INDETERMINATE QUARTIC EQUATIONS 67 

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



x-5y)*+ (Sx+ 3</) 4 + (5x+ 8y) 

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



ar=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=5ql, y=3q2 so that 

 3x 5?/=7, and we derive 



(5q- 1) 4 + (3g-2) 4 + (8?-3) 4 = 1 4 + (7?-3) 4 + (7g-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 

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



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

 3 =2, r=l gives ! 4 +ll 4 +12 4 =4 4 +9 4 +13 4 . 

 g=3, r=2 gives 4 4 +9 4 +13 4 +18 4 +19 4 =7 4 +ll 4 +12 4 +14 4 +21*. 

 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' 1 4+PY+(2P) 4 . (1) 



Putting P^a+b, P z =c-d, P\=a-b, P' 2 =c+eZ, P=ax, (I) 

 becomes 



a s b+ab 3 =c s d+cd 3 +2a*x* (2) 



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



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



so that rf=6 4 (c-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' 2 =2c 4 a: 4 +6 4 c 8 -6 12 , 2P=26c(6 8 -c 8 )a:. 

 Thus we have the identity 

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



+ [2c 4 a; 4 +6 8 (c 8 -6 8 )] 4 +[26c(6 8 -c 8 )a:] 4 . (3) 



