CHAP, xxii] EQUATIONS OF DEGREE FOUR. 669 



From the linear relation between p 4 and p 3 q, we get 



p : q = Sn(l-n 2 ) : 5-10n 2 +n 4 . 

 J. A. Euler 309 treated his father's 308 problem. Multiply 



by 4<? 2 and the like identity in q by (p 2 +l) 2 and add. Thus 



Hence we have three squares whose sum is a square. The sum of their 

 products by twos is 4<? 2 times 



This is to be made a square. Set A = (p 2 +l) 2 , B = 4p(p--l). Then 

 (2 2 -l) 2 .A 2 +g 2 2 is to be a square, say (Aq*+v) z . Then q 2 = (A 2 -v z )/d, 

 where d=2A 2 -B*+2Av. Take v 2 = A 2 -B 2 . Then d is the square of 

 A+v. Now A 2 - 2 =(p 4 -6p 2 +l) 2 . Hence 



B 



Hence, after multiplication by (p 2 I) 2 , we have the solution 



For p = 2, we get 35, 72, 96. Next (p. 47) let x = am, y = bm, z = cn, where 

 c 2 , m = 2pq, n = p 2 q 2 . Then 



The latter is the square of m(a 4 +b 4 ) if p = a, q = b. Then 



z = 2a 2 5, y = 2ab 2 , 2 = c(a 2 -6 2 ) (a 2 +6 2 = c 2 ) 



is a solution. It may be obtained by using his father's notations and 

 assuming that p 2 +# 2 = c 2 . Then the condition (1) becomes 



which is satisfied if r = cp/q, since the left member becomes C 2 (p 4 +g 4 ) 2 /g 4 . 



The problem 3090 to find four integers whose sum is a biquadrate and sum 

 of any two a square reduces to finding a biquadrate n 4 which is a sum of 

 two squares in three ways. Take as n a product of two or more primes 

 4k+l. 



J. Cunliffe 3096 noted that the problem to find three positive integers 

 whose sum is a square and sums by twos are biquadrates is evidently 

 equivalent to that to find three biquadrates half of whose sum is a square 

 and the sum of any two exceeds the remaining one. Half the sum of the 

 fourth powers of m+n+sv, m+rv, n+v(r+s) is A 2 +2Bv-{ +V, where 



809 Acta Acad. Petrop., pro anno 1779, I, 1782, Mem., pp. 40-48. 

 3090 New Series of Math. Repository (ed., T. Leybourn), 1, 1806, I, 59-61. 

 3096 Ibid., 2, 1809, I, 178-9. If we wave the condition that the numbers be positive, we may 

 use the biquadrates m 4 , n 4 , (m+n) 4 , half of whose sum is (m 2 +mn+n 2 ) 2 . 



