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 



P =^ 2 +v 2 , P 1 =u 2 -v 2 f 

 then P 4 -P 1 4 =(w 2 +v 2 ) 4 -(^ 2 -v 2 ) 4 =8^V(^ 4 +t; 4 ). (2) 



Now every biquadrate is of the form 5n or 5w+l ; hence 

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

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

 from (2) 8wV(w 4 +v 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 2 v 2 must therefore be divisible 

 by 5 4 . Since u 2 v 2 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 2 y 2 ) 2 , or on 

 analogy with the results of Question 2, it=(# 4 +42/ 4 ) 2 , 

 v=(a; 4 4f/ 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 =(* 2 +2/ 2 ) 4 +(* 2 -2/ 2 ) 4 > P 1 =(* 2 +2/ 2 ) 4 ~(* 2 -2/ 2 ) 4 (3) 



or P =(z 4 +42/ 4 ) 4 +(z 4 -42/ 4 ) 4 , P 1 =(x 4 +4i/ 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 x 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-\-P^+P 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 2 =2(x*-y*)[2xy(x 2 +y 2 )], 



P 3 =2(x*-y*)[2xy(x 2 -y 2 )l or P 3 = 



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



