5 always. If we assume that P t is the root prime to 5, then 

 from (2) 8wV(M 4 +*; 4 ) must be divisible by 5 4 . But this is 

 impossible so long as u and v are both prime to 5. Hence, 

 since P and P l are both prime to 5, one of u, v is a multiple of 

 5 and the other prime to 5, and u z v z must therefore be divisible 

 by 5 4 . Since u z v z is always to be divisible by 5 4 , this suggests 

 that possibly u and v are both squares, the one always divisible 

 by 5, and the other always prime to 5. This again suggests 

 as a suitable transformation u=(x 2 +y 2 ) 2 , v=(x z y 2 ) 2 , or on 

 analogy with the results of Question 2, M=(z 4 +4/ 4 ) 2 , 

 v=(x 4 4/ 4 ) 2 , each of which manifestly satisfies the required 

 condition so long as x and y are both prime to 5. Hence we 

 are led to the assumptions 



P =(a; 2 +i / 2 ) 4 +(a; 2 -i/ 2 ) 4 , P 1 =(x 2 + 2 / 2 ) 4 -(x 2 - 2 / 2 ) 4 (3) 



or P =(* 4 +4t/ 4 ) 4 +(a; 4 -42/ 4 ) 4 , P 1 =(a; 4 +4?/ 4 ) 4 -(a: 4 -42/ 4 ) 4 (4) 



of which the latter is merely the specialised form obtained from 

 the former by writing in it a; 2 instead of x and 2y 2 instead of y, 

 It now remains to choose P 2 , P 3 , P 4 in such a way as to 

 make P 2 4 +P 3 4 +P 4 4 a homogeneous symmetrical function of 

 x and y of the 32nd or 64th degree according as we take 

 assumption (3) or (4). This we may do in a variety of ways, 

 but probably the simplest forms which present themselves 

 would be 



P z =2(x*-y*)[2xy(x 2 +y*)],] 



P= 



2 



or P s =2(x*-y*)(xy*) 

 and the forms obtained by replacing x by x z and y by 2y 2 . 



