82 ON THE ALGEBRAICAL SOLUTION OF 



COR. We may now give algebraical solutions of the 

 equation 



where r=6+2ra. 



For we have the identity 1 



m .j.,.* - [6 ( &< ~ &')]' + [ c '( < ' 



(&'+')' 



_[6 3 (6 4 -3c 4 )] 4 + [c 3 (c 4 -36 4 )] 4 + [26c(6 4 -c 4 )] 4 (6 4 +c 4 ) 



(6 4 +c 4 ) 4 



Put now b=u 1 z , c=2v x 2 and this becomes 



) 4 (7) 



Again, equation (5) is 

 [Z 4 + T 4 J 4 = [X 4 - 7 4 ] 4 + [2Z F 3 ] 4 + [4a;t/Z 2 7J 4 (x 8 + 16/ 8 ) 



+ [4ajyZ F 2 ] 4 (a; 8 + 16y 8 ) (8) 



If then we put x=u 1 3 , y=2w 1 3 , a; 8 + 16i/ 8 becomes M 1 24 +2 12 v 1 24 

 which is expressible rationally as the sum of four biquadrates. 

 Making these substitutions in (8) we see that if X' and T' 

 denote the new values of X and Y, then [X'*+ 7' 4 ] 4 is ex- 

 pressible simultaneously as the sum of 6, 8, or of 10 rational 

 biquadrates which may all be made integral by multiplying 

 each root by (% 8 + IB^ 8 ). As a matter of fact we have 



(% 12 -64u 1 12 ) 4 ] 4 = [( 



2 ) 34 + [8 W( 

 [% 24 + 2 12 V 4 ] 

 2 -64t; 1 12 )] 4 [w 1 24 + 2 1 V 4 ] 

 = [K 12 + 64V 2 ) 4 - (M 1 12 



1 See 3 of Part I. 



