CHAP. XXII] SUM OF BlQUADRATES A BlQUADRATE. 649 



could give five biquadrates with a biquadrate as sum. He 167 again re- 

 marked that he was trying to find four such biquadrates. 



Euler 191 gave an incomplete discussion of the " difficult ' problem to 

 find four biquadrates whose sum is a biquadrate. Evidently 



for 



A 2 = (p 2 +g 2 -f r 2 -s 2 )/7i, B* = 2psfn, C 2 = 2qsfn, 



D z = 2rsln, E 2 = (p 2 +g 2 +r 2 +s 2 )/n. 



These five functions are to be made squares. This will be true of the first 



and last if 



(1) (p 2 +2 2 +r 2 )/n = a 2 +& 2 , s 2 /n = 2ab. 



Then s 2 = 2a6n= D if 2w = a/3, a = a/ 2 , b = (3g 2 , whence s = a(3fg. Next, 

 2ps 2qs 2rs 



71 71 



if p = x*l(fg), q = y z l(fg), r = z z l(fg). Substitute these values into (1 1) ; we get 



But no discussion of this final condition is given. 

 D. S. Hart 192 employed the sum 



of n consecutive biquadrates I 4 , , w 4 , and 



Thus (s+w) 4 can be expressed as a sum of biquadrates if a t is. Evi- 



dently n>8. For n = 9, s = 14, w = l, (7-^ = 3124 = ! 4 +2 4 +3 4 +5 4 +7 4 , 



yieldhig 



(2) 4 4 +6 4 +8 4 +9 4 +14 4 = 15 4 . 



For n = 20, s = 30, m = 4, 34 4 is the sum of the fourth powers of 1, 3, 4, 5, 9, 

 10, 11, 12, 14, 15, 16, 17, 18, 19, 30. 



A. Martin 193 gave (2). 



A. Martin 194 started with the identity 



(1 +4m 4 ) 4 = 1 4 + (2m) 4 +96(m 2 ) 4 + (4m 3 ) 4 + (4m 4 ) 4 . 



But 96 = 3 4 +2 4 I 4 . Hence the new right member has six positive bi- 

 quadrates and the term (m 2 ) 4 . For m = 2, the latter cancels (2m) 4 and 

 we get 



! 4 +8 4 +12 4 +32 4 +64 4 = 65 4 , 



which was communicated to him by D. S. Hart. For m = 3, 

 where 



191 Opera postuma, 1, 1862, 216-7 (about 1772). 



192 Math. Quest. Educ. Times, 14, 1871, 86-7. 



193 Ibid., 20, 1873, 55. L'interme'diaire des math., 1, 1894, 26. 

 1M Math. Magazine, 2, 1896, 173-184. 



