INDETERMINATE QUARTIC EQUATIONS 67 

 Hence, subtracting (7) from (3) we derive 



-5i/) 4 + (Sx+ 32/) 4 

 x=y gives 3 4 +5 4 +8 4 +14 4 =2 4 +ll 4 +13 4 ; 



z=2, y=l gives 2 4 +ll 4 +13 4 +21 4 =l 4 +7 4 +18 4 +19 4 . 



Cor. 3. In equation (3) put x=5qI, y=3q2 so that 

 3x 5y=7, and we derive 



(5g-l) 4 +(3g-2) 4 +(8^-3) 4 =l 4 +(7^-3) 4 +(7^-2) 4 (8) 

 .-. (5r-l) 4 +(3r-2) 4 +(8r-3) 4 =l 4 +(7r-3) 4 +(7r-2) 4 (9) 



Hence, subtracting (8) and (9) we have 



+ (3r-2) 4 + (8r-3) 4 + (7^-3) 4 + (7?-2) 4 

 q=2, r =l gives l 4 +H 4 +12 4 =4 4 +9 4 +13 4 . 

 0=3, r=2 gives 4 4 +9 4 +13 4 +18 4 +19 4 =7 4 +H 4 +12 4 +14 4 +21 4 . 

 3. We shall now show how the equation 



may be solved by a single formula which holds for all values 

 of n and r except the case n=r=0. 

 Let us first solve the equation 



P 1 4 + P 2 4 = p' i 4+P / 2 4 + (2P) 4 . (1) 



Putting P^a-f 6, P 2 =c-d, P\=a-b, P' 2 =c+d, P=ax, (1) 

 becomes 



a 36+ a& 3 =c 3 d+ cd*+ 2a*x* (2) 



If now we take d=ab 3 /c s , (2) is satisfied by taking 



i.e. a=6(c 8 -6 8 )/2z 4 c 8 , 



so that ^=6 4 (c 8 -6 8 )/2a; 4 c 11 . 



Hence omitting the common denominator, and replacing 

 x throughout by x/c 2 , we have as a solution of (1) 



\ P\=bc*(c 8 -2x*-1>*) 9 



Thus we have the identity 

 [6c 3 (6 8 -c 8 +2a: 4 )] 4 +[2c% 4 -& 8 (c 8 -6 8 )] 4 =[6c 3 (c 8 -6 8 -2a; 4 )] 4 



+ [2c 4 z 4 + 6 8 (c 8 -6 8 )] 4 + [26c(6 8 -c 8 )x] 4 . (3) 



