86 ON THE ALGEBRAICAL SOLUTION OF 



in one only of the two places in which it occurs on the right- 

 hand side of (6), say the second, write equation (6) as 

 [(x 2 + 3i/ 2 ) 2 +4z 4 ] 4 = [(x 2 + 3t/ 2 ) 2 -4z 4 ] 4 + (4i/2) 4 [(a; 2 + 3y 2 ) 4 + (2z 2 ) 4 ] 



This is therefore an algebraical solution, by means of a 

 single formula, of the equation 



P *=PS+P Z *+ . . . +P, 4 , 

 for all values of s greater than 6. 



As before, the defect of all these solutions is the unnecessary 

 peculiarity possessed by one (or more) of the roots of the 

 biquadrates on the right-hand side of the equation, viz., 

 that it is always equal to the sum of the roots of two of the 

 remaining biquadrates. 



5. From the foregoing results other solutions may be 

 obtained for various values of n. For if we have solved, 

 arithmetically or algebraically, the equations 



then each of the biquadrates on the right-hand side of the 

 equation 



may be expressed as the sum of s biquadrates so that this 

 equation gives a biquadrate equal to the sum of r+k(s 1) 

 biquadrates, where fc=0, 1, 2, 3, . . ., r. This is obvious and 

 calls for no special exemplification. 



Also, for particular values of n, other independent formulae 

 may be found. 



For example, let w=16. Then since 



(2u+ v)*+ (u+ 2v)*+ (u- v)*= 18(u?+ uv+ v 2 ) 2 , 

 we have, on putting u=2xy+y z , v=x 2 y z , 



so that (x 2 +4:xy+y z )*+(2x 2 +2xy-y 2 )*+(x 2 -2xy-2y z ) 



