CHAP, xxiv] INTEGERS WITH EQUAL SUMS OF LIKE POWERS. 711 



0. Birck (p. 182) took x+y+z = Q and 



= ixky, r] = iykz, $ = izkx, Tr = iykx, K = izky, p = ixkz. 

 Then 

 ft+> n, n+1], n rj, n+, n = n-\-ir, n ir, n-\-n, n n, n-\-p, np. 



O. Birck 41 noted that 



for 



subject to the condition A; 3 i 3 +(k %i)x 2 f% 2 = 0. From one solution 

 (i, k, x, y) of the latter he derived two or more new solutions. 

 A. Ge"rardin 42 noted that 



p(p+a+b), p*+2p(a+b)+2al), p(a+b)+2ab 



= ap, bp, p z +p(a+2b)+2ab, p*+p(2a+b)+2ab. 

 E. B. Escott 43 noted that (4), for n = 2, 4, has the solutions 



x = m 2 +mn+3n 2 , y= 2m 2 4mn w 2 , z= 3m 2 2n 2 , 

 w = 3m 2 mn+ri 2 , v= m 2 +4mn+2?i 2 , w=2m 2 +3n 2 , 



where m, n are odd, and gave two analogous solutions. Gerardin gave 

 (ibid.) a process to obtain solutions. 



Crussol 44 treated the last problem with the restriction y+z = v+w. 

 The equations can be written in the form 



(x+pn) k +(y+pm) k +(zpm) k =(xpri) k +(y-pm) k +(z+pm) k , k = 2, 4, 

 where m, n are relatively prime. Thus 



xn = m(z-y], 4p 2 n 2 (n 2 -m 2 )==3n? 

 Set s = 3a 2 -/5 2 (n 2 -4m 2 ). Then the solution is 



Crussol 45 noted that the system 



+(x-a) k +(y+b) k +(y-b) k = (z+a) k +(z-a) k + (t+b*) k + 



k = 2,4, 6, 

 is equivalent to x 2 -\-y 2 =z 2 +t 2 and 



6(a 2 -6 2 ) =y 2 +t 2 -x*-z 2 , 10(a 2 +6 2 ) =y 2 +t 2 +x 2 +z 2 . 

 Set X = aq j3p, y = ap+fiq, z = aq-{-j3p } t ap pq. Thus 



3(<z 2 -6 2 ) = (a 2 -(3 2 )(p 2 -q 2 ), 5(a 2 +6 2 ) = (a 2 +/3 2 )O 2 +? 2 ), 

 a 2 +^ 2 = 5(7 2 +5 2 ), a = 25+7, P = 2y-d, a = yp+dq, b = yq-dp, 



3(y 2 -d 2 )(p 2 -q 2 )=2yd(2p+q)(p-2q). 



The discriminant of this quadratic in 7, 5 must be a square. The first of 



41 L'intermediaire des math., 20, 1913, 273-7. 



42 Sphinx-Oedipe, 8, 1913, 134; correction, 157. 

 Ibid., 141-2. Cf. papers 206-7 of Ch. XXII. 

 44 Ibid., 175-6. 



Ibid., 189. 



