go ON THE ALGEBRAICAL SOLUTION OF 



7. Finally we may solve the equation 



P 9 *=PS+PS+ . . . +P. 4 (1) 



for all values of n greater than 2, when a particular solution 

 is known. 



To do this we may employ the method of Diophantos, and 

 take 



We have then to solve the equation 

 (y Q r 2 +Xor+a )*=(x 1 r+a 1 )*+(x 2 r+a 2 )*+ +(x n r+a n )*. (2) 



Now if a 2 =a 1 4 +a 2 4 + . . . +a, 4 , 



it is clear that on expansion of the square and biquadrates in 

 (2) we can make the coefficients of r and r 2 both vanish by 

 solving linear equations for x and y Q) and therefore on division 

 by r 3 , (2) becomes a linear equation in r. The value of r found 

 from this equation with the values of x and y Q already found 

 substituted in it makes (2) an identity and it is an algebraical 

 solution which clearly presents on integralisation the roots of 

 the biquadrates as functions of the fourth degree in the n 

 variables x l9 x z , . . ., x n . 



It is further clear that the equation 



*o^o 2 =^i 4 +^2 4 + +MV (3) 



is in general soluble by the same method if a particular solution 

 is known ; thus if X =n, X 1 =X 2 = . . . =X W =1, we may take 

 as a particular solution P =a 2 , P 1 =P 2 = . . . =P n =a. 



For particular values of n, other special solutions of (1) 

 and (3) may be found which will give in general different 

 numerical results from the foregoing. 



Thus take the case of (1) when n=3. We have identically 

 ) 2 +(6 2 w 2 -aV) 2 =(aw) 4 +(6^) 4 +(av) 4 +(6v) 4 . (4) 



Let us now choose a, 6, u, v so that 



To do this we have necessarily 



a=x 2 ?/ 2 , b=2xy 

 so that 2xyu=(x 2 -\-y*)v. 



