INDETERMINATE QUARTIC EQUATIONS 85 



simply by taking as a particular case ca, d=b, in which case 

 it becomes 



[(a 2 +a6+6 2 ) 4 +4] 4 =[(a 2 +a6+6 2 ) 4 -4] 4 +(8a6+46 2 ) 4 -f(4a 2 -46 2 ) 4 

 + (4a 2 +8a6) 4 +2 4 (a 2 +a6+6 2 ) 8 [(a 2 -6 2 ) 4 +(2a6+6 2 ) 4 +(2a6+a 2 ) 4 ]. 

 For example a=2, 6=1 gives 



2405 4 = 2397 4 + 784 4 + 490 4 + 294 4 + 32 4 + 20 4 + 1 2 4 . 

 Another solution, closely allied but giving smaller results, 

 may be obtained thus. We have identically 



(a 2 +46 2 ) 4 =(a 2 -46 2 ) 4 +2-2% 2 6 2 (a 4 + 166 4 ). (4) 



Now put 6=z 2 , a=x 2 +3y z , and this becomes 

 |> 2 + 3t/ 2 ) 2 +4z 4 ] 4 =[(z 2 + 3*/ 2 ) 2 -4z 4 ] 4 + (2z) 4 [(a; 2 + 3*/ 2 ) 4 + (2z 2 ) 4 ] x 



[(x+y)*+(x-y)*+(2y)*\ (5) 

 For example, we have when 



x=y=z, 5 4 =4 4 +4 4 +3 4 +2 4 +2 4 , 



x=0, y=z, 13 4 =12 4 +8 4 +6 4 +6 4 +5 4 +4 4 +4 4 , 



=3, y=z=l, 37 4 =35 4 +24 4 +12 4 +12 4 +4 4 +2 4 +2 4 , 

 z=2, y=z=I, 53 4 =45 4 +42 4 +28 4 +14 4 +12 4 +8 4 +4 4 . 

 Also, since the equation 



can be solved algebraically for all values of r greater than 2 

 (see 7 infra), it follows that by putting 6=z 2 , a=Q in (4) that 

 we can get algebraical solutions of the equation 



where s=7+2w. 



Again, write equation (5) in the form 

 [(a; 2 + 3?/ 2 ) 2 +4z 4 ] 4 = [(z 2 + 3i/ 2 ) 2 -4z 4 ] 4 + (4yz)*[( 

 + (2z)*(x*+3y*)*[(x+y)*+(x-y)*]+(4:z*)*[(x+y)*+(x-y)*] (6) 

 Now, an algebraical solution of the equation 



1 4 +g 2 4 =P 1 4 +P 2 4 + . . . +P r 4 (7) 



has been found 1 for all values of r greater than 2 by means of 

 a single formula. If then, we choose x and y so that 



we can, on replacing (x+y)*+(xy)* by P^+P 2 *+ . . . +P r 4 , 



1 See 3 of Part II., Section I. 



