650 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xxn 



so that we get 6 or 7 biquadrates whose sum is a biquadrate. Multiplying 

 (2) by 2 4 and by 5 4 and eliminating 30 4 , we see that 75 4 is the sum of the 

 fourth powers of 8, 12, 16, 18, 20, 28, 40, 45, 70. Finally, he tabulated the 

 values of S= 1 4 + +w 4 for 71^=285 to use in seeking by trial to express 

 S 6 4 as a sum of distinct biquadrates ^n 4 . Example in Martin, 68 Ch. 

 XXIII. 



E. Fauquembergue 195 gave the identity 



(4z 4 +?/ 4 ) 4 = (4x 4 -?/ 4 ) 4 + (4x 3 yY+ (4x*yY+(2xy z y+ (2xy*) 4 , 



which becomes 5 4 = 3 4 +4 4 +4 4 +2 4 +2 4 foTx = y = l. 



C. B. Haldeman 196 noted that a 4 +6 4 +(a+6) 4 =2(a 2 +ab+6 2 ) 2 [Proth 227 ]. 

 Hence on adding d 4 +e 4 , the sum will be a biquadrate if a?-\-ab-\-tf = de and 

 d 2 +e z = D. To satisfy the latter, take e= (d 4 -4z 4 )/(4dz 2 ) ; then the former 

 condition gives 



a = , t = ^-4z 4 -36V. 



2z 



Take t = d 2 z 2 , whence d! 2 = |(36 2 +52 2 ). Since b=z makes d rational, set 

 b = y+z, and take d=2z-\-sy/t, whence we find y and then b, d. Or we 



may take d = 2, 2 = 1, whence t= V12 36 2 ; set & = H-1, ^ = sf/^+3, whence 

 we get v and 



(3) S(2s 2 12^-6^) 4 +S(4s 2 Tl2^ 2 ) 4 +(3s 2 +9^) 4 =(5s 2 +15^ 2 ) 4 . 



Or, finally, take d = 9, e = 4, a 2 +a&+& 2 = 4-37 since 2(4-37) 2 +9 4 +4 4 = 15 4 . 

 Since 6 = 6 gives a rational value for a, set 6 = 6+^. Then 



(2a+fe) 2 = 592-36 2 = -3r 2 -36r+484= ( j+22 Y, 



by choice of r rationally in s, L Hence the sum of the fourth powers of 

 8s 2 +40s-24 2 , 6s 2 -44s-18 2 , 14s 2 -4s^-42i 2 , 9s 2 +27 2 , 4s 2 +12^ 2 equals 

 (15s 2 +45i 2 ) . 4 For s = 1, t = 0, we get (2), which is believed to be the solution 

 in least integers. 



For six biquadrates, add e 4 +/ 4 to each member of his 239 identity (1) and 

 take 3(3a 2 + 2 ) 2 = e/. It remains to make e 2 +/ 2 =D, say the square of 

 1201(3a 2 + 2 )/140, whence e = 7(3a 2 + 2 )/20. Or we may take the sum of 

 three of the six to be 



(4) Q a>6 =(2a) 4 +(a+6) 4 +(a-6) 4 = 2(3a 2 +6 2 ) 2 



and the others to be the fourth powers of 6, 12, 13 or 26, 27, 42 and the 

 sum of the six to be 15 4 or 45 4 . 



For seven biquadrates, take Q a ,b+d 4 +e 4 +(2gY+g 4 = (30) 4 , 3a 2 +6 2 = de, 

 whence d 2 +e 2 = 8# 2 , which holds if e = +7d, g=-5d/2. Take d = y+a, 

 b = ry{t+2a. Then y = 2a(7P 2rt)/(r 2 7t 2 ) and we have an answer. Or 

 use Qa.b+Qd,e+3 4 = 5 4 , which is satisfied if 3a 2 +6 2 = 4, 3d 2 +e 2 =16; taking 

 b = 2 asft, e = 4: dv/z, we get a, b in terms of s, t, and d, e in terms of v, z. 

 Next, Q a , b +Q d< e +2 4 +l 4 = 3 4 if 3a 2 +6 2 = 4 = 3d 2 +e 2 (like preceding case). 



195 L'intermSdiaire des math., 5, 1898, 33. 



196 Math. Magazine, 2, 1904, 288-296. The editor Martin noted (p. 349 and in his 1900 



paper below) that this MS. had been long in the editor's hands. 



