INDETERMINATE QUARTIC EQUATIONS 91 

 Hence we may take 



ii := x ~T~y > v=^2ixy 



and with these values equation (4) becomes 

 (x 8 + 14# 4 2/ 4 +y 8 ) 2 =(a: 4 y t )*+[2xy(x 2 +y z )~\*+[2xy(x z / 2 )] 4 . (5) 

 Thus x=2, y=l gives 



481 2 =12 4 +15 4 +20 4 . 



Cor. The foregoing result evidently amounts to this, that 

 if 2 6 2 +6 2 c 2 =c% 2 , then 



(a 2 -6 2 +c 2 ) 2 =a 4 +6 4 +c 4 . 

 Again, to solve (3) when X =2, n=3, X 1 =X 2 =X 3 =1, since 



*+ & 4 + (a+ 6) 4 = 2(a 2 + ab+ 6 2 ) 2 , 

 replacing a and b, each by its square, we have 



a*+ 6 8 + ( 2 + b 2 )*=$[a*+ 6 4 +a 2 +6 2 ] 2 . 

 If then we put a=x z y 2 , b=2xy, this becomes 



; 4 ] (6) 

 by (5) above. Thus a:= 2, y=l gives 



3 8 +4 8 +5 8 =2(12 4 +15 4 +20 4 )=2-481 2 . 

 The last part of (6), which is simply a solution of 



a 8 + 6 8 + c 8 = 2(a 4 6 4 + 6 4 c 4 + c% 4 ), 



is immediately obtained otherwise. For if a, b, c be integers 

 connected by the equation a 2 +6 2 =c 2 , we have, on squaring 

 each side of this equation 



and on again squaring we obtain 



a 8 + & 8 + c 8 = 2(a 4 6 4 + 6 4 c 4 + c 4 a 4 ) 



which is the required result if we put ax z y z , b=2xy, 

 c=x 2 +y 2 to make 2 +6 2 =c 2 . Corresponding results may 

 naturally be obtained by squaring any identity of the form 



+ \iy 2 =vz 2 . 



But for all values of n greater than 4, a single algebraical 



